1.3: Mathematical Treatment of Measurement Results
-
- Last updated
- Save as PDF
- Scott Van Bramer
- Widener University
Learning Objectives
- Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities.
- Describe how to use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties.
- Convert between the three main temperature units: Fahrenheit, Celsius, and Kelvin.
Math Support
If you have any concerns about your math background or just want to make sure you have all the math tools needed to succeed in this course, review the short "Math Minutes" videos from Dr. Martin in the Appendix: 11.1: Dr Martin's Math Minutes . You may also want to review Appendix: 11.3: Essential Mathematics .
Introduction
It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:
\[\mathrm{speed=\dfrac{distance}{time}}\]
An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of
\[\mathrm{\dfrac{100\: m}{10\: s}=10\: m/s}\]
Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:
\[\mathrm{time=\dfrac{distance}{speed}}\]
The time can then be computed as:
\[\mathrm{\dfrac{25\: m}{10\: m/s}=2.5\: s}\]
Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”
These calculations are examples of a versatile mathematical approach known as dimensional analysis, or the factor-label method . Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers . This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.
Conversion Factors
A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor . For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,
\[\mathrm{\dfrac{2.54\: cm}{1\: in.}\:(2.54\: cm=1\: in.)\: or\: 2.54\:\dfrac{cm}{in.}}\]
Several other commonly used conversion factors are given in Table \(\PageIndex{1}\). A more extensive list is in Appendix 11.6: Units and Conversion Factors .
| Length | Volume | Mass |
|---|---|---|
| 1 m = 1.0936 yd | 1 L = 1.0567 qt | 1 kg = 2.2046 lb |
| 1 in. = 2.54 cm (exact) | 1 qt = 0.94635 L | 1 lb = 453.59 g |
| 1 km = 0.62137 mi | 1 ft 3 = 28.317 L | 1 (avoirdupois) oz = 28.349 g |
| 1 mi = 1609.3 m | 1 tbsp = 14.787 mL | 1 (troy) oz = 31.103 g |
When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:
\[\mathrm{34\: \cancel{in.} \times \dfrac{2.54\: cm}{1\:\cancel{in.}}=86\: cm}\]
Since this simple arithmetic involves quantities , the premise of dimensional analysis requires that we multiply both numbers and units . The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield
\[\mathrm{\dfrac{in.\times cm}{in.}}.\]
Just as for numbers, a ratio of identical units is also numerically equal to one,
\[\mathrm{\dfrac{in.}{in.}=1}\]
and the unit product thus simplifies to cm . (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.
Example \(\PageIndex{1}\): Using a Unit Conversion Factor
The mass of a competition Frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table \(\PageIndex{1}\)).
Solution
Since we have the conversion factor, we can determine the mass in ounces using an equation similar the one used for converting length from inches to centimeters.
\[x\:\mathrm{g=125\: g\times unit\: conversion\: factor}\nonumber\]
We write the unit conversion factor in its two forms:
\[\mathrm{\dfrac{1\: oz}{28.349\: g}\:and\:\dfrac{28.349\: g}{1\: oz}}\nonumber\]
The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.
\[\begin{align*}
x\:\ce{oz}&=\mathrm{125\:\cancel{g}\times \dfrac{1\: oz}{28.349\:\cancel{g}}}\\
&=\mathrm{\left(\dfrac{125}{28.349}\right)\:oz}\\
&=\mathrm{4.41\: oz\: (three\: significant\: figures)}
\end{align*}\]
Exercise \(\PageIndex{1}\)
Convert a volume of 9.345 qt to liters.
- Answer
-
8.844 L
Conversion Factors and SI Prefixes
The same problem solving strategy can be used to convert units within the SI system using SI prefixes. The prefix is a conversion factor that allows conversion between units. For example when we multiply a quantity (such as distance given in cm) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in meters). For example, since centi = 10 -2 , a basketball player’s vertical jump of 86 cm can be converted to meters by:
\[\mathrm{86\: \cancel{cm.} \times \dfrac{1\: cm}{10^{-2}\:\cancel{m.}}=0.86\: m}\]
Example \(\PageIndex{2}\): Using SI Prefixes for Unit Conversion
The mass of a competition Frisbee is 125 g. Convert its mass to milligrams using the unit conversion factor derived from the relationship milli = 10 -3 .
Solution
If we have the conversion factor, we can determine the mass in milligrams using an equation similar the one used for converting length from inches to centimeters.
\[x\:\mathrm{g =125\: g\times unit\: conversion\: factor}\nonumber\]
We write the unit conversion factor in its two forms:
\[\mathrm{\dfrac{1\: mg}{10^{-3}\: g}\:and\:\dfrac{10^{-3}\: g}{1\: mg}}\nonumber\]
The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.
\[\begin{align*}
x\:\ce{g}&=\mathrm{125\:\cancel{g}\times \dfrac{1\: mg}{10^{-3}\:\cancel{g}}}\\
&=\mathrm{\left(\dfrac{125}{10^{-3}}\right)\:mg}\\
&=\mathrm{1.25 \times 10^5\: mg\: (three\: significant\: figures)}
\end{align*}\]
Exercise \(\PageIndex{2}\)
Convert a volume of 25.4 milliliters to liters.
- Answer
-
0.0254 L
Conversion Factors and Dimensional Analysis
Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.
Example \(\PageIndex{3}\): Computing Quantities from Measurement Results
What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.
Solution
Since \(\mathrm{density=\dfrac{mass}{volume}}\), we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A \(\times\) unit conversion factor. The necessary conversion factors are given in Table 1.7.1: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:
\[\mathrm{9.26\:\cancel{lb}\times \dfrac{453.59\: g}{1\:\cancel{lb}}=4.20\times 10^3\:g}\nonumber \]
We need to use two steps to convert volume from quarts to milliliters.
- Convert quarts to liters.
\[\mathrm{4.00\:\cancel{qt}\times\dfrac{1\: L}{1.0567\:\cancel{qt}}=3.78\: L}\nonumber\]
- Convert liters to milliliters.
\[\mathrm{3.78\:\cancel{L}\times\dfrac{1000\: mL}{1\:\cancel{L}}=3.78\times10^3\:mL}\nonumber\]
Then,
\[\mathrm{density=\dfrac{4.20\times10^3\:g}{3.78\times10^3\:mL}=1.11\: g/mL}\nonumber\]
Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:
\[\mathrm{\dfrac{9.26\:\cancel{lb}}{4.00\:\cancel{qt}}\times\dfrac{453.59\: g}{1\:\cancel{lb}}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\:\cancel{L}}{1000\: mL}=1.11\: g/mL}\nonumber\]
Exercise \(\PageIndex{3}\)
What is the volume in liters of 1.000 oz of antifreeze, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?
- Answer
-
\(\mathrm{2.956\times10^{-2}\:L}\)
Example \(\PageIndex{4}\): Computing Quantities from Measurement Results
While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.
- What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
- If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?
Solution
(a) We first convert distance from kilometers to miles:
\[\mathrm{1250\: km\times\dfrac{0.62137\: mi}{1\: km}=777\: mi}\nonumber\]
and then convert volume from liters to gallons:
\[\mathrm{213\:\cancel{L}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\: gal}{4\:\cancel{qt}}=56.3\: gal}\nonumber\]
Then,
\[\mathrm{(average)\: mileage=\dfrac{777\: mi}{56.3\: gal}=13.8\: miles/gallon=13.8\: mpg}\nonumber\]
Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:
\[\mathrm{\dfrac{1250\:\cancel{km}}{213\:\cancel{L}}\times\dfrac{0.62137\: mi}{1\:\cancel{km}}\times\dfrac{1\:\cancel{L}}{1.0567\:\cancel{qt}}\times\dfrac{4\:\cancel{qt}}{1\: gal}=13.8\: mpg}\nonumber \]
(b) Using the previously calculated volume in gallons, we find:
\[\mathrm{56.3\: gal\times\dfrac{$3.80}{1\: gal}=$214}\nonumber \]
Exercise \(\PageIndex{4}\)
A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).
- What (average) fuel economy, in miles per gallon, did the Prius get during this trip?
- If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?
- Answer a
-
51 mpg
- Answer b
-
$62
Unit Conversion Video Example
This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects. Also, these videos are meant to act as a learning resource for all General Chemistry students .
Video Topics
Dimensional analysis: Units can be multiplied, divided, and canceled just like an algebraic quantity. A conversion factor is a fractional quantity used to convert one unit to another.
Link to Video
C onverting Between Units: https://youtu.be/wSaOh48k8Wg
Significant Digits in Measurement
Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number . If we count eggs in a carton, we know exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used.
The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid.
Refer to the illustration in Figure \(\PageIndex{1}\). The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark, and so a reasonable estimate of the liquid’s volume would be 21.6 mL. In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL.
This concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of ± 0.01 gram. If we weigh the quarter on a more precise balance, we may find that its mass is 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty , which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant digits or significant figures . Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.
Whenever you make a measurement properly, all the digits in the result are significant. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them.
Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant digits in the measurement unless the last digit is a trailing zero lying to the left of the decimal point.
Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located.
The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as 8.32407 \(\times\) 10 −3 ; then the number 8.32407 contains all of the significant digits, and 10 −3 locates the decimal point.
The number of significant digits is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3 \(\times\) 10 3 (two significant digits), 1.30 \(\times\) 10 3 (three significant digits, if the tens place was measured), or 1.300 \(\times\) 10 3 (four significant digits, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant.
When determining significant digits, be sure to pay attention to reported values and think about the measurement and significant digits in terms of what is reasonable or likely—that is, whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant digits, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 317 million, or \(3.17 \times 10^8 \) people.
Significant Digits in Calculations
A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. We must take the uncertainty in our measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant digits, which is determined by the following three rules for rounding numbers:
- When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction).
- When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant digits (the least precise value in terms of multiplication and division).
- If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is more than 5, we “round up” and increase the retained digit by 1; if the dropped digit is 5, we round up or down, whichever yields an even value for the retained digit. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)
The following examples illustrate the application of this rule in rounding a few different numbers to three significant digits:
- 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
- 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
- 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even)
- 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)
Let’s work through these rules with a few examples.
Example \(\PageIndex{5}\): Rounding Numbers
Round the following to the indicated number of significant digits:
- 31.57 (to two significant digits)
- 8.1649 (to three significant digits)
- 0.051065 (to four significant digits)
- 0.90275 (to four significant digits)
Solution
- 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
- 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)
- 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
- 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)
Exercise \(\PageIndex{5}\)
Round the following to the indicated number of significant digits:
- 0.424 (to two significant digits)
- 0.0038661 (to three significant digits)
- 421.25 (to four significant digits)
- 28,683.5 (to five significant digits)
- Answer a
-
0.42
- Answer b
-
0.00387
- Answer c
-
421.2
- Answer d
-
28,684
Example \(\PageIndex{6}\): Addition and Subtraction with Significant Digits Rule:
When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).
- Add 1.0023 g and 4.383 g.
- Subtract 421.23 g from 486 g.
Solution
(a)
\[\begin{align*}
&\mathrm{1.0023\: g}\\ +\: &\underline{\mathrm{4.383\: g}\:\:}\\ &\mathrm{5.3853\: g}
\end{align*}\]
Answer is 5.385 g (round to the thousandths place; three decimal places)
(b)
\[\begin{align*}
&\mathrm{486\: g}\\ -\: &\underline{\mathrm{421.23\: g}}\\ &\mathrm{\:\:64.77\: g}
\end{align*}\]
Answer is 65 g (round to the ones place; no decimal places)
Exercise \(\PageIndex{6}\)
- Add 2.334 mL and 0.31 mL.
- Subtract 55.8752 m from 56.533 m.
- Answer a
-
2.64 mL
- Answer b
-
0.658 m
Example \(\PageIndex{7}\): Multiplication and Division with Significant Digits
Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant digits (the least precise value in terms of multiplication and division).
- Multiply 0.6238 cm by 6.6 cm.
- Divide 421.23 g by 486 mL.
Solution
(a)
\[\mathrm{0.6238\: cm\times6.6\:cm=4.11708\:cm^2\rightarrow result\: is\:4.1\:cm^2}\:\textrm{(round to two significant digits)}\]
\[\textrm{four significant digits}\times \textrm{two significant digits}\rightarrow \textrm{two significant digits answer}\]
(b)
\[\mathrm{\dfrac{421.23\: g}{486\: mL}=0.86728...\: g/mL\rightarrow result\: is\: 0.867\: g/mL} \: \textrm{(round to three significant digits)}\]
\[\mathrm{\dfrac{five\: significant\: digits}{three\: significant\: digits}\rightarrow three\: significant\: digits\: answer}\]
Exercise \(\PageIndex{7}\)
- Multiply 2.334 cm and 0.320 cm.
- Divide 55.8752 m by 56.53 s.
- Answer a
-
0.747 cm 2
- Answer b
-
0.9884 m/s
In the midst of all these technicalities, it is important to keep in mind the reason why we use significant digits and rounding rules—to correctly represent the certainty of the values we report and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation.
Example \(\PageIndex{8}\): Calculation with Significant Digits
One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters. Give your answer with the appropriate number of significant digits.
Solution
\[\begin{align*}
V&=l\times w\times d\\ &=\mathrm{13.44\: dm\times 5.920\: dm\times 2.54\: dm}\\ &=\mathrm{202.09459...dm^3}\:\textrm{(value from calculator)}\\ &=\mathrm{202\: dm^3,}\textrm{ or 202 L (answer rounded to three significant digits)}
\end{align*}\]
Exercise \(\PageIndex{8}\):
What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm 3 ? Give your answer with the appropriate number of significant digits.
- Answer
-
1.034 g/mL
Example \(\PageIndex{9}\): Determining Density by Water Displacement
A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.
- Use these values to determine the density of this piece of rebar.
- Rebar is mostly iron. Does your result in (a) support this statement? How?
Solution
The volume of the piece of rebar is equal to the volume of the water displaced:
\[\mathrm{volume=22.4\: mL-13.5\: mL=8.9\: mL=8.9\: cm^3}\nonumber\]
(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)
The density is the mass-to-volume ratio:
\[\mathrm{density=\dfrac{mass}{volume}=\dfrac{69.658\: g}{8.9\: cm^3}=7.8\: g/cm^3}\nonumber\]
(rounded to two significant digits, per the rule for multiplication and division)
The density of iron is 7.9 g/cm 3 , very close to that of rebar, which lends some support to the fact that rebar is mostly iron.
Exercise \(\PageIndex{9}\)
An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.
- Use these values to determine the density of this material.
- Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.
- Answer a
-
19 g/cm 3
- Answer b
-
It is likely gold; it has the right appearance for gold and very close to the density given for gold.
Accuracy and Precision
Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the accepted or true value. Precise values agree with each other; accurate values agree with an accepted value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure \(\PageIndex{2}\)).
Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table \(\PageIndex{2}\).
| Dispenser #1 | Dispenser #2 | Dispenser #3 |
|---|---|---|
| 283.3 | 298.3 | 296.1 |
| 284.1 | 294.2 | 295.9 |
| 283.9 | 296.0 | 296.1 |
| 284.0 | 297.8 | 296.0 |
| 284.1 | 293.9 | 296.1 |
Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL).
Summary
Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.
Quantities can be exact or measured. Measured quantities have an associated uncertainty that is represented by the number of significant digits in the measurement. The uncertainty of a calculated value depends on the uncertainties in the values used in the calculation and is reflected in how the value is rounded. Measured values can be accurate (close to the accepted value) and/or precise (showing little variation when measured repeatedly).
Key Equations
- \(T_\mathrm{^\circ C}=\dfrac{5}{9}\times T_\mathrm{^\circ F}-32\)
- \(T_\mathrm{^\circ F}=\dfrac{9}{5}\times T_\mathrm{^\circ C}+32\)
- \(T_\ce{K}={^\circ \ce C}+273.15\)
- \(T_\mathrm{^\circ C}=\ce K-273.15\)
Glossary
- dimensional analysis
- (also, factor-label method) versatile mathematical approach that can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities
- Fahrenheit
- unit of temperature; water freezes at 32 °F and boils at 212 °F on this scale
- unit conversion factor
- ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit
- uncertainty
- estimate of amount by which measurement differs from accepted value
- significant digits
- (also, significant figures) all of the measured digits in a determination, including the uncertain last digit
- rounding
- procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation
- precision
- how closely a measurement matches the same measurement when repeated
- exact number
- number derived by counting or by definition
- accuracy
- how closely a measurement aligns with a correct value
Contributors and Attributions
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).
-
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>).


7.6: Method of coordinates
The following exercise is important; it shows that our axiomatic definition agrees with the model definition.
Let \(\ell\) and \(m\) be perpendicular lines in the Euclidean plane. Given a point \(P\), let \(P_{\ell}\) and \(P_m\) denote the foot points of \(P\) on \(\ell\) and \(m\) respectively.
- Show that for any \(X \in \ell\) and \(Y \in m\) there is a unique point \(P\) such that \(P_{\ell} = X\) and \(P_m = Y\).
- Show that \(PQ^2 = P_{\ell}Q_{\ell}^2 + P_m Q_m^2\) for any pair of points \(P\) and \(Q\).
- Conclude that the plane is isometric to \((\mathbb{R}^2, d_2)\).
- Hint
-
(a). Use the uniqueness of the parallel line ( Theorem 7.1.1 ).
(b). Use Lemma 7.5.1 and the Pythagorean theorem ( Theorem 6.2.1 )
Once this exercise is solved, we can apply the method of coordinates to solve any problem in Euclidean plane geometry. This method is powerful and universal; it will be developed further in Chapter 18.
Use the Exercise \(\PageIndex{1}\) to give an alternative proof of Theorem 3.5.1 in the Euclidean plane.
That is, prove that given the real numbers \(a, b\), and \(c\) such that
\(0 < a \le b \le c \le a + b\),
there is a triangle \(ABC\) such that \(a = BC\), \(b = CA\), and \(c = AB\).
- Hint
-
Set \(A = (0, 0), B = (c, 0)\), and \(C = (x, y)\). Clearly, \(AB = c\), \(AC^2 = x^2 + y^2\) and \(BC^2 = (c - x)^2 + y^2\).
It remains to show that there is a pair of real numbers \((x, y)\) that satisfy the following system of equations:
\(\begin{cases} b^2 = x^2 + y^2 \\ a^2 = (c- x)^2 + y^2 \end{cases}\)
if \(0 < a \le b \le c \le a + c\).
Consider two distinct points \(A = (x_A, y_A)\) and \(B = (x_B, y_B)\) on the coordinate plane. Show that the perpendicular bisector to \([AB]\) is described by the equation
\(2 \cdot (x_B - x_A) \cdot x + 2 \cdot (y_B - y_A) \cdot y = x_B^2 + y_B^2 - x_A^2 - y_B^2\).
Conclude that line can be defined as a subset of the coordinate plane of the following type:
- Solutions of an equation \(a \cdot x + b \cdot y = c\) for some constants \(a, b\), and \(c\) such that \(a \ne 0\) or \(b \ne 0\).
- The set of points \((a \cdot t + c, b \cdot t + d)\) for some constants \(a, b, c\), and \(d\) such that \(a \ne 0\) or \(b \ne 0\) and all \(t \in \mathbb{R}\).
- Hint
-
Note that \(MA = MB\) if and only if
\((x - x_A)^2 + (y - y_A)^2 = (x - x_B)^2 + (y - y_B)^2\)
where \(M = (x, y)\). To prove the first part, simplify this equation. For the remaining parts use that any line is a perpendicular bisector to some line segment.


4.6E: Exercises for Section 12.5
-
- Last updated
- Save as PDF
Finding Components of Acceleration & Kepler's Laws
1) Find the tangential and normal components of acceleration for \(\vecs r(t)=t^2\,\hat{\mathbf{i}}+2t \,\hat{\mathbf{j}}\) when \(t=1\).
- Answer:
- \(a_\vecs{T}=\sqrt{2}, \quad a_\vecs{N}=\sqrt{2}\)
In questions 2 - 8, find the tangential and normal components of acceleration.
2) \(\vecs r(t)=⟨\cos(2t),\,\sin(2t),1⟩\)
3) \(\vecs r(t)=⟨e^t \cos t,\,e^t\sin t,\,e^t⟩\). The graph is shown here:
- Answer:
- \(a_\vecs{T}=\sqrt{3}e^t, \quad a_\vecs{N}=\sqrt{2}e^t\)
4) \(\vecs r(t)=⟨\frac{2}{3}(1+t)^{3/2}, \,\frac{2}{3}(1-t)^{3/2},\,\sqrt{2}t⟩\)
5) \(\vecs r(t)=\left\langle 2t,\,t^2,\,\dfrac{t^3}{3}\right\rangle\)
- Answer:
- \(a_\vecs{T}=2t, \quad a_\vecs{N}=2\)
6) \(\vecs r(t)=t^2\,\hat{\mathbf{i}}+t^2\,\hat{\mathbf{j}}+t^3\,\hat{\mathbf{k}}\)
7) \(\vecs r(t)=⟨6t,\,3t^2,\,2t^3⟩\)
- Answer:
- \(a_\vecs{T}=\dfrac{6t +12t^3}{\sqrt{1+t^2+t^4}}, \quad a_\vecs{N}=6\sqrt{\dfrac{1+4t^2+t^4}{1+t^2+t^4}}\)
8) \(\vecs r(t)=3\cos(2πt)\,\hat{\mathbf{i}}+3\sin(2πt)\,\hat{\mathbf{j}}\)
- Answer:
- \(a_\vecs{T}=0, \quad a_\vecs{N}=12\pi^2\)
9) Find the tangential and normal components of acceleration for \(\vecs r(t)=a\cos(ωt)\,\hat{\mathbf{i}}+b\sin(ωt)\,\hat{\mathbf{j}}\) at \(t=0\).
- Answer:
- \(a_\vecs{T}=0, \quad a_\vecs{N}=aω^2\)
10) Suppose that the position function for an object in three dimensions is given by the equation \(\vecs r(t)=t\cos(t)\,\hat{\mathbf{i}}+t\sin(t)\,\hat{\mathbf{j}}+3t\,\hat{\mathbf{k}}\).
a. Show that the particle moves on a circular cone.
b. Find the angle between the velocity and acceleration vectors when \(t=1.5\).
c. Find the tangential and normal components of acceleration when \(t=1.5\).
- Answer:
- c. \(a_\vecs{T}=0.43\,\text{m/sec}^2, \quad a_\vecs{N}=2.46\,\text{m/sec}^2\)
11) The force on a particle is given by \(\vecs f(t)=(\cos t)\,\hat{\mathbf{i}}+(\sin t)\,\hat{\mathbf{j}}\). The particle is located at point \((c,0)\) at \(t=0\). The initial velocity of the particle is given by \(\vecs v(0)=v_0\,\hat{\mathbf{j}}\). Find the path of the particle of mass \(m\). (Recall, \(\vecs F=m\vecs a\).)
- Answer:
- \(\vecs r(t)=\left(\dfrac{-\cos t}{m}+c+\frac{1}{m}\right)\,\hat{\mathbf{i}}+\left(\dfrac{−\sin t}{m}+\left(v_0+\frac{1}{m}\right)t\right)\,\hat{\mathbf{j}}\)
12) An automobile that weighs \(2700\) lb makes a turn on a flat road while traveling at \(56\) ft/sec. If the radius of the turn is \(70\) ft, what is the required frictional force to keep the car from skidding?
13) Using Kepler’s laws, it can be shown that \(v_0=\sqrt{\dfrac{2GM}{r_0}}\) is the minimum speed needed when \(\theta=0\) so that an object will escape from the pull of a central force resulting from mass \(M\). Use this result to find the minimum speed when \(\theta=0\) for a space capsule to escape from the gravitational pull of Earth if the probe is at an altitude of \(300\) km above Earth’s surface.
- Answer:
- \(10.94\) km/sec
14) Find the time in years it takes the dwarf planet Pluto to make one orbit about the Sun given that \(a=39.5\) A.U.
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .


22.4: Neutral Thermal Environment
By the end of this section, you will be able to:
- Illustrate the importance of maintaining a neutral thermal environment when caring for a newborn patient
- Compare and contrast all four ways that a newborn can lose heat and verbalize ways a nurse can help maintain a neutral thermal environment
- Discuss the importance of the Ten Step Warm Chain and how it impacts the care given to a neonate
- Summarize the signs and symptoms of cold stress and how to care for a newborn with cold stress
One of the most vital goals for a nurse caring for a neonate is providing support to maintain their temperature while they transition from intrauterine to extrauterine life.
Thermoregulation in the Newborn
The thermal balance created by the loss of heat to the environment at a rate equal to the heat being produced is called thermoregulation . Neonates have a poor ability to maintain thermoregulation because they have very little brown adipose tissue that stores the energy useful for creating heat. The neonate is also unable to shiver, which is a reflex helpful in raising temperatures. Maintaining a thermoneutral environment, between 36.5° C (97.7° F) and 37° C (98.6° F), throughout the transitional period is essential for a neonate. An appropriate temperature environment can be maintained via radiant warmer , swaddling (Figure 22.8), and skin-to-skin contact with the birthing person, the partner, or available extended family (Fraser, 2014).
Heat Loss in the Newborn
Neonates can experience heat loss in the hospital or home environment in four ways: radiation, evaporation, convection, and conduction. Neonates are homeothermic , meaning that they attempt to stabilize their core internal temperature despite significant temperature variations in their environment (Fraser, 2014; Trevisanuto & Sedin, 2016). A neutral thermal environment is the minimum temperature range necessary for heat production to maintain normal body temperature (Fraser, 2014; Trevisanuto & Sedin, 2016) (Figure 22.9). Term and postterm neonates require higher environmental temperatures than adults to maintain a neutral thermal environment, between 20° C and 22° C (68° F and 72° F).
Several neonatal characteristics affect thermal stability:
- Heat transfer from the newborn’s organs to skin surface is increased compared to that of adults because of the neonate’s decreased subcutaneous fat and large body surface–to–weight ratio (Fraser, 2014; Trevisanuto & Sedin, 2016).
- The blood vessels in the neonate are closer to the skin than those of an adult. This causes the circulating blood to be more readily influenced by environmental temperatures, thus influencing the hypothalamic temperature-regulating center in the brain (Trevisanuto & Sedin, 2016).
- Newborns rely on nonshivering thermogenesis for heat production via metabolism of brown adipose tissue (Fraser, 2014).
- At birth, the newborn prefers to stay in a fixed posture, decreasing the surface area exposed to the environment, reducing heat loss (Trevisanuto & Sedin, 2016).
Evaporation
One of the most common ways that a neonate experiences heat loss in the birthing room and on the postpartum unit is evaporation , which is heat loss that occurs when liquid is converted to vapor. The amniotic fluid and/or blood on the infant holds water that cools the neonate when exposed to the cooler environment (Figure 22.10). Evaporation can also occur if the neonate is tachypneic and exhaled moisture or water is escaping from the respiratory tract. Radiant warming beds and phototherapy can also cause evaporation heat loss (Fraser, 2014; Interprofessional Education and Research Committee of the Champlain Maternal Newborn Regional Program, 2013). Neonates are at particular risk of heat loss via evaporation at the time of birth and during their newborn bath. In the birthing room, the nurse can use warmed blankets to dry the neonate, discarding wet ones and replacing them with dry ones as needed. At the newborn bath, the nurse exposes only the necessary areas of the neonate and ensures that they have had stable temperatures before bathing.
Convection
The loss of heat from the body surface to the surrounding air by a current, like a fan in a room, is called convection . This can be caused by a fan, an air-conditioned room, removing a neonate from an incubator for a procedure or feeding, or even unwarmed oxygen in a mask. The impact on the neonate depends on the amount of time they are exposed to the cooled air, the velocity of the moving air, the temperature difference between the neonate’s skin and the air, and the amount of body surface exposed. (Lubkowska et al., 2019).
Conduction
The loss of heat when in direct contact with a cooler surface is called conduction . Nurses caring for a newborn can prevent this loss of heat by warming their hands, using warm blankets on scales and examination tables before placing the newborn on them, warming stethoscopes, and prewarming the warmer or incubator (Isolette) prior to placing the infant on the bed. Placing a neonate skin-to-skin with the birthing person, their other parent, or an extended family member is a very effective way to warm the neonate and prevent conductive heat loss.
Radiant Heat Loss
Another way that a neonate experiences heat loss is through radiation , which is heat loss occurring from transfer of heat to cool solid objects not in direct contact with the newborn. The walls or windows in a room, for example, can transfer heat from the neonate’s body if the crib or incubator is placed too close to a cold space (Fraser, 2014; Interprofessional Education & Research Committee, 2013). If a nurse places a bag of ice near the newborn in the crib or incubator while preparing to draw blood gases, that will increase radiant heat losses. Placing a hat on the newborn conserves heat, decreasing radiant heat loss. The nurse and parents must be very aware of where the newborn is placed because radiant heat loss for a newborn can happen quickly.
Figure 22.11 summarizes the methods of heat loss.
South African Cultural Attitudes about Neonatal Baths
In the Vhembe District, Limpopo province of South Africa, indigenous people do not bathe newborns only for hygienic purposes. They mix indigenous herbs with the bath water for protection from evil spirits carried by people, as well as general well-being and relaxation. The traditional bath that occurs in the first three to four days of life, gumululo , contains three traditional medicines that are boiled with water and added to the bath in place of soap. The bath is given to neonates to make them strong and protect them. At the end of the bath, they offer a small amount of the bath water to the neonate to drink (Tuelo & Mulaudzi, 2021).
Ten-Step Warm Chain
The ten-step warm chain was developed in 1997 by the World Health Organization in advocating for a more standardized approach to support thermal stability in the newborn (World Health Organization, 1997). While not the standard of care in the United States, the ten-step model is still widely used in low-resource countries. The ten steps include
- warm delivery room,
- immediate drying,
- skin-to-skin contact,
- breast-feeding,
- bathing and weighing postponed,
- appropriate clothing and bedding,
- mother and baby together,
- warm transportation,
- warm resuscitation , and
- training/awareness-raising (World Health Organization, 1997).
According to the ten-step warm chain, the delivery room should be at least 25° C (77° F) with no drafts, with no fans moving air around, and with all supplies needed to keep the newborn warm prepared ahead of time (World Health Organization, 1997). At no time should the birthing person, visitors, or nurses change the temperature of the delivery room. Immediately after the birth, the nurse can prevent heat loss in the neonate by drying the newborn with a warm blanket or cloth to prevent evaporation (World Health Organization, 1997) After drying the infant, the nurse will place the newborn on a warm surface, such as skin-to-skin with the birthing person’s chest or abdomen or in the radiant warmer to prevent conductive heat losses. This is when the nurse should encourage the birthing parent to initiate breast-feeding, according to the ten-step warm chain (World Health Organization, 1997). Bathing must wait until at least 6 hours post birth or by 24 hours of age and when the newborn’s temperature has been consistently stable (Association of Women’s Health, Obstetric and Neonatal Nurses [AWHONN], 2019; World Health Organization, 1997). Bathing should be done quickly with warm water. The newborn should be dried thoroughly, dressed in appropriate clothing, and given to the birthing person to be kept warm. If transportation is necessary, skin-to-skin with the birthing person is safest and recommended. If that is not possible, the next best option is a fully clothed newborn, swaddled, and in the arms of another adult or a transport device (World Health Organization, 1997).
Cold Stress in the Newborn
When a neonate loses more heat in a period than they can reproduce, they experience cold stress (AWHONN et al., 2023), which requires them to use compensatory mechanisms (increased respirations and nonshivering thermogenesis) to maintain core body temperature. When a neonate experiences excessive heat loss, they use stores of energy to compensate and maintain core temperature in the normal range. An increase in metabolic heat production above the basal metabolism that is not associated with muscle activity is called nonshivering thermogenesis (NST) . This occurs mostly through metabolizing brown fat and, to a much lesser degree, through the metabolization of skeletal muscle, the liver, the brain, and white fat. Brown fat mass peaks at birth and decreases in size over the first few weeks of life. Its metabolism is initiated with a catecholamine release. A newborn’s responsiveness to the release of catecholamines is increased by cutting the umbilical cord. Cutting the cord, release of catecholamines, and the stimulation of brown fat tissue are the driving factors in neonatal thermogenesis (Dattani & Gevers, 2016).
The metabolic consequences of cold stress are increased oxygen and glucose requirements and decreased cardiac output causing a decreased pulmonary blood flow leading to pulmonary hypertension and decreased surfactant production. These consequences result from activation of the sympathetic nervous system and are potentially fatal if not caught and reversed in time. Because of these major metabolic changes, the neonate goes into respiratory distress, sometimes experiences intraventricular hemorrhage, and has a higher risk of mortality than newborns who have not experienced cold stress (AWHONN et al., 2023). Preterm neonates and those who experienced intrauterine growth restriction are at increased risk for cold stress and higher mortality due to their decreased adipose tissue, brown fat, and glycogen available for metabolism (Gardner, 2020).
Causes of Cold Stress
As discussed earlier, heat loss can occur in a neonate from four different mechanisms: evaporation, convection, conduction , and radiation. The ways in which a neonate can lose heat during the immediate care period are numerous, and only some of the most frequent are listed here. Through evaporation, the newborn loses heat quickly immediately post birth in the birthing room and must be dried vigorously by the nurse, wrapped in a warm blanket, or placed skin-to-skin with the birthing person. Heat loss can also be caused by newborn baths, and for this reason, it is generally recommended that the bath be given between 6 hours and 24 hours after birth, when the newborn has had multiple temperatures over 37.5° C (99.5° F). Baths should use warm water and be as short as possible. The nurse should plan to use appropriate rewarming measures after bathing, including skin-to-skin contact (New, 2019). Placing the crib too close to a drafty window or a cold wall accelerates neonatal heat loss, as does having a ceiling fan or circulating fan in the room.
Signs and Symptoms of Cold Stress
It is critical that the nurse be aware of the signs and symptoms of cold stress developing in the neonate so that they can recognize subtle changes. The most obvious sign is an axillary temperature reading below 36.5° C (97.7° F). (See Chapter 23 Newborn Assessment for a discussion of taking axillary temperatures in the newborn.) Neonates should have temperatures monitored often during the first hour after birth and every 4 hours after that. Beyond a low axillary temperature, a neonate experiencing cold stress will appear acrocyanotic, irritable, and hypotonic (Gardner, 2020). Additionally, the neonate will become hypoglycemic due to the increased metabolic demand on their body brought on by the cold stress.
Neonatal Cold Stress: Management
Management of hypothermia in a neonate is vital to keep them thermodynamically stable. The nurse caring for the neonate needs to know the steps to take to correct hypothermia before it escalates to cold stress.
1. Radiant Warmer : Warmers limit heat loss because of ease of access and constant temperature monitoring and control. During the immediate care period, neonates are subjected to many interventions that require them to be away from the birthing person and away from skin-to-skin temperature protection. Therefore, anytime the neonate is away from that protection, they should be wrapped in warm blankets, wear a hat, and be placed under the radiant warmer .
2. Handling and Temperature: Neonate temperatures should be monitored hourly for the first 4 hours post birth and then every 4 hours after that if they are stable. If they are not stable, temperatures need to be monitored more frequently while corrective measures are being taken. Temperatures should be taken before an intervention is done and then after. If the nurse uses corrective measures to increase a hypothermic neonate’s temperature, they should always recheck the temperature and document any change.
3. Glucose monitoring: Infants that are small-for-gestational age or large-for-gestational-age, will need blood glucose monitoring every 4 hours. Knowing that metabolic requirements increase for neonates in hypothermic states, the nurse will recognize that if a neonate is hypoglycemic, the nurse should also check their temperature.
4. Clothing: Neonates are sensitive to changes in airflow and evaporation, as well as changes in environment, and should be dressed and swaddled appropriately in the first 28 days. Hats, T-shirts or footed pajamas, and warm blankets are the very least they should be dressed in to ensure temperature regulation.
5. Respiratory Changes: Two of the first signs of cold stress are irritability and changes in respiration, so thorough and accurate assessment of the respiratory system is key to recognizing subtle changes in the neonate. Assessment of the respiratory system will be done hourly in the first 4 hours and after that every 4 hours until discharge. (Fellows, 2010)
Prevention and Management of Cold Stress
Prevention of cold stress is the focus for the nurse and includes keeping the neonate dry after birth and wrapped in warm blankets or in skin-to-skin contact with the birthing person if possible. If the newborn’s temperature is unstable with skin-to-skin contact, the nurse will swaddle the newborn in warm blankets and place them under the radiant warmer, rechecking their temperature in 30 minutes. If the neonate is already experiencing cold stress, the nurse will check the blood glucose of the newborn with a point of care machine and work diligently to rewarm the baby with swaddling and radiant warmer. Other interventions include ensuring that the newborn is dry and has a hat on their head, that all steps of the ten-step warm chain have been followed, and that a health-care provider has been notified of this critical change for the newborn.
When to Call for Help
The nurse will call the pediatrician or the NICU for help if, upon recheck in 30 minutes, the newborn’s temperature is still below 36.5°C; the infant loses consciousness; or the infant experiences respiratory distress signs, such as grunting, nasal flaring, retractions, tremors, or jitteriness along with inconsolable crying. Close glucose monitoring may be required for at-risk neonates because symptoms can be subtle.


5.4: Multiple Bonds
-
- Last updated
- Save as PDF
Learning Objectives
By the end of this section, you will be able to:
- Describe multiple covalent bonding in terms of atomic orbital overlap
- Relate the concept of resonance to π-bonding and electron delocalization
The hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of σ and π bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, C 2 H 4 , shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.
The three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the σ bonds from each carbon atom are formed using a set of sp 2 hybrid orbitals that result from hybridization of two of the 2 p orbitals and the 2 s orbital (Figure \(\PageIndex{1}\)).
These orbitals form the C–H single bonds and the σ bond in the Figure \(\PageIndex{2}\). The π bond in the C=C double bond results from the overlap of the third (remaining) 2p orbital on each carbon atom that is not involved in hybridization. This unhybridized p orbital (lobes shown in red and blue in Figure \(\PageIndex{2}\)) is perpendicular to the plane of the sp2 hybrid orbitals. Thus the unhybridized 2p orbitals overlap in a side-by-side fashion, above and below the internuclear axis (Figure \(\PageIndex{2}\)) and form a π bond.
In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of sp 2 hybrid orbitals tilted relative to each other, the p orbitals would not be oriented to overlap efficiently to create the π bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between σ and π bonds; rotation around single (σ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the σ bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the π bonding orbitals, essentially breaking the π bond.
In molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom (Figure \(\PageIndex{3}\)). We find this situation in acetylene, \(\ce{H−C≡C−H}\) (Figure \(\PageIndex{4}\)), The remaining sp orbitals form σ bonds with hydrogen atoms. The two unhybridized p orbitals per carbon are positioned such that they overlap side by side and, hence, form two π bonds. The two carbon atoms of acetylene are thus bound together by one σ bond and two π bonds, giving a triple bond.
Hybridization involves only σ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of π bonds are possible. Since the arrangement of π bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.
For example, molecule benzene has two resonance forms (Figure \(\PageIndex{5}\)). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is sp 2 . The electrons in the unhybridized p orbitals form π bonds. Neither resonance structure completely describes the electrons in the π bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory. (See the next module.)
Example \(\PageIndex{1}\): Assignment of Hybridization Involving Resonance
Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, SO 2 , is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in SO 2 ?
Solution
The resonance structures of SO 2 are
The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp 2 .
Exercise \(\PageIndex{1}\)
Another acid in acid rain is nitric acid, HNO 3 , which is produced by the reaction of nitrogen dioxide, NO 2 , with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO 2 ? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)
- Answer
-
sp 2


2 .
A full neurologic assessment of a newborn includes evaluation of each newborn reflex. List the newborn reflexes and the expected finding for each.
3 .
A newborn is feeding and acting fussy when the nurse walks into the room to complete a Ballard Assessment. What should the nurse know about timing the Ballard Assessment?
4 .
Explain why it is important that the nurse not perform the Ballard Assessment on a newborn that is possibly withdrawing from substances.
5 .
Summarize why the nurse must wear gloves the entire time they are completing an assessment on a newborn that has not yet had a bath post birth.
6 .
Describe two reasons the nurse would want to perform a behavioral assessment on a newborn before discharge home.
7 .
After completing a physical assessment and noting weight and measurements, the nurse completes a Ballard Assessment on a newborn.
|
SIGN
|
SCORE
|
|
Posture
|
3
|
|
Square Window
|
2
|
|
Arm Recoil
|
3
|
|
Popliteal Angle
|
3
|
|
Scarf Sign
|
3
|
|
Heel to Ear
|
2
|
|
TOTAL SCORE
|
|
Table 23.4
Neuromuscular Maturity Scoring
|
SIGN
|
SCORE
|
|
Skin
|
2
|
|
Lanugo
|
2
|
|
Plantar Surface
|
3
|
|
Breast
|
2
|
|
Eye/Ear
|
2
|
|
Genitals (Male/Female)
|
2 female
|
|
TOTAL SCORE
|
|
Table 23.5
Physical Maturity Score
Based on the indicated scores, what overall maturity score would this newborn receive?


11.7: The Huge Impact of Advertising
How are Advertising Media Selected?
Most Americans are bombarded daily with advertisements to buy things. Traditional advertising is any paid form of nonpersonal presentation by an identified sponsor. It may appear on television or radio; in newspapers, magazines, books, or direct mail; or on billboards or transit cards. In the United States, children between the ages of two and 11 are exposed to more than 25,600 exposures to advertising through TVs and online exposures a year. Adults are exposed to three times as many—more than two million commercials in a lifetime.
The money that big corporations spend on advertising is mind-boggling. Total advertising expenses in this country were estimated at more than $206 billion in 2017. Global advertising expenditures are approximately $546 billion annually. General Motors is America’s largest advertiser, spending over $3.1 billion annually. This is slightly over $350,000 per hour, seven days a week, 24 hours per day. America’s biggest global spender on advertising is Procter & Gamble at $4.6 billion.
Nissan was a sponsor of the 2016 Rio Olympic Games and provided 5,000 vehicles for the events. Ads for the 2018 Super Bowl cost between $5 million and $5.5 million for a 30-second commercial. A 30-second spot on NBC’s Sunday Night Football costs about $650,000.
The Impact of Technology and the Internet on Traditional Advertising
Many new media are not hardwired or regulated, and digital technology is delivering content anytime, anywhere. Cable, satellite, and the internet have highly fragmented audiences, making them tougher than ever to reach. In the late 1950s, Gunsmoke on CBS captured a 65 percent share of the TV audience nearly every Saturday night. Only one event, the Super Bowl, has a chance to do that now.
Traditional forms of entertainment are being rapidly digitized. Magazines, books, movies, shows, and games can be accessed through a laptop or a cell phone. In 2017, 93 million U.S. homes have broadband connections—nearly as many as the 119.6 million that now have cable and satellite hookups.
Technology is driving many of the changes, but so is consumer behavior. Advertiser questions abound. How do you market a product to young people when millions of them are glued to video game screens instead of TVs? How do you reach TV audiences when viewers can TiVo their way past your ads? How do you utilize social media to get the word out about your product, and once you do, how do you control the message if something goes viral? What role do influencers play in promoting products and services via various electronic platforms? What should you make of blogs? How do you plan a website that fosters sales and continually provides information and other forms of value for your customers? Product placements in films and streaming content? Podcasts? We will touch on each of these later in the chapter.
Choosing Advertising Media
The channels through which advertising is carried to prospective customers are the advertising media . Both product and institutional ads appear in all the major advertising media. Each company must decide which media are best for its products. Two of the main factors in making that choice are the cost of the medium and the audience reached by it.
Advertising Costs and Market Penetration
Cost per contact is the cost of reaching one member of the target market. Naturally, as the size of the audience increases, so does the total cost. Cost per contact enables an advertiser to compare media vehicles, such as television versus radio or magazine versus newspaper, or, more specifically, Forbes versus The Wall Street Journal. An advertiser debating whether to spend local advertising dollars for TV spots or radio spots could consider the cost per contact of each. The advertiser might then pick the vehicle with the lowest cost per contact to maximize advertising punch for the money spent. Often costs are expressed on a cost per thousand (CPM) contacts basis.
Reach is the number of different target consumers who are exposed to a commercial at least once during a specific period, usually four weeks. Media plans for product introductions and attempts at increasing brand awareness usually emphasize reach. For example, an advertiser might try to reach 70 percent of the target audience during the first three months of the campaign. Because the typical ad is short-lived and often only a small portion of an ad may be perceived at one time, advertisers repeat their ads so consumers will remember the message. Frequency is the number of times an individual is exposed to a message. Average frequency is used by advertisers to measure the intensity of a specific medium’s coverage.
Media selection is also a matter of matching the advertising medium with the product’s target market. If marketers are trying to reach teenage females, they might select Seventeen magazine. If they are trying to reach consumers over 50 years old, they may choose AARP: The Magazine. A medium’s ability to reach a precisely defined market is its audience selectivity . Some media vehicles, such as general newspapers and network television, appeal to a wide cross section of the population. Others—such as Brides, Popular Mechanics, Architectural Digest, MTV, ESPN, and Christian radio stations—appeal to very specific groups. Marketers must also consider utilizing various social media platforms and which platforms are most likely to reach the targeted market.
CONCEPT CHECK
- How is technology impacting the way advertisers reach their markets?
- What are the two main factors that should be considered when selecting advertising media?


43: Animal Reproduction and Development
Animal reproduction is necessary for the survival of a species. In the animal kingdom, there are innumerable ways that species reproduce. Asexual reproduction produces genetically identical organisms (clones), whereas in sexual reproduction, the genetic material of two individuals combines to produce offspring that are genetically different from their parents.
-
- 43.0: Prelude to Animal Reproduction and Development
- During sexual reproduction the male gamete (sperm) may be placed inside the female’s body for internal fertilization, or the sperm and eggs may be released into the environment for external fertilization. Seahorses provide an example of the latter. Following a mating dance, the female lays eggs in the male seahorse’s abdominal brood pouch where they are fertilized. The eggs hatch and the offspring develop in the pouch for several weeks.
-
- 43.1: Reproduction Methods
- During sexual reproduction the genetic material of two individuals is combined to produce genetically diverse offspring that differ from their parents. The genetic diversity of sexually produced offspring is thought to give species a better chance of surviving in an unpredictable or changing environment. Species that reproduce sexually must maintain two different types of individuals, males and females, which can limit the ability to colonize new habitats as both sexes must be present.
-
- 43.2: Fertilization
- Sexual reproduction starts with the combination of a sperm and an egg in a process called fertilization. This can occur either inside (internal fertilization) or outside (external fertilization) the body of the female. Humans provide an example of the former whereas seahorse reproduction is an example of the latter.
-
- 43.3: Human Reproductive Anatomy and Gametogenesis
- As animals became more complex, specific organs and organ systems developed to support specific functions for the organism. The reproductive structures that evolved in land animals allow males and females to mate, fertilize internally, and support the growth and development of offspring.
-
- 43.4: Hormonal Control of Human Reproduction
- The human male and female reproductive cycles are controlled by the interaction of hormones from the hypothalamus and anterior pituitary with hormones from reproductive tissues and organs. In both sexes, the hypothalamus monitors and causes the release of hormones from the pituitary gland. When the reproductive hormone is required, the hypothalamus sends a gonadotropin-releasing hormone (GnRH) to the anterior pituitary.
-
- 43.5: Human Pregnancy and Birth
- Pregnancy begins with the fertilization of an egg and continues through to the birth of the individual. The length of time of gestation varies among animals, but is very similar among the great apes: human gestation is 266 days, while chimpanzee gestation is 237 days, a gorilla’s is 257 days, and orangutan gestation is 260 days long. The fox has a 57-day gestation. Dogs and cats have similar gestations averaging 60 days.
-
- 43.7: Organogenesis and Vertebrate Formation
- Gastrulation leads to the formation of the three germ layers that give rise, during further development, to the different organs in the animal body. This process is called organogenesis. Organogenesis is characterized by rapid and precise movements of the cells within the embryo.
Thumbnail: Views of a Foetus in the Womb, detail. (Public Domain; Leonardo da Vinci via Wikimedia Commons )


1.3: Uncertainty in Measurement
3
Harper College Chemistry Department
By the end of this section, you will be able to:
- Define accuracy and precision
- Distinguish exact and uncertain numbers
- Correctly represent uncertainty in quantities using significant figures
- Apply proper rounding rules to computed quantities
Accuracy and Precision
Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision and accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure 1).
Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense
10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results in Table 1.
| Volume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers | ||
|---|---|---|
| Dispenser #1 | Dispenser #2 | Dispenser #3 |
| 283.3 | 298.3 | 296.1 |
| 284.1 | 294.2 | 295.9 |
| 283.9 | 296.0 | 296.1 |
| 284.0 | 297.8 | 296.0 |
| 284.1 | 293.9 | 296.1 |
Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL).
Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number. If we count eggs in a carton, we know exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used.
Significant Figures in Measurement
The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid.
Refer to the illustration in Figure 2. The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark and so a reasonable estimate of the liquid’s volume would be 21.6 mL. Thus, in the number 21.6, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL.
This concept holds true for all measurements, even if you do not actively make an estimation. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams with a nominal uncertainty in the measurement of ± 0.01 gram. If we weigh the quarter on a more sensitive balance, we may find that its mass is 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty, which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits. Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.
Whenever you make a measurement properly, all the digits in the result are significant. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them.
Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point. Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located.
In the above left example, all four digits are significant since all four are nonzero. On the right, the trailing zero is to the right of the decimal, hence all three digits are significant.
When zeros are captive (in between nonzero numbers), they are considered significant. In the the example of 70.607 mL, all digits are significant because both zeros are captive. The leading zeros in the left example are not significant. We could use exponential notation (review here ) and express the number as 8.32407 × 10 −3, then the number 8.32407 contains all of the significant figures and 10 −3 locates the decimal point.
The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3 × 10 3 (two significant figures), 1.30 × 10 3 (three significant figures, if the tens place was measured), or 1.300 × 10 3 (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant.
When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely when evaluating whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 3.17 × 10 8 people.
Key Concepts and Summary
Quantities can be exact or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the measurement. Measured values can be accurate (close to the true value) and/or precise (showing little variation when measured repeatedly).
Chemistry End of Chapter Exercises
1. Express each of the following numbers in scientific notation with correct significant figures:
(a) 711.0
(b) 0.239
(c) 90743
(d) 134.2
(e) 0.05499
(f) 10000.0
(g) 0.000000738592
2. Express each of the following numbers in exponential notation with correct significant figures:
(a) 704
(b) 0.03344
(c) 547.9
(d) 22086
(e) 1000.00
(f) 0.0000000651
(g) 0.007157
3. Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty:
(a) the number of eggs in a basket
(b) the mass of a dozen eggs
(c) the number of gallons of gasoline necessary to fill an automobile gas tank
(d) the number of cm in 2 m
(e) the mass of a textbook
(f) the time required to drive from San Francisco to Kansas City at an average speed of 53 mi/h
4. Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty:
(a) the number of seconds in an hour
(b) the number of pages in this book
(c) the number of grams in your weight
(d) the number of grams in 3 kilograms
(e) the volume of water you drink in one day
(f) the distance from San Francisco to Kansas City
5. How many significant figures are contained in each of the following measurements?
(a) 38.7 g
(b) 2 × 10 18 m
(c) 3,486,002 kg
(d) 9.74150 × 10 −4 J
(e) 0.0613 cm 3
(f) 17.0 kg
(g) 0.01400 g/mL
6. How many significant figures are contained in each of the following measurements?
(a) 53 cm
(b) 2.05 × 10 8 m
(c) 86,002 J
(d) 9.740 × 10 4 m/s
(e) 10.0613 m 3
(f) 0.17 g/mL
(g) 0.88400 s
7. The following quantities were reported on the labels of commercial products. Determine the number of significant figures in each.
(a) 0.0055 g active ingredients
(b) 12 tablets
(c) 3% hydrogen peroxide
(d) 5.5 ounces
(e) 473 mL
(f) 1.75% bismuth
(g) 0.001% phosphoric acid
(h) 99.80% inert ingredients
8. Round off each of the following numbers to two significant figures:
(a) 0.436
(b) 9.000
(c) 27.2
(d) 135
(e) 1.497 × 10 −3
(f) 0.445
9. Round off each of the following numbers to two significant figures:
(a) 517
(b) 86.3
(c) 6.382 × 10 3
(d) 5.0008
(e) 22.497
(f) 0.885
12. Consider the results of the archery contest shown in this figure.
(a) Which archer is most precise?
(b) Which archer is most accurate?
(c) Who is both least precise and least accurate?
13. Classify the following sets of measurements as accurate, precise, both, or neither.
(a) Checking for consistency in the weight of chocolate chip cookies: 17.27 g, 13.05 g, 19.46 g, 16.92 g
(b) Testing the volume of a batch of 25-mL pipettes: 27.02 mL, 26.99 mL, 26.97 mL, 27.01 mL
(c) Determining the purity of gold: 99.9999%, 99.9998%, 99.9998%, 99.9999%
Glossary
- accuracy
- how closely a measurement aligns with a correct value
- exact number
- number derived by counting or by definition
- meniscus
- the curve in the upper surface of a liquid close to the surface of the container, caused by surface tension
- precision
- how closely a measurement matches the same measurement when repeated
- rounding
- procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation
- significant figures
- (also, significant digits) all of the measured digits in a determination, including the uncertain last digit
- uncertainty
- estimate of amount by which measurement differs from true value


1.2.2: Documenting a Process
Documenting a Process
Every day, each of us will conduct many processes without even thinking about them: getting ready for work, using an ATM, reading our e-mail, etc. But as processes grow more complex, they need to be documented. For businesses, it is essential to do this, because it allows them to ensure control over how activities are undertaken in their organization. It also allows for standardization: McDonald’s has the same process for building a Big Mac in all of its restaurants.
The simplest way to document a process is to simply create a list. The list shows each step in the process; each step can be checked off upon completion. For example, a simple process, such as how to create an account on eBay, might look like this:
- Go to ebay.com.
- Click on “register.”
- Enter your contact information in the “Tell us about you” box .
- Choose your user ID and password.
- Agree to User Agreement and Privacy Policy by clicking on “Submit.”
Documenting a process in list format may be beneficial for minor uses or quick fixes; however, if the process is more extensive, listing every detail will cause disorganization. It is best to invest in resources that help manage all of the activities and resources that are required. The ability of a business to effectively document processes and activities can determine the overall efficiency of the business.
There are a great number of office applications (both free and paid) that can help efficiently document processes and business activities. Microsoft offers many resources that are designed for businesses to manage their business as efficiently as possible, inclusive towards developing efficient information systems. There are too many to list in this section, however; you can do research to see what other resources will be beneficial for your company and its needs.
For processes that are not so straightforward, documenting the process as a checklist may not be sufficient. For example, here is the process for determining if an article for a term needs to be added to Wikipedia:
- Search Wikipedia to determine if the term already exists.
- If the term is found, then an article is already written, so you must think of another term. Go to 1.
- If the term is not found, then look to see if there is a related term.
- If there is a related term, then create a redirect.
- If there is not a related term, then create a new article.
This procedure is relatively simple – in fact, it has the same number of steps as the previous example – but because it has some decision points, it is more difficult to track with as a simple list. In these cases, it may make more sense to use a diagram to document the process:


5.7: Predicting Solubility Trends
The solubility of many simple ionic compounds can be predicted by applying the set of rules shown below.
- Salts of the alkali metal ions and the ammonium ion, Li + , Na + , K + , and NH 4 + are almost always soluble .
- Virtually all metal nitrates and metal acetates are soluble.
- Metal halides are generally soluble , except for salts of Ag + , Pb 2 + , Cu + and Hg + .
- Metal sulfates are generally soluble , except for salts of Ba 2 + , Pb 2 + and Ca 2 + .
- With exception of the alkali metal ions and ammonium ( Rule 1 ), the following salts are generally insoluble : metal carbonates (CO 3 2 - ), metal phosphates (PO 4 3 - ) and metal chromates (CrO 4 2 - ).
- Metal hydroxides and metal sulfides are generally insoluble , except for those covered by Rule 1 and Ca 2 + , Sr 2 + and Ba 2 + .
Applying these rules to the reaction between lead nitrate and potassium iodide, the reactants are both soluble (Rule 1 and Rule 2). In the products, potassium nitrate will be soluble (Rule 2) and lead iodide will be insoluble , based on Rule 3.
Mixing each of the following salt solutions results in the formation of a precipate. In each case, identify the insoluble salt.
- NaCl + Pb(NO 3 ) 2
- Fe(C 2 H 3 O 2 ) 3 + KOH
- Ca(NO 3 ) 2 + K 2 SO 4
- Li 2 S + CuSO 4
- Co(C 2 H 3 O 2 ) 2 + LiOH
For each of the ionic compounds given below, determine whether or not the compound will be soluble in water, according to the trends given above.
- AgNO 3 ☐ soluble ☐ insoluble
- MgCl 2 ☐ soluble ☐ insoluble
- Na 2 SO 4 ☐ soluble ☐ insoluble
- AgCl ☐ soluble ☐ insoluble
- Ba(NO 3 ) 2 ☐ soluble ☐ insoluble
- PbI 2 ☐ soluble ☐ insoluble
- Mg(NO 3 ) 2 ☐ soluble ☐ insoluble
- BaSO 4 ☐ soluble ☐ insoluble
- FeCl 3 ☐ soluble ☐ insoluble
- Pb(CH 3 COO) 2 ☐ soluble ☐ insoluble


27.7: Sample proposal review (Measuring g using a pendulum)
-
- Last updated
- Save as PDF
Summary and Goal
The authors propose to measure the value of \(g\) to within \(10\)% by measuring the period of a simple pendulum, using the SHM equations and theory. The proposal is reasonably clear, but lacks some details in how to measure the initial angle of the pendulum. The authors propose to use a an amplitude of \(90^{\circ}\) for the pendulum, but at such a large angle, the motion is not expected to be SHM, since it is only so at small angles. By using a smaller angle, the experiment has a good chance of being successful in the proposed timeline.
Review
The experimental methods are described clearly and succinctly, with most information clearly stated. For the materials list, it is stated that “a mass” must be used. Here, it should be stated that a small, solid, non-deformable mass should be used to minimize drag and to act as a point mass. The authors refer to a “measuring device” when determining the amplitude of the pendulum, but this is not described. Anyhow, the amplitude of the oscillations in irrelevant for a pendulum in SHM, as long as the amplitude is small.
Most equations are described in the theory section, but it is incorrectly assumed that the period of a pendulum is independent of the drop angle for all angles. The small angle approximation is not expected to apply with an oscillation amplitude of \(90\).
No justification is provided for the use of \(20\) oscillations prior to measuring the period - it may be necessary to iterate on the reason why \(20\) oscillations was chosen.
The equipment can be easily obtained and is fairly inexpensive. Adequate resources are available to the group to perform this experiment. A clear troubleshooting plan is described and a method for evaluating success is included.
Timeline and team
This experiment is fairly simple and the equipment/setup is not difficult to handle. The proposed team should be qualified to perform this experiment in the proposed amount of time, although I worry a little bit about Dennis, as he seems to be a bit of a menace.
Overall Rating of the Proposal
Good - this proposal was clearly explained and is scientifically sound, apart from the use of a large angle for the oscillations. It was succinctly written, and most components of the experiment were clearly described. A little more detail in the justification for using \(20\) oscillations is necessary.


23.4: The induced electric field and eddy currents
-
- Last updated
- Save as PDF
So far, we have described electromagnetic induction in terms of the voltage that is induced by a changing magnetic field. This voltage is related to an electric field, which we discuss in this section. In Faraday’s Law, the voltage is induced across a closed loop (and can be thought of as an ideal battery placed in the loop). This is illustrated in Figure \(\PageIndex{1}\) which shows a loop in the plane of the page, and a magnetic field out of the plane of the page.
Exercise \(\PageIndex{1}\)
In Figure \(\PageIndex{1}\), with the induced voltage as shown, is the magnetic field increasing or decreasing?
- The magnetic field is increasing.
- The magnetic field is decreasing.
- Answer
-
As you recall, the electric potential difference between two points, \(A\) and \(B\) , is obtained from the electric field:
\[\begin{aligned} \Delta V = \int_A^B \vec E\cdot d\vec l\end{aligned}\]
In the case of an induced voltage across a loop, the points \(A\) and \(B\) are the same. The integral is thus over a closed path:
\[\begin{aligned} \Delta V = \oint \vec E\cdot d\vec l\end{aligned}\]
We can include this into Faraday’s Law by using the electric field instead of the potential difference:
\[\begin{aligned} \Delta V=-\frac{d\Phi_{B}}{dt} \end{aligned}\]
\[\therefore\oint\vec E\cdot d\vec l=-\frac{d\Phi_{B}}{dt}\]
where the last line is a more general form of Faraday’s Law. Note that in the case of electrostatics, where the electric field is produced by a distribution of charges, the integral \(\oint \vec E\cdot d\vec l\) must be zero, since the electric force is conservative; the work done by the electric field on a charge \(q\) over a closed path, which is just a charge \(q\) multiplied by that integral, must be zero. The force from an electric field that is induced by a time-varying magnetic field is not conservative!
Faraday’s Law as expressed with the electric field is much more general, and implies that a time-varying magnetic field will induce an electric field. This is true, independently of there existing a physical wire to carry the induced current.
Example \(\PageIndex{1}\)
A circular region with radius, \(R\) , of space contains a magnetic field that is uniform, and decreasing in magnitude with time:
\[\begin{aligned} \vec B(t) = B_0(1-at)\hat z\end{aligned}\]
where \(a\) and \(B_0\) are positive constants. Determine the electric field at a distance, \(r\) , from the center of the region, inside and outside of the region with the magnetic field.
Solution
Figure \(\PageIndex{2}\) shows the circular region of magnetic field, as well as a circular path of radius, \(r\) , that defines the region over which we calculate the flux of the magnetic field.
First, we consider the induced electric field in the region with a magnetic field, where \(r<R\) . We choose a circle of radius \(r\) to calculate the flux of the magnetic field. Since the magnetic field is uniform within that region, the flux is given by:
\[\begin{aligned} \Phi_B = \vec B \cdot \vec A = BA = B_0(1-at) \pi r^2\end{aligned}\]
The circulation of the electric field is easily found, since the electric field forms concentric circles (by symmetry):
\[\begin{aligned} \oint \vec E \cdot d\vec l = \oint Edl = E \oint dl = E(2\pi r)\end{aligned}\]
Applying Faraday’s Law, the electric field is found to be:
\[\begin{aligned} \oint \vec E\cdot d\vec l &= -\frac{d\Phi_B}{dt}\\[4pt] E(2\pi r) &= -\frac{d}{dt} B_0(1-at) \pi r^2\\[4pt] 2E &= B_0ar\\[4pt] \therefore E&=\frac{B_0a}{2}r\quad\text{(inside the region of magnetic field)}\end{aligned}\]
and we see that, inside the region with the magnetic field, the strength of the induced electric field is proportional to the distance from the center of the region (i.e. it increases linearly with \(r\) ).
For the region where the magnetic field is zero, we again calculate the circulation of the electric field around a circular loop of radius \(r>R\) :
\[\begin{aligned} \oint \vec E \cdot d\vec l = \oint Edl = E \oint dl = E(2\pi r)\end{aligned}\]
The flux of the magnetic field through that loop is however related to the area of the region with the magnetic field (of radius, \(R\) ):
\[\begin{aligned} \Phi_B = \vec B \cdot \vec A = BA = B_0(1-at) \pi R^2\end{aligned}\]
Again, applying Faraday’s Law:
\[\begin{aligned} \oint \vec E\cdot d\vec l &= -\frac{d\Phi_B}{dt}\\[4pt] E(2\pi r) &= -\frac{d}{dt} B_0(1-at) \pi R^2\\[4pt] 2Er&= B_0aR^2\\[4pt] \therefore E&=\frac{B_0aR^2}{2r}\quad\text{(outside the region of magnetic field)}\end{aligned}\]
Outside the region with a magnetic field, the magnitude of the electric field decreases with the distance from the center of the region.
Discussion
In this example, we determined the electric field that is induced by a varying magnetic field. In this case, the electric field lines form closed circles and result in a non-conservative force. When the electric field is formed by a distribution of electric charges, the field lines begin and end on charges, which is not the case for an induced electric field.
Magnetic braking
When a conducting material moves into a region of magnetic field, an electric field forming closed loops is induced in the material, thus inducing small current loops, called “eddy currents”. The magnetic field can then exert a force on those currents, effectively resulting in a force on the material. This is the principle behind magnetic braking, which is used in some trains and in other applications.
Figure \(\PageIndex{3}\) illustrates how a magnetic brake can be used to slow a rotating wheel made of a conducting material (the material must conduct or the induced electric field will not produce any current). A magnetic field is produced (e.g. by a fixed permanent magnet) in a direction perpendicular to the wheel, over a small area (shown at the bottom of the wheel in Figure \(\PageIndex{3}\)).
For material located at the bottom left of the wheel, the magnetic flux is increasing, since the material is moving from a region with no magnetic field into a region with a magnetic field. In that part of the region, clockwise eddy currents will form, as those result in a magnetic field into the page, to counter the increasing magnetic flux (Lenz’s Law). The bottom right side of the wheel is leaving the magnetic field, and will thus have eddy currents in the opposite direction. The currents from both sides add up in the center, resulting in a net downwards current. The magnetic force on that downwards current is to the left, resulting in a torque that slows the wheel. This is magnetic braking.
Again, this is no more than conservation of energy at play. Since we induce currents by making the wheel move into/out of a region of magnetic field, the electrical energy in those currents must come from somewhere (either we do work to keep the wheel rotating, or the wheel loses kinetic energy). Any time that we try to move a conductor through a magnetic field, in a way that current is induced, we will have to exert a force and do work. In the case of magnetic braking, the wheel will convert its rotational kinetic energy into heat (the eddy currents will heat up the wheel). The main issue with magnetic braking is that one needs to be able to dissipate the heat. The main advantage is that there are no parts that wear out, as opposed to braking with friction. In addition, magnetic braking is very smooth, and only acts when there is motion. As soon as the wheel stops rotating, the magnetic flux is constant everywhere and the eddy currents disappear.
Exercise \(\PageIndex{1}\)
Suppose that the magnetic field in Figure \(\PageIndex{3}\) pointed into the page. Would the magnetic break still work?
- Yes.
- No.
- Answer


Skip to main content
-
absorption
-
the taking in of broken-down substances into the cells
-
Bitot’s spots
-
oval or triangular patches on the conjunctiva of built-up keratin
-
chyme
-
gastric juices and digested food particles
-
digestion
-
the process by which the body breaks down what is ingested into absorbable substances
-
icterus
-
another word for jaundice, a yellowish pigmentation of the skin
-
ingestion
-
the taking in of substances into the body through swallowing or transdermal absorption
-
koilonychia
-
soft nails with a concave indentation, making them appear “scooped out”
-
mastication
-
digestion of food using mostly mechanical and some chemical processes
-
metabolism
-
the chemical processes in the body that convert absorbed substances to energy
-
xanthelasmas
-
yellow plaques on or near the eyelids
-
xanthomas
-
localized lipid deposits
-
xeroderma
-
dry skin
-
xerosis
-
extreme dryness of skin or tissue; in corneal xerosis, the cornea may appear hazy