11.8: Oxidation of Alkenes to Vicinal Diols
After completing this section, you should be able to
- write the equation for the hydroxylation of an alkene using osmium tetroxide, and draw the structure of the cyclic intermediate.
- draw the structure of the diol formed from the reaction of a given alkene with osmium tetroxide.
- identify the alkene, the reagents, or both, that must be used to prepare a given 1,2-diol.
Make certain that you can define, and use in context, the key terms below.
- diol
- glycol
- hydroxylation
The previous section discussed the reduction of a double bond, so adding hydrogen to the the double bond. This section will discuss oxidation. In organic chemistry, this is a reaction that where the carbon atom loses electron density, which happens when new bond to a more electronegative atom occurs or when a double bond is broken between a carbon and a less electronegative atom. A simplified to say this is in organic chemistry a reduction is more bonds to hydrogen and oxidation is more bonds to oxygen often.
Syn Dihydroxylation
A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons. Osmium tetroxide oxidizes alkenes to give glycols through a syn addition, resulting in cis vicinal diols. Next semester in the chapter on ethers, you'll learn how to make trans vicinal diols.
Introduction
The reaction with \(OsO_4\) is a concerted process that has a cyclic intermediate and no rearrangements. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give meso products and trans alkenes give racemic mixtures .
\(OsO_4\) is formed slowly when osmium powder reacts with gasoues \(O_2\) at ambient temperature. Reaction of bulk solid requires heating to 400 °C:
\[Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4\]
Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO 4 and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO 4 was potassium permanganate, \(KMnO_4\). Although syn diols will result from the reaction of KMnO 4 and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of overoxidation .
Mechanism
-
Electrophilic attack on the alkene
- Pi bond of the alkene acts as the nucleophile and reacts with osmium (VIII) tetroxide (OsO 4 )
-
2 electrons from the double bond flows toward the osmium metal
- In the process, 3 electron pairs move simultaneously
- Cyclic ester with Os (VI) is produced
-
Reduction
-
H
2
S reduces the cyclic ester
- NaHSO 4 with H 2 O may be used
- Forms the syn-1,2-diol (glycol)
-
H
2
S reduces the cyclic ester
Example: Dihydroxylation of 1-ethyl-1-cycloheptene
Hydroxylation of alkenes
Dihydroxylated products (glycols) are obtained by reaction with aqueous potassium permanganate (pH > 8) or osmium tetroxide in pyridine solution. Both reactions appear to proceed by the same mechanism (shown below); the metallocyclic intermediate may be isolated in the osmium reaction. In basic solution the purple permanganate anion is reduced to the green manganate ion, providing a nice color test for the double bond functional group. From the mechanism shown here we would expect syn-stereoselectivity in the bonding to oxygen, and regioselectivity is not an issue.
When viewed in context with the previously discussed addition reactions, the hydroxylation reaction might seem implausible. Permanganate and osmium tetroxide have similar configurations, in which the metal atom occupies the center of a tetrahedral grouping of negatively charged oxygen atoms. How, then, would such a species interact with the nucleophilic pi-electrons of a double bond? A possible explanation is that an empty d-orbital of the electrophilic metal atom extends well beyond the surrounding oxygen atoms and initiates electron transfer from the double bond to the metal, in much the same fashion noted above for platinum. Back-bonding of the nucleophilic oxygens to the antibonding π*-orbital completes this interaction. The result is formation of a metallocyclic intermediate, as shown above.
Outside links
References
- Dehestani, Ahmad et al. (2005). Ligand-assisted reduction of osmium tetroxide with molecular hydrogen via a [3+2] mechanism. Journal of the American Chemical Society, 2005, 127 (10), 3423-3432.
- Sorrell, Thomas, N. Organic Chemistry . New York: University Science Books, 2006.
- Vollhardt, Peter, and Neil E. Schore. Organic Chemistry: Structure and Function . 5th Edition. New York: W. H. Freeman & Company, 2007.
1. Give the major product.
2. What is the product in the dihydroxylation of (Z)-3-hexene?
3. What is the product in the dihydroxylation of (E)-3-hexene?
4. Draw the intermediate of this reaction.
5. Fill in the missing reactants, reagents, and product.
- Answer s
-
1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side.
2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction.
3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products.
4. A cyclic osmic ester is formed.
5. The Diels-Alder cycloaddition reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol.
Contributors and Attributions
- Shivam Nand
-
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
- Kristen Perano
- Layne Morsch (University of Illinois Springfield)
- Lauren Reutenauer (Amherst College)


6.9: Documentation
Documentation Tips:
Documentation should include all data described in the pre, during, and post-chest tube removal areas. Record the date/time of the chest tube removal, any drainage not recorded in the collection chamber, and the appearance of the dressing and wound if possible. Note the patient’s response to the procedure. Include vital signs and respiratory assessment. Document patient teaching and patient’s level of understanding.
Sample Documentation:
06/27/20xx 1430
Chest tube to right lateral lower chest wall intact. Water seal chest tube drainage system in upright position and below the level of the client’s chest. Tubing is free of kinks and patent. Suction is set at prescribed -20 mmHg. Tidaling present in water seal system. No air leak identified. 50 mL of serosanguineous drainage noted in collection chamber over 8 hours without clots present. Respiratory rate 18 and pulse oximetry reading 96% on room air. Respirations are symmetrical and unlabored. Breath sounds are diminished in the posterior right lower lobe. No adventitious breath sounds or subcutaneous emphysema noted. Trachea is midline. Dressing is dry and intact. Client rates pain at 2/10 and at a tolerable level. Denies sputum. Continue to encourage deep breathing and coughing hourly. Two clamps and a bottle of sterile water are at the bedside.
Hector Ramos, RN


7.12: Rigoletto
To study an aria, you simply have to know the larger story of which it is a part. Our aria, “La donna e mobile,” is heard in the beginning of act 3.
Synopsis
Place: Mantua
Time: the Sixteenth century
Act 1
Scene 1: A room in the palace
At a ball in his palace, the Duke sings of a life of pleasure with as many women as possible: “Questa o quella” (“This woman or that”). He has seen an unknown beauty in church and desires to possess her, but he also wishes to seduce the Countess of Ceprano. Rigoletto, the Duke’s hunchbacked court jester, mocks the husbands of the ladies to whom the Duke is paying attention, and advises the Duke to get rid of them by prison or death. Marullo, one of the guests at the ball, informs the noblemen that Rigoletto has a “lover”, and the noblemen cannot believe it. The noblemen resolve to take vengeance on Rigoletto. Subsequently Rigoletto mocks Count Monterone, whose daughter the Duke had seduced. Count Monterone is arrested at the Duke’s order and curses the Duke and Rigoletto. The curse genuinely terrifies Rigoletto.
Scene 2: A street, with the courtyard of Rigoletto’s house
Thinking of the curse, Rigoletto approaches his house and is accosted by the assassin Sparafucile, who walks up to him and offers his services. Rigoletto considers the proposition but finally declines; Sparafucile wanders off, after repeating his own name a few times. Rigoletto contemplates the similarities between the two of them: “Pari siamo!” (“We are alike!”); Sparafucile kills men with his sword, and Rigoletto uses “a tongue of malice” to stab his victims. Rigoletto opens a door in the wall and returns home to his daughter Gilda. They greet each other warmly: “Figlia!” “Mio padre!” (“Daughter!” “My father!”). Rigoletto has been concealing his daughter from the Duke and the rest of the city, and she does not know her father’s occupation. Since he has forbidden her to appear in public, she has been nowhere except to church and does not even know her own father’s name.
When Rigoletto has gone, the Duke appears and overhears Gilda confess to her nurse Giovanna that she feels guilty for not having told her father about a young man she had met at the church. She says that she fell in love with him, but that she would love him even more if he were a student and poor. As she declares her love, the Duke enters, overjoyed. Gilda, alarmed, calls for Giovanna, unaware that the Duke had sent her away. Pretending to be a student, the Duke convinces Gilda of his love: “È il sol dell’anima” (“Love is the sunshine of the soul”). When she asks for his name, he hesitantly calls himself Gualtier Maldè. Hearing sounds and fearing that her father has returned, Gilda sends the Duke away after they quickly trade vows of love: “Addio, addio” (“Farewell, farewell”). Alone, Gilda meditates on her love for the Duke, whom she believes is a student: “Gualtier Maldè!… Caro nome” (“Dearest name”).
Later, a preoccupied Rigoletto returns: “Riedo!… perché?” (“I’ve returned!… why?”), while the hostile noblemen outside the walled garden (believing Gilda to be the jester’s mistress, unaware she is his daughter) get ready to abduct the helpless girl. Convincing Rigoletto that they are actually abducting the Countess Ceprano, they blindfold him and use him to help with the abduction: “Zitti, zitti” (“Softly, softly”). With her father’s unknowing assistance Gilda is carried away by the noblemen. Upon realizing that it was in fact Gilda who was carried away, Rigoletto collapses, remembering the curse.
Act 2
The Duke’s Palace
The Duke is concerned that Gilda has disappeared: “Ella mi fu rapita!” (“She was stolen from me!”) and “Parmi veder le lagrime” (“I seem to see tears”). The noblemen then enter and inform him that they have captured Rigoletto’s mistress. By their description, he recognizes it to be Gilda and rushes off to the room where she is held: “Possente amor mi chiama” (“Mighty love beckons me”). Pleased by the Duke’s strange excitement, the courtiers now make sport with Rigoletto, who enters singing. He tries to find Gilda by pretending to be uncaring, as he fears she may fall into the hands of the Duke. Finally, he admits that he is in fact seeking his daughter and asks the courtiers to return her to him: “Cortigiani, vil razza dannata” (“Accursed race of courtiers”). Rigoletto attempts to run into the room in which Gilda is being held, but the noblemen beat him. Gilda rushes in and begs her father to send the people away. The men leave the room, believing Rigoletto has gone mad. Gilda describes to her father what has happened to her in the palace: “Tutte le feste al tempio” (“On all the blessed days”). In a duet Rigoletto demands vengeance against the Duke, while Gilda pleads for her lover: “Sì! Vendetta, tremenda vendetta!” (“Yes! Revenge, terrible revenge!”).
Act 3
A street outside Sparafucile’s house
Listen: “Bella figlia dell’amore”
Please listen to the following audio file to hear the 1907 Victor Records recording with Enrico Caruso, Bessie Abott, Louise Homer and Antonio Scotti’s
An audio element has been excluded from this version of the text. You can listen to it online here: http://pb.libretexts.org/map/?p=300
A portion of Sparafucile’s house is seen, with two rooms open to the view of the audience. Rigoletto and Gilda, who still loves the Duke, arrive outside. The Duke’s voice can be heard singing “La donna è mobile” (“Woman is fickle”), laying out the infidelity and fickle nature of women. Rigoletto makes Gilda realize that it is the Duke who is in the assassin’s house attempting to seduce Sparafucile’s sister, Maddalena: “Bella figlia dell’amore” (“Beautiful daughter of love”).
 He orders his daughter to put on a man’s clothes to prepare to leave for Verona and states that he plans to follow later. With falling darkness, a thunderstorm approaches and the Duke determines to remain in the house. Sparafucile assigns to him the ground floor sleeping quarters.
Gilda, who still loves the Duke despite knowing him to be unfaithful, returns dressed as a man. She overhears Maddalena begging for the Duke’s life, and Sparafucile promises her that if by midnight another can be found in place of the Duke, he will spare the Duke’s life. Gilda resolves to sacrifice herself for the Duke and enters the house. She is immediately mortally wounded and collapses.
At midnight, when Rigoletto arrives with money, he receives a corpse wrapped in a sack, and rejoices in his triumph. Weighting it with stones, he is about to cast the sack into the river when he hears the voice of the Duke singing a reprise of his “La donna è mobile” aria. Bewildered, Rigoletto opens the sack and, to his despair, discovers his mortally wounded daughter. For a moment, she revives and declares she is glad to die for her beloved: “V’ho ingannato” (“Father, I deceived you”). She dies in his arms. Rigoletto’s wildest fear materializes when he cries out in horror: “La maledizione!” (“The curse!”)
Contributors and Attributions
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu . License : CC BY: Attribution
- Rigoletto. Provided by : Wikipedia. Located at : http://en.Wikipedia.org/wiki/Rigoletto . License : CC BY-SA: Attribution-ShareAlike


1.7.2 Cache Memory - Locality of reference
Locality of Reference
Locality of reference refers to a phenomenon in which a computer program tends to access same set of memory locations for a particular time period. In other words, Locality of Reference refers to the tendency of the computer program to access instructions whose addresses are near one another. The property of locality of reference is mainly shown by:
- Loops in program cause the CPU to repeatedly execute a set of instructions that constitute the loop.
- Subroutine calls, cause the set of instructions are fetched from memory each time the subroutine gets called.
- References to data items also get localized, meaning the same data item is referenced again and again.
Even though accessing memory is quite fast, it is possible for repeated calls for data from main memory can become a bottleneck. By using faster cache memory, it is possible to speed up the retrieval of frequently used instructions or data.
In the above figure, you can see that the CPU wants to read or fetch the data or instruction. First, it will access the cache memory as it is near to it and provides very fast access. If the required data or instruction is found, it will be fetched. This situation is known as a cache hit. But if the required data or instruction is not found in the cache memory then this situation is known as a cache miss. Now the main memory will be searched for the required data or instruction that was being searched and if found will go through one of the two ways:
- The inefficent method is to have the CPU fetch the required data or instruction from main memory and use it. When the same data or instruction is required again the CPU again has to access the main memory to retrieve it again .
- A much more efficient method is to store the data or instruction in the cache memory so that if it is needed soon again in the near future it could be fetched in a much faster manner.
Cache Operation:
This concept is based on the idea of locality of reference. There are two ways in which data or instruction are fetched from main memory then get stored in cache memory:
-
Temporal Locality
Temporal locality means current data or instruction that is being fetched may be needed soon. So we should store that data or instruction in the cache memory so that we can avoid again searching in main memory for the same data. -
When CPU accesses the current main memory location for reading required data or instruction, it also gets stored in the cache memory which is based on the fact that same data or instruction may be needed in near future. This is known as temporal locality. If some data is referenced, then there is a high probability that it will be referenced again in the near future.
-
Spatial Locality
Spatial locality means instruction or data near to the current memory location that is being fetched, may be needed by the processor soon. This is different from the temporal locality in that we are making a guess that the data/instructions will be needed soon. With temporal locality we were talking about the actual memory location that was being fetched.
Adapted From:
"Locality of Reference and Cache Operation in Cache Memory"
by
balwant_singh
,
Geeks for Geeks
is licensed under
CC BY-SA 4.0


Skip to main content
-
More precisely, in the case of a complex wavefunction φ, the probability is the product of φ and its complex conjugate φ*
-
Hoffmann, R. (1963). "An Extended Hückel Theory. I. Hydrocarbons.".
J. Chem. Phys.
39
(6): 1397–1412. doi:
10.1063/1.1734456
. Bibcode:
1963JChPh..39.1397H
.
-
Cotton, F. A.; Harris, C. B. Inorg. Chem., 1965, 4 (3), 330-333. DOI|10.1021/ic50025a015
-
C. F. Bender and H. F. Schaefer III, New theoretical evidence for the nonlinearity of the triplet ground state of methylene, J. Am. Chem. Soc. 92, 4984–4985 (1970).
-
Hoffmann, R. (1982). "Building Bridges Between Inorganic and Organic Chemistry (Nobel Lecture)".
Angew. Chem. Int. Ed.
21
(10): 711–724. doi:10.1002/anie.198207113.
-
I. Naumov and R. J. Hemley, Acc. Chem. Res. 47, 3551–3559 (2014)
dx.doi.org/10.1021/ar5002654
.


4.26: Double Entry Bookkeeping System
Learning Outcomes
- Describe the double entry bookkeeping system
Double-Entry Bookkeeping
Double-entry bookkeeping, in accounting, is a system of bookkeeping so named because every entry to an account requires a corresponding and opposite entry to a different account. The double entry has two equal and corresponding sides known as debit and credit. The left-hand side is debit and right-hand side is credit. For instance, recording a sale of $100 might require two entries: a debit of $100 to an account named “Cash” and a credit of $100 to an account named “Revenue.”
Debits and Credits, Left and Right
You may hear accountants talk in terms of debits and credits. Debit literally means left, and credit means right. Accountants use a two-column journal to record transactions, so the above bonus would be recorded like this in 2017:
| Account | Debit | Credit |
|---|---|---|
| Bonus (expense) | 5,000 | |
| Accrued Bonus (liability) | 5,000 |
In this entry, the bonus is recognized before the cash payment to the salesperson actually occurs, along with a liability in the same amount. In the following month, when the company pays the bonus, it would record the following entry:
| Account | Debit | Credit |
|---|---|---|
| Accrued Bonus (liability) | 5,000 | |
| Checking Account (asset) | 5,000 |
The cash balance declines as a result of paying the commission, which also eliminates the liability. The reason your debit card is called a debit card is because the bank shows your balance as a liability because they owe your money to you—in essence, they are just holding it for you. A liability usually has a credit balance (balance on the right side of the ledger) and so when you spend money and they pay it out on your account, they debit your account (a debit offsets a credit.) The figure below illustrates how debits and credit affect various types of accounts.
The Accounting Cycle
The accounting cycle begins with transactions and ends with completed financial statements. The first step is to journalize the transaction. The journal is a chronological list of each accounting transaction and includes at a minimum the date, the accounts affected, and the amounts to be debited and credited.
Periodically, depending on the business, journal entries are posted to the general ledger. The general ledger is the exact same information as the journal, but sorted by account.
Once all journal entries are posted, the ledger balances are posted to a trial balance, which is a list of all the accounts and balances, checking to see that the total of the debit column is equal to the total of the credit column. Each account is then checked for accuracy and adjusted, if need be, before the financial statements are run. This process is referred to as the “Accounting Cycle.”
There are no debits or credits on the financial statements—they are informational reports, not data, so they are built with the user in mind
Summary
The double-entry bookkeeping system is summarized below:
- Assets. Items of financial value that the business controls (“owns”) for the purpose of producing income for the owners.
- Liabilities. Monies that the business owes to non-owners.
- Owners Equity. The theoretical value of the business that would be distributed to the owners after the assets were sold and the liabilities paid.
- Revenue. Payments made to the business by customers for the goods and/or services provided by the business.
- Expenses. Costs incurred by the business in providing the goods and/or services purchased by the customers.
Practice questions


4.7: Stoichiometric calculations
Conversion factors from a chemical equation
Stoichiometry pronounced as “stɔɪkiˈɒmɪtri” is the calculation of the amount of reactants and products in a chemical reaction. It is based on the fact that a balanced chemical equation is also a set of mole-to-mole equalities between the reactants and the products. Each equality gives two conversion factors that allow calculating the mole of one substance from the given mole of any other substance in the equation.
Fig. 4.6.1 lists the chemical equation for photosynthesis reaction, the mole-to-mole equalities from the equation, and the two conversion factors from each of the equality, as an example. The conversion factors are used to calculate the unknown quantity in the mole from the known quantity in the mole of any other reactant or product in the same chemical equation, as explained in the following examples.
Mole-to-mole conversion from a chemical equation
Calculation of the mole of the desired substance from the given mole of any reactant or product needs one conversion factor. The steps are:
- write the given quantity and the desired quantity,
- write the balanced chemical equation,
- write the equality between the given and the desired substances,
- right the conversion factor that has the given substance in the denominator and the desired substance in the numerator,
- multiply the given quantity with the conversion factor. Double-check to make sure that it cancels the given substance and leaves the desired substance in the answer.
Calculate the moles of glucose produced from 3.0 moles of carbon dioxide in the photosynthesis reaction?
Solution
i. Given: 3.0 mole CO 2 , Desired: ? moles of C 6 H 12 O 6
ii. Chemical equation: \(\ce{6CO2 + 6H2O -> C6H12O6 + 6O2}\)
iii. The desired equality: 6 mol CO 2 = 1 mol C 6 H 12 O 6 ,
iv. Desired conversion factor: \(\frac{1 \mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{6 \mathrm{~mol} \mathrm{~CO_2}}\)
v. Calculations:\[3.0 \cancel{\mathrm {~mol~CO_2}} \times \frac{1 \mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{6 \cancel{\mathrm{~mol} \mathrm{~CO_2}}}=0.50 \mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\nonumber\]
Magnesium reacts with HCl by this reaction: \(\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)^}\). Calculate the moles of Mg needed to produce 3 moles of H 2 ?
Solution
i. Given: 3 mol of H 2 . Desired: ? Moles of Mg.
ii. The chemical equation is given.
iii. The desired equality: 1 mol Mg = 1 mol H 2 .
iv. The desired conversion factor: \(\frac{1 ~mol~Mg}{1 ~mol{~H}_{2}}\)
v. Calculation: \[3 \cancel{\mathrm{~mol} \mathrm{~H}_{2}} \times \frac{1 \mathrm{~mol} \mathrm{~Mg}}{1 \cancel{\mathrm{~mol} \mathrm{~H}_{2}}}=3 \operatorname{~mol} ~M g\nonumber\]
Mole-to-mass conversion from a chemical equation
Calculations described in the previous two examples show calculating moles of the desired substance from the moles of the given substance using a conversion factor from a chemical equation. The molar mass is a conversion factor from a mole-to-gram of the substance. So, add a molar mass of the desired substance as a second conversion factor, as explained in the following examples. Double-check that all the units cancel out, leaving the mass unit of the desired substance in the final answer.
Calculate grams of AgCl precipitate formed from 2.0 moles of CaCl 2 consumed in the following reaction: \[\ce{2AgNO3 + CaCl2 -> Ca(NO3)2 + 2AgCl(s)(v)}\nonumber\]
Solution
i. Given: 2.0 moles of CaCl 2 . Desired: ? g AgCl
ii. Molar mass of AgCl = 1x107.87 g Ag.mol -1 + 1x35.45 g C.mol -1 = 143.3 g AgCl.mol -1
iii. The chemical equation is given.
iv. The desired equality: 1 mol CaCl 2 = 2 mol AgCl, and 1 mol AgCl =143.3 g AgCl.
v. The desired conversion factors: \[\frac{2 \mathrm{~mol} \mathrm{~AgCl}}{1 \mathrm{~mol} \mathrm{~CaCl}_{2}} \quad\text { and }\quad \frac{143.3 \mathrm{~g} \mathrm{~AgCl}}{1 \mathrm{~mol} \mathrm{~AgCl}}\nonumber\]
vi. Calculation: \[2.0\cancel{\mathrm{~mol}\mathrm{~CaCl}_{2}}\times\frac{2 \cancel{\mathrm{~mol} \mathrm{~AgCl}}}{1 \cancel{\mathrm{~mol} \mathrm{~CaCl}_{2}}} \times\frac{143.3 \mathrm{~g} \mathrm{~AgCl}}{1 \cancel{\mathrm{~mol} \mathrm{~AgCl}}}=573.3 \mathrm{~g} \mathrm{~AgCl}\nonumber\]
How many grams of carbon dioxide are needed to react with 2 moles of water in the photosynthesis reaction?
Solution
i. Given: 2.0 moles of H 2 O Desired: ? g CO 2 .
ii. Molar mass of CO 2 = 1x12.011 g C.mol -1 + 2x15.999 g O.mol -1 = 44.009 g CO 2 .mol -1
iii. The chemical equation: : 6CO 2 + 6 H 2 O → C 6 H 12 O 6 + 6O 2 .
iv. The desired equality: 6 mol CO 2 = 6 mol H 2 O, and 1 mol CO 2 = = 44.009 g CO 2 .
v. The desired conversion factors: \[\frac{6 \mathrm{~mol} \mathrm{~CO}_{2}}{6 \mathrm{~mol} \mathrm{~H}_{2}}\mathrm{O}\quad \text { and }\quad \frac{44.009 \mathrm{~g} \mathrm{~CO}_{2}}{1 \mathrm{~mol} \mathrm{~CO}_{2}}\nonumber\]
vi.  If the given substance is in grams and the desired substance is also in grams, then two additional conversion factors based on the molar masses are needed. That is, the following conversions are needed:
Mass of given substance⇒mole of the given substance⇒mole of the desired substance⇒grams of the desired substance.
The reciprocal molar mass of the given substance is the first conversion factor, the mole to mole conversion factor from the chemical equation is the second conversion factor, and the molar mass of the desired substance is the third conversion factor needed. Make sure that each conversion factor cancels the denominator unit of its multiplier to the right, and the desired unit is left in the answer. The following examples explain these calculations.
How many grams of Mg are needed to produce 1.01 g of H 2 gas in this reaction: \[\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)^}\nonumber\]
Solution
i. Given: 1.01 g H 2 Desired: ? g Mg.
ii. Molar mass of H 2 = 2 x1.008 g H.mol -1 = 2.016 g H 2 .mol -1 , and molar mass = 24.305 g Mg.mol -1 .
iii. The chemical equation is given in the problem.
iv.  Calculate by multiplying the given quantity consecutively with the three desired conversion factors from the equalities:
\[1.01 \cancel{\mathrm{~g ~H_{2}}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~H}_{2}}}{2.016 \cancel{\mathrm{~g} \mathrm{~H}_{2}}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~Mg}}}{1 \cancel{\mathrm{~mol} \mathrm{~H}_{2}}} \times \frac{24.305 \mathrm{~g} \mathrm{~Mg}}{1 \cancel{\mathrm{~mol} \mathrm{~Mg}}}=11.9 \mathrm{~g~Mg}\nonumber\]
How many grams of glucose are produced if 22.0 g of carbon dioxide is consumed in the photosynthesis reaction?
Solution
i. Given: 22.0 g CO 2 Desired: ? g C 6 H 12 O 6 .
ii. Molar masses: of CO 2 = 1x12.011 g C.mol -1 + 2x15.999 g O.mol -1 = 44.009 g CO 2 .mol -1 , and molar mass of C 6 H 12 O 6 = 6x12.011 g C.mol -1 + 12x1.008 g H.mol -1 + 6x15.999 g O.mol -1 = 180.156 g C 6 H 12 O 6 .mol -1 .
iii. The chemical equation: \(\ce{6CO2 + 6H2O -> C6H12O6 + 6O2}\)
iv. The desired equalities: 6 mol CO 2 = 1 mol C 6 H 12 O 6 , 1 mol CO 2 = 44.009 g CO 2 , 1 mol C 6 H 12 O 6 = 180.156 g C 6 H 12 O 6 .
v. Calculate by multiplying the given quantity consecutively with the three desired conversion factors from the equalities:
\[22.0 \cancel{\mathrm{~g} \mathrm{~CO}_{2}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~CO}_{2}}}{44.009 \cancel{\mathrm{~g} \mathrm{~CO}_{2}}} \times \frac{1 \cancel{\mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}}{6 \cancel{\mathrm{~mol} \mathrm{~CO}_{2}}} \times \frac{180.156 \mathrm{~g} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}{1 \cancel{\mathrm{~mol} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}}=15.0 \mathrm{~g} \mathrm{~C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\nonumber\]
Fig. 4.6.2 illustrates what each conversion factor does in the case of the above example number 4.6.5


9.4: Return on Assets
- Calculate a return on assets (ROA) for a sample retailer
One way a firm can reflect its efficiency is return on assets (ROA). ROA refers to how much income is produced by its use of assets. Again, for a retail business, assets are inventory available for sale and the dollars generated by sales. Generally, return on assets is calculated as ROA = net profit / total assets. But, it can also be calculated by multiplying the operating profit margin × asset turnover.
Of course, it’s important to understand that retailers manage their operations to meet their own financial goals. Thus, factors that influence operating profit margin and asset turnover may vary between retailers. For example, consider Daisy Donuts and Jasmine Jewelers, which are neighbors in the same strip mall, but have much different financial goals. As such, they have much different operating profit margin and asset turnover, despite ROAs of 15%.
-
Daisy Donuts
- Asset turnover: 5.0, given donuts are perishable and are priced to sell quickly to avoid spoilage
- Operating profit margin: 3%.
-
Jasmine Jewelers
- Asset turnover: 1.0, given jewelry is expensive and a considered purchase
- Operating profit margin: 15%.
As you can see, both outlets have the same ROA, but the components that make up that ROA are vastly different, reflective of each’s retailer’s financial objectives.
Contributors and Attributions
- Return on Assets. Authored by : Patrick Williams. Provided by : Lumen Learning. License : CC BY: Attribution
- Financial Statements: Return on Assets. Authored by : ProfAlldredge. Located at : https://youtu.be/5HQfgKsJUV0 . License : CC BY: Attribution


3.10: Additional Resources
Alternate Translations
A modernized version of Beowulf: A Verse Translation with Treasures of the Ancient North can be downloaded as a PDF from the University of Oxford website.
Part 1 of Beowulf: A Verse Translation with Treasures of the Ancient North
Part 2 of Beowulf: A Verse Translation with Treasures of the Ancient North
Please note that the PDFs come through sideways. You can rotate them in your PDF viewer.
Audio Files
“Beowulf Lecture” by Dr. Francis Leneghan from University of Oxford
Podcast lecture of the background and import of Beowulf in British Literature.
- Survey of English Literature I. Authored by : Wendy Howard Gray. Provided by : Reynolds Community College. Located at : http://www.reynolds.edu/ . License : CC BY: Attribution


UT Arlington Library has six libraries as part of its system, including three full-service libraries.
Full-Service Libraries
Central Library
UT Arlington Library’s largest facility.
-
Open 24 hours a day, 7 days a week.
-
Food and drink allowed in most places.
-
Enhanced group study rooms with electronic white boards. Reservations required. Other group study spaces available.
-
Quiet study floors on fourth and fifth floors.
-
Research assistance on second floor.
Architecture and Fine Arts Library (AFA)
-
Located in room 104 of the Architecture Building, serving students studying art, film, mu- sic, and design.
-
Quiet study area.
-
Conference room with electronic whiteboard and laptop hookups available for reservation.
-
Equipment for viewing VHS and DVDs and for listening to CDs, LPs, and cassettes.
UT Arlington Central Library
Science and Engineering Library (SEL)
-
Located in the basement of Nedderman Hall serving students studying sciences, math, and engineering.
-
Enhanced group study rooms with electronic white boards. Reservations required.
-
Quiet study space with individual study carrels.
Features and Services
The Library sponsors a wealth of resources and information for you.
Computers
Hundreds of PCs available for student use, each loaded with Microsoft Office applications. To log in, use your NetID and password.
Laptops
The Library makes a limited number of laptops available for students to check out at the circulation desk of any full-service library. Loan times vary. Laptops may leave the building and can access the campus wireless network.
Books
-
With over 1 million books, the Library supports all areas of study.
-
You can return books to any full-service library.
-
A drive-by book return is located behind the library on Planetarium Place.
-
Your MavID card is your library card.
-
Undergraduates can check out up to 50 books for 30 days.
-
You may renew each item twice. Renewals can be done in person or online.
-
Log in to your account through the catalog and see what books you have checked out.
Databases
The library offers hundreds of databases, each containing thousands of articles covering a range of subjects and eras.
-
All databases are available on all library computers.
-
To find a database relevant to your work, look at the subject guides on the library’s home page.
-
Use your Net ID to access databases from home or elsewhere.
-
Read many database articles online.
-
Print, save, or e-mail articles from databases.
Expert Help
Paper’s Due Drop Inn
Offered Monday–Thursday during Fall and Spring semesters. Librarians provide assistance with:
-
Research assignments
-
Selecting/narrowing a topic
-
Identifying appropriate research tools
-
Citing sources
No appointments needed. A tutor from the Writing Center is available to assist with writing and citation.
Writing Center
Located on the fourth floor of Central Library, the Writing Center helps students understand the writing process and become better editors of their own work.
Digital Media Studio (DMS)
-
A self-service learning and multimedia production facility located in the basement of Central Library
-
Access to Adobe Creative Suite products
-
Audio and video editing
-
3D modeling with Maya and Lightwave
Reference Desk
All UT Arlington libraries offer friendly, knowledgeable librarians available to help with finding and using appropriate print or electronic resources. Get research help in person, by phone, or online.
Phone:<PHONE_NUMBER> or toll-free at<PHONE_NUMBER>
AIM, Yahoo IM, Google Talk: utalibrary
Text: 66746. Start your question with ASKUTA
Live Help chat: available through the UT Arlington Library Web page


27.1: The process of science and the need for scientific writing
selected template will load here
This action is not available.
Conducting experiments that test a scientific theory is integral to the advancement of science and to the refining of scientific theories. In practice, scientists do not have a lab full of equipment ready to go and to be used for testing whichever theory suits their fancy. Instead, they need to write a “proposal” for conducting a particular experiment to a funding source (e.g. a funding agency). That funding source will then select a panel of experts in the field to review whether the proposal is feasible and useful in advancing science, to decide whether it should be funded. If the scientist is awarded with funds, they are then expected to carry out their experiment and report on the results in a peer-reviewed scientific journal. Again, before the results are published, the scientific journal will ask a panel of experts to review the results to ensure that they are scientifically valid and interesting.
In order for a proposal to be funded, it must thus propose an experiment that is well-thought out and feasible. For example, the reviewers will want to make sure that the proposed experiment is designed in the best possible way to test a theory. Often, this means that thought has been put into designing an experiment that minimizes the uncertainty on the result, so that the test of the theory is as stringent as possible.
A proposal needs to be well-written and precise. We generally call this type of writing “scientific writing”, and it is a style of writing that takes some practice. Similarly, when reporting on the results of an experiment, the report will need to be clear and precise as well. For example, in scientific writing, one avoids giving opinions or using sentences that do not add necessary information or that are not factual.
This chapter provides some guidelines for scientific writing, writing proposals, and writing reports. In addition to this, guidelines for reviewing others’ proposals and reports are also presented. Not only is it important to develop the ability to critically evaluate others’ work, but it is also helpful in learning to reflect and improve on one’s own work.


1.12.2: Heat Capacity- Isobaric- Partial Molar- Solution
Equilibrium isobaric heat capacities of solutions can be treated for most purposes as extensive variables. Thus for an aqueous solution prepared using \(\mathrm{n}_{1}\) moles of solvent (water) and \(\mathrm{n}_{j}\) moles of a simple solute \(j\) the isobaric (equilibrium) heat capacity of the solution \(\mathrm{C}_{\mathrm{p}}(\mathrm{aq})\) can be related to the composition of the solution using equations (a) and (b) [1].
\[\mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{A}=0)=\mathrm{n}_{1} \, \mathrm{C}_{\mathrm{p} 1}(\mathrm{aq})+\mathrm{n}_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}}(\mathrm{aq}) \nonumber \]
where
\[\mathrm{C}_{\mathrm{pl}}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{A}=0)}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{j})} \quad \text { and } \mathrm{C}_{\mathrm{pj}}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{A}=0)}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{l})} \nonumber \]
Similar equations are encountered in a discussion of the partial molar enthalpies but with reference to these properties we develop a number of strategies because it is not possible to determine the enthalpy of a solution. In the present case the outlook is much more favourable because it is possible to measure isobaric heat capacities of solutions [1-3]. The fact that we can measure the temperature dependence of the equilibrium enthalpy of a solution but not the actual enthalpy is an interesting philosophical point. Nevertheless it is informative to develop the analysis starting from equations relating partial molar enthalpies and compositions of solutions.
A given aqueous solution is prepared using \(\mathrm{n}_{1}\) moles of solvent (water) and \(\mathrm{n}_{j}\) moles of solute. The partial molar enthalpies are related to the composition of the solution by the following equations. Thus
\[\mathrm{H}_{1}(\mathrm{aq})=\mathrm{H}_{1}^{*}(\ell)+\mathrm{R} \, \mathrm{T}^{2} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \,(\partial \phi / \partial \mathrm{T})_{\mathrm{p}} \nonumber \]
and
\[\mathrm{H}_{\mathrm{j}}(\mathrm{aq})=\mathrm{H}_{\mathrm{j}}^{\infty}(\mathrm{aq})-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{T}\right)_{\mathrm{p}} \nonumber \]
At all \(\mathrm{T}\) and \(\mathrm{p}\),
\[\operatorname{limit}\left(m_{j} \rightarrow 0\right) \phi=1 \text { and } \gamma_{j}=1 \nonumber \]
By definition,
\[\mathrm{C}_{\mathrm{p} 1}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}}{\partial \mathrm{n}_{1}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{j})}=\left(\frac{\partial \mathrm{H}_{1}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial^{2} \mathrm{H}}{\partial \mathrm{n}_{1} \, \partial \mathrm{T}}\right)_{\mathrm{p}, \mathrm{n}(\mathrm{j})} \nonumber \]
And,
\[\mathrm{C}_{\mathrm{pj}}(\mathrm{aq})=\left(\frac{\partial \mathrm{C}_{\mathrm{p}}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(1)}=\left(\frac{\partial \mathrm{H}_{\mathrm{j}}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial^{2} \mathrm{H}}{\partial \mathrm{n}_{\mathrm{j}} \, \partial \mathrm{T}}\right)_{\mathrm{p}, \mathrm{n}(1)} \nonumber \]
The latter two equations trace the story from the enthalpy of the solution to partial molar isobaric heat capacities. Using equation (f) in conjunction with equation (c) we obtain an equation for dependence for \(\mathrm{C}_{\mathrm{p}1}(\mathrm{aq})\) on molality \(\mathrm{m}_{j}\).
\[\mathrm{C}_{\mathrm{pl} 1}(\mathrm{aq})=\mathrm{C}_{\mathrm{pl} 1}^{*}(\ell)+2 \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \,(\partial \phi / \partial \mathrm{T})_{\mathrm{p}}+\mathrm{R} \, \mathrm{T}^{2} \, \mathrm{M}_{1} \, \mathrm{m}_{\mathrm{j}} \,\left(\partial^{2} \phi / \partial \mathrm{T}^{2}\right)_{\mathrm{p}} \nonumber \]
Similarly using equations (d) and (g), we obtain an equation relating \(\mathrm{C}_{\mathrm{p}j}(\mathrm{aq})\) and molality \(\mathrm{m}_{j}\); the origin of the two minus signs is the Gibbs - Helmholtz Equation.
\[\mathrm{C}_{\mathrm{pj}}(\mathrm{aq})=\mathrm{C}_{\mathrm{pj}}^{\infty}(\mathrm{aq})-2 \, \mathrm{R} \, \mathrm{T} \,\left[\partial \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{T}\right]_{\mathrm{p}}-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial^{2} \ln \left(\gamma_{\mathrm{j}}\right) / \partial \mathrm{T}^{2}\right]_{\mathrm{p}} \nonumber \]
We consider a solution prepared using \(1 \mathrm{~kg}\) of water and \(\mathrm{m}_{j}\) moles of solute \(j\) [4].
\[\mathrm{C}_{\mathrm{p}}\left(\mathrm{aq} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)=\left(1 / \mathrm{M}_{1}\right) \, \mathrm{C}_{\mathrm{pl}}(\mathrm{aq})+\mathrm{m}_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}} \text { (aq) } \nonumber \]
If the thermodynamic properties of the solution are ideal, from the definitions of both practical osmotic coefficient \(\phi\) and activity coefficient \(\gamma_{j}\), the last two terms in equations (h) and (i) are zero.
\[\mathrm{C}_{\mathrm{p}}\left(\mathrm{aq} ; \mathrm{id} ; \mathrm{w}_{1}=1 \mathrm{~kg}\right)=\left(1 / \mathrm{M}_{1}\right) \, \mathrm{C}_{\mathrm{p} 1}^{*}(\ell)+\mathrm{m}_{\mathrm{j}} \, \mathrm{C}_{\mathrm{pj}}^{\infty}(\mathrm{aq}) \nonumber \]
This is an interesting equation because two experiments yield \(\mathrm{C}_{\mathrm{p}}(\mathrm{aq} ; \mathrm{id})\) and \(\mathrm{C}_{\mathrm{p} 1}^{*}(\ell)\). Hence, granted the ideal conditions, we obtain an estimate of \(\mathrm{C}_{\mathrm{pj}}^{\infty}(\mathrm{aq})\), limiting isobaric heat capacity of solute \(j\) in solution. Unfortunately the assumption concerning ideal properties of a solution is often unrealistic. Nevertheless equation (k) offers a reference against which we can examine the properties of real solutions.
Footnotes
[1] An important technological development was the design of the Picker flow calorimeter; P. Picker, P.-A. Leduc, P. R. Philip and J. E. Desnoyers, J. Chem. Thermodyn.,1971, 3 ,631.
[2] For details of calibration of the Picker calorimeter; D. E.White and R. H. Ward, J. Solution Chem.,1982, 11 ,223.
[3] For extension to measurement of thermal expansion coefficients; J. F. Alary, M. N. Simard, J. Dumont and C. Jolicoeur, J. Solution Chem.,1982, 11 ,755.
[4] \(\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~kg}^{-1}\right]=\left[\mathrm{kg} \, \mathrm{mol}^{-1}\right]^{-1} \,\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right]+\left[\operatorname{mol~kg}{ }^{-1}\right] \,\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right]\)


18.9: International Diversity
- Describe the benefits of international diversity
- Describe the challenges of international diversity
Diversity in the workplace is extremely helpful in presenting differing viewpoints and opinions. However, diversity can also be a challenge as differing personalities may clash and create conflict. International businesses are presented with additional challenges when navigating diversity, including language barriers and cultural differences. So, if diversity presents a number of challenges, how beneficial can it be? The answer is, it can be very beneficial—although diversity presents challenges, the benefits outweigh them. Let’s dive into some of the benefits and challenges of international diversity.
Benefits
Each person’s experiences and cultural backgrounds influence the way in which they see the world. When environments, such as the workplace, bring together a wide variety of individuals, a wide variety of viewpoints are introduced. When diverse groups of individuals work together, they challenge their peers and coworkers to view their world from differing perspectives. This can open people up to new ideas and ways of doing things. Diversity helps to spark innovation by combining a large range of experiences, cultures, and areas of expertise.
Companies that value and instill diversity in their organization are more attractive to job applicants. A survey conducted by GlassDoor found that two-thirds of people consider diversity important when deciding where to work. [1] Therefore, diverse companies have access to a larger talent pool since they appeal to a broader group of applicants. Especially in a global marketplace where recruiting efforts are more competitive, a diverse workplace can help to ensure companies are attracting and hiring the right candidates for the job. In addition to helping to hire quality employees, a diverse workplace also helps retain them. Employees who work in diverse work environments often feel a stronger sense of loyalty to their company because they feel valued and understood.
International companies can greatly benefit from their local employees in foreign countries. These international employees have valuable insight and knowledge about their home country’s market and culture. The information international employees can provide to their companies can help make them more competitive in foreign markets and therefore more profitable.
Challenges
Whenever differing personalities and cultures are working together, there are bound to be some challenges. Cultural differences especially can be difficult to understand and overcome.
Differences in business etiquette and nonverbal communication account for the majority of culturally related challenges. Let’s take a look at a few primary areas of difference and potential miscommunication: [2]
- Clothing: managing the first impression
- Conversation: appropriate business and ice-breaker conversation
- Greeting: how people greet each other differs from one culture to the next. Local customs and expectations, including greeting style—the distinctions that inspired the title of the best-selling guide to business etiquette and practices, Kiss, Bow or Shake Hands
- Forms of address: level of formality and use of titles and degrees
- Time: interpretations of “on time,” may vary on the country you’re in. Additionally, the length of average workdays may be different, or how long people take for lunch may differ.
- Space: Personal space and eye contact are other examples of cultural nuances that need to be considered when engaging in international business.
Navigating these differences may pose a challenge, but is an important part of creating a strong and unified team.
Language barriers can also present challenges. When the workplace has employees who speak a variety of languages, clear and concise communication can be difficult to attain. Even in an international setting where everyone is capable of speaking the same language, certain language barriers may present themselves through differing cultural terminologies, idioms, and slang terms. The way in which employees present their message is also different from one culture to the next. Some may prefer sending emails, where others require face-to-face conversations in certain situations. Understanding these types of language and cultural differences can help employees to maintain a respectful and professional work environment.
Resources
Reynolds, Katie. “ 13 Benefits and Challenges of Cultural Diversity in the Workplace .” Hult International Business School, July 11, 2019.
-
“
Two-Thirds of People Consider Diversity Important When Deciding Where to Work, Glassdoor Survey
.” Glassdoor About Us, November 17, 2014. ↵
-
Magloff, Lisa. "
Cross-Cultural Business Etiquette
,"
Chron
. Web. 26 June 2018. ↵
Contributors and Attributions
- International Diversity. Authored by : Freedom Learning Group. Provided by : Lumen Learning. License : CC BY: Attribution
- Authored by : Kyle Glenn. Provided by : Unsplash. Located at : https://unsplash.com/photos/nXt5HtLmlgE . License : CC0: No Rights Reserved . License Terms : Unsplash License
- City architecture and skyscrapers near waterfront. Authored by : Pedro Lastra. Provided by : Unsplash. Located at : https://unsplash.com/photos/Nyvq2juw4_o . License : CC0: No Rights Reserved . License Terms : Unsplash License


1.E: Essential Ideas of Chemistry- Homework
-
- Last updated
- Save as PDF
- Scott Van Bramer
- Widener University
For Chapter 1 you MUST know the following:
- The names and symbols for elements 1-18
- The SI Base units, the measured property, and the symbol
- The SI Prefixes, their abbreviation, and their meaning (M, k, c, m, u , n, p)
Turn in your answers for the following questions - show your work
-
Perform the following unit conversions. Express your answers in scientific notation with the appropriate number of significant figures.
- Convert 78.01 inches into: feet, meters, and centimeters.
- Convert 15.42 meters into kilometers, centimeters, and millimeters.
- Convert 98.6 °F into °C and K.
- Convert 75 miles per hour into: km per hour and m s -1 .
 A graduated cylinder initially contains 48.6 mL water. When the jewelry is submerged in the graduated cylinder, the total volume increases to 61.2 mL.
- Determine the density of this piece of jewelry.
- Assuming that the jewelry is made from only one substance, what substance is it likely to be? Explain.
- In the movie Goldfinger , James Bond foils a plot to break into Fort Knox. 007 does some quick mental calculations to determine the feasibility of removing the gold. If the price of gold is $14.00 per troy ounce (31.1035 grams) at the time the movie was produced, what is the mass (in kg) of 1 million dollars of gold? What is the volume of 1 million dollars of gold in L. How much would this gold be worth today if the price of gold is $9068.77 per kg?
- The following experiment is performed with an unknown liquid. The liquid is added to a graduated cylinder with a mass of 54.6789 grams. After 20.00 mL of the liquid is added to the cylinder (at 298 K), the mass is 74.6215 grams. Is the liquid water? How do you know? If it is not water, what could it be?
- The experiment above is repeated with a second liquid, after 21.3 mL of this liquid (at 20 °C) is added the mass of the cylinder is 72.7364 grams. Is the second liquid ethanol? How do you know? If it is not ethanol, what could it be?
The Following Questions are for your practice - Do Not Turn In. They include answers so you can check your work
-
Perform the following calculations and report each answer with the correct number of significant figures.
- 628 \(\times\) 342
- (5.63 \(\times\) 10 2 ) \(\times\) (7.4 \(\times\) 10 3 )
- \(\dfrac{28.0}{13.483}\)
- 14.98 + 27,340 + 84.7593
- 42.7 + 0.259
- Answer (a)
- (2.15 \(\times\) 10 5 )
- Answer (b)
- 4.2 \(\times\) 10 6
- Answer (c)
- 2.08
- Answer (d)
- 27,440
- Answer (e)
- 43.0
-
Use scientific (exponential) notation to express the following quantities in terms of the SI base units in
- 0.13 g
- 86.3 mg
- 37.6 cm
- 54 μm
- 27 ns
- Answer (a)
- 1.3 \(\times\) 10 −4 kg
- Answer (b)
- 8.63 \(\times\) 10 −5 kg
- Answer (c)
- 3.76 \(\times\) 10 −1 m
- Answer (d)
- 5.4 \(\times\) 10 −5 m
- Answer (e)
- 1 \(\times\) 10 9 s
-
Make the conversion indicated in each of the following:
- the length of a soccer field, 120 m (three significant figures), to feet
- the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers
- the area of an 8.5 t 11-inch sheet of paper in cm 2
- the estimated mass of the atmosphere, 5.6 t 10 15 tons, to kilograms
- the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)
- Answer (a)
- 394 ft
- Answer (b)
- 5.9634 km
- Answer (c)
- 6.0 \(\times\) 10 2
- Answer (d)
- 5.1 \(\times\) 10 18 kg
- Answer (e)
- 324 mg
-
Calculate these masses.
- What is the mass of 6.00 cm 3 of mercury, density = 13.5939 g/cm 3 ?
- What is the mass of 25.0 mL octane, density = 0.702 g/cm 3 ?
- Answer (a)
- 81.6 g
- Answer (b)
- 17.6 g
-
Calculate these volumes the volume of 25 g iodine, density = 4.93 g/cm
3
?
- Answer
- 5.1 mL
-
Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and kelvin.
- Answer
- 5371 °F, 3239 K
-
Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and kelvin.
- Answer
- −23 °C, 250 K
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>).


8.4: Assisting Clients to Transfer
It is important to note that most injuries that happen to clients and staff occur when clients are being transferred. Safety is an integral component of moving clients and should receive the highest priority. Special consideration should be given to these items to prevent injury from occurring:
- Gait belt fit and placement
- Brakes on the bed
- Brakes on the wheelchair
- Brakes released on the lift
- Placement of nonskid footwear
- Resident’s proximity to the lift
- Objects in the room that may be a hazard during the movement
Nursing assistants should always review a client’s care plan for their current transfer status before moving them. Transfer status refers to how a resident moves from one place to the other, such as from a bed to wheelchair or wheelchair to toilet. Physical therapists (i.e., health specialists who evaluate and treat movement disorders) assess clients and make recommendations for how clients should be safely transferred. Transfer status orders are determined by how much body weight a client can independently bear and how much weight an assistant is required to support. Transfer status orders include these types of orders:
- Independent: The client requires no assistance to move from one place to another.
- Contact-Guard-Assist (CGA): One assistant must have their hand on the client at all times to steady their balance.
- 1 assist (1A): One assistant is required to help the resident transfer (with a gait belt in place).
- 2 assist (2A): Two assistants are required to help the resident transfer (with a gait belt in place).
- Sit-to-Stand: A sit-to-stand mechanical lift is required to help a resident transfer. A sit-to-stand mechanical lift provides support as the client stands while allowing them to bear some body weight and maintain joint mobility and leg strength. (Using a gait belt would require extensive assistance and could cause injury to the client or the staff.) Sit-to-stands may be completed with one or two assistants, as determined by the physical therapist and agency policy.
- Full-Body Mechanical Lift: A full-body mechanical lift is required when the resident cannot bear any of their weight when transferred from bed to chair and back. (“Full-body mechanical lift” is a generic term. The facility or organization may refer to this type of lift by the manufacturer of the lift, such as a “Hoyer lift” or “PAL lift.”) Full-body mechanical lifts may be portable or attached to the ceiling. Two assistants are always required for transfer with full-body mechanical lifts for safety purposes. Due to federal liability laws, the health care professional moving the lift must be 18 years of age or older.
Assisting to Seated Position or Dangling
When transferring a client using a 1A, 2A, or sit-to-stand transfer method, first assist the resident to move to a seated position on the side of their bed. Residents who can transfer with one of these methods are able to bear some or most of their weight and should be able to move partially on their own. Use your hands on the person’s limbs to direct the movement, and use the lift sheet (similar to when boosting a client up in bed).
Due to heart and circulation changes that occur with age, orthostatic hypotension can occur when a person moves suddenly from a lying to sitting position or from a sitting to standing position. Orthostatic hypotension is a sudden drop in blood pressure that can cause clients to feel dizzy and increase their risk for falls with position changes. Some clients may experience vertigo , a sensation that the room is spinning. To prevent orthostatic hypotension and these symptoms, tell the person to dangle (i.e., sit up on the edge of the bed) for a few moments before continuing with the transfer. Dangling gives the cardiovascular system time to regulate blood pressure and blood flow to the brain, thus preventing dizziness and falls. Ask the client if they are feeling dizzy before you proceed with transferring. See Figure 8.11 [1] for illustration of the steps to safely assist a client to a seated position [2] :
- Lock the brakes on the bed.
- Raise the bed to a working height.
- Stand facing the head of the bed at a 45-degree angle with your feet apart, with one foot in front of the other. Stand next to the waist of the resident.
- Ask the resident to turn onto their side, facing you, as they move closer to the edge of the bed. Use the lift sheet to assist the resident if needed.
- Place one hand behind the resident’s shoulders, supporting their neck and vertebrae.
- Place the other hand around the resident’s knees.
- On the count of three, instruct the resident to use their elbows to push up against the bed and then grasp the side rail. Support their shoulders as they move to a seated position. Shift your weight from the front foot to the back foot as you assist them to sit. Do not allow the resident to place their arms around your shoulders because this can cause serious back injuries.
- As you shift your weight, gently grasp the resident’s outer thighs with your other hand and help them slide their feet off the bed to dangle or touch the floor. This step helps the resident sit and move their legs off the bed at the same time.  Move the resident as one entire unit (rather than the upper body followed by the lower body).
- Lower the bed so the resident’s feet touch the floor.
- Observe the resident for symptoms of orthostatic hypotension or vertigo. Ask if they are feeling dizzy before attempting any further movement.
- Check that the resident is wearing nonskid footwear before transferring.
If the resident is having difficulty moving during this procedure, use the lift sheet to pull them closer to you or to assist them from a lying to a seated position. The head of the bed can be raised before they turn on their side to support their core strength and to reduce the weight the assistant must bear. During this entire process, do not use the resident’s limbs to move them but rather move them with the trunk of their body to prevent shear and injury to their limbs and skin.
After the person is sitting upright and states they are not experiencing dizziness, apply a gait belt for a 1A or 2A transfer. The gait belt should be placed around their waist while considering the location of their breasts and abdominal folds. See Figure 8.12 [3] for an image of applying a gait belt. The fit of the gait belt should be snug, but you should be able to put your fingers underneath the gait belt for support. As the resident stands and their core muscles contract, the gait belt can loosen and tend to slide up, so it is important for it to be snug. If the belt is too long, it can get caught in the patient’s legs during transfer, so tuck the excess length back into the belt. Gait belts should not be used for clients with abdominal wounds or some types of heart conditions; a different transfer method should be in the care plan. Contact the nurse if you have concerns about using a gait belt based on the client’s condition.
Place nonskid footwear on the resident before transferring them from the bed to the chair. These preparations should be done before the wheelchair or sit-to-stand is brought closer to the bed to reduce the risk of injury from the resident inadvertently hitting the lift or chair while moving.
The steps to complete a one person assist (1A) are listed in the “ Transfer From Bed to Chair With a Gait Belt ” Skills Checklist. If a two-person assist (2A) is required, the same steps are used, but the assistants stand on each side of the resident to provide additional support during the transfer.
View the following video showing a transfer of a patient from a bed to a regular chair [4] : Assisting From Bed to Chair With a Gait Belt or Transfer Belt .
Transferring with Mechanical Lifts
Mechanical lifts include sit-to-stand lifts and full-body lifts. Some facilities have full-body mechanical lifts that are attached to the ceiling of the room. See Figure 8.13 [5] of an image of a sit-to-stand lift and Figure 8.14 [6] of an image of a full-body mechanical lift and mechanical swing. NAs should be aware of agency policy regarding transferring clients using mechanical lifts; for safety purposes, most agencies require two NAs or a nurse and an NA to transfer clients using a mechanical lift.
The legs of portable full-body mechanical lifts can be placed in a closed or open position. The open position provides the greatest stability due to a wide base of support. See Figure 8.15 [7] for an image comparing the legs of a portable full-body mechanical lift in a closed and open position.
When transferring a client from their bed to wheelchair using a sit-to-stand or portable full-body mechanical lift, the wheelchair should be positioned near the bed while also allowing enough room for the lift to rotate towards the chair. Because the lift will have to slide underneath the bed, check for any cords or equipment under the bed that can cause the lift to get tangled. Raising the bed height just before placing the lift under the bed should also alleviate potential problems.
The lift has greatest stability when its legs are open with a wide base of support, but some beds do not have enough space underneath to allow the legs to be open. If this is the case, open the legs as soon as possible when moving the lift from under the bed to provide a stable base.
Sit-to-stand and full-body lifts have brakes, but brakes should not be applied when the resident is standing in the sit-to-stand or raised off the bed in a full-body lift. (If the client’s weight shifts while the brakes are on, it can cause the lift to tip and endanger the resident, as well as the assistants.)
Before initiating a transfer, verify that the lift will support the weight of the resident. Most mechanical lifts have a weight capacity of 400 pounds. Bariatric lifts are used to support a client weighing 600 or more pounds. See Figure 8.16 [8] for locating the weight capacity on a lift.
Full-body mechanical lifts have different types of slings used to lift the client. Slings may be full-body or split-leg (butterfly). The type of sling used is determined by the physical therapist, based on the client’s strength and mobility, and should be noted in the resident’s care plan. A sling has various loops to connect it to the lift and are often color-coded to ensure the resident’s body is in proper position for transferring. See Figure 8.17 [9] for images of a full-body sling and Figure 8.18 [10] for images of a split-leg (butterfly) sling.
The top of a full-body sling should be placed above the resident’s head and should end just above the knee joint to avoid hyperextending the knees when they are suspended in the lift. The top of a split-leg (butterfly) sling should be placed at shoulder height, and the bottom of the sling should be around the buttocks. Depending on hip mobility, the split-leg sling may be crossed between the client’s legs or placed around their legs (often referred to as a “basket”). See Figure 8.19 [11] for an image of a mannequin prepared to transfer using a crossed sling and Figure 8.20 [12] for an image of transferring a mannequin with the sling wrapped around their legs (i.e., a “basket” approach).
The handles of the sling should face away from the client, allowing the assistants to position the client in their chair or bed. (Do not move clients by directly contacting their limbs because this can cause injury.)
After transferring a client from a bed to a chair using a full-body sling, the sling should remain under the resident while they are seated in a wheelchair or other chair. When placing the resident back in bed, the sling is then removed. However, when transferring a client using a split-leg (butterfly) sling, it can be removed from underneath the client in the chair and replaced before they are transferred again.
See the Skills Checklists “ Transfer From Bed to Chair With Sit-to-Stand ” and “ Transfer From Bed to Chair With Mechanical Lift ” for steps for providing safe transfers with both types of lifts. Each brand of sit-to-stand and mechanical lift has some variances; the facility where you work will provide specific training on their lifts.
Watch the following YouTube video for a demonstration of moving a resident with a sit-to-stand [13] : Aidacare Training Video – Manual Handling – Sit To Stand.
Explore the following YouTube video [14] on a mechanical lift completed with a butterfly or split-leg sling: Aidacare Training Video – Manual Handling – Lie To Sit
- “Book-pictures-2015-5851.jpg,” “Book-pictures-2015-587.jpg,” and “Book-pictures-2015-588.jpg” by unknown authors are licensed under CC BY 4.0 . Access for free at https://opentextbc.ca/clinicalskills/chapter/3-5-positioning-a-patient-on-the-side-of-a-bed/ ↵
- This work is a derivative of Nursing Fundamentals by Chippewa Valley Technical College and is licensed under CC BY 4.0 ↵
- “Sept-22-2015-119.jpg” and “Sept-22-2015-121-001.jpg” by unknown authors are licensed under CC BY 4.0 . Access for free at https://opentextbc.ca/clinicalskills/chapter/3-5-positioning-a-patient-on-the-side-of-a-bed/ ↵
- Thompson Rivers University Open Learning. (n.d.). Assisting from bed to chair with a gait belt or transfer belt [Video]. Thompson Rivers University Open Learning. All rights reserved. https://barabus.tru.ca/nursing/assisting_from_bed.html ↵
- " Transferring Client Using a Sit to Stand " by Landon Cerny is licensed under CC BY 4.0 ↵
- " Transferring a Client Using a Mechanical Lift and a Full Body Sling ", " Transferring a Client Using a Mechanical Lift and a Split Leg Sling ", and " Transferring a Client Using a Mechanical Lift and Split Leg Sling in a Basket Approach " by Landon Cerny are licensed under CC BY 4.0 ↵
- “Portable Full-Body Mechanical Lift With Legs in Closed Position” and “Portable Full-Body Mechanical Lift With Legs in Open Position” by Myra Reuter for Chippewa Valley Technical College are licensed under CC BY 4.0 ↵
- “Weight Capacity of a Mechanical Lift” by Myra Reuter for Chippewa Valley Technical College is licensed under CC BY 4.0 ↵
- “Front of Full-Body Sling,” “Back of Full-Body Sling,” and “Loops” by Myra Reuter for Chippewa Valley Technical College are licensed under CC BY 4.0 ↵
- “Front of Full-Body Sling,” “Back of Full-Body Sling,” and “Loops” by Myra Reuter for Chippewa Valley Technical College are licensed under CC BY 4.0 ↵
- “Preparing to Transfer With Crossed Sling” and “Suspended in a Crossed Sling” by Myra Reuter for Chippewa Valley Technical College are licensed under CC BY 4.0 ↵
- “Preparing to Transfer With a ‘Basket’ Approach” and “Suspended in a Sling With a ‘Basket’ Approach” by Myra Reuter for Chippewa Valley Technical College are licensed under CC BY 4.0 ↵
- Aidacare. (2017, July 5). Aidacare training video - Manual handling - Sit To stand [Video]. YouTube. All rights reserved. https://youtu.be/L914lkoub6E ↵
- Aidacare. (2017, July 5). Aidacare training video – Manual handling – Lie to sit [Video]. YouTube. All rights reserved. https://youtu.be/3GOgp_HX4JQ ↵


6.4: Principles of Nonverbal Communication
selected template will load here
This action is not available.
In this section, we will address the five principles of nonverbal communication: our nonverbal communication is continuous, multichanneled, conscious/intentional and unconscious/unintentional, ambiguous, and can occur both face-to-face and through mediated platforms.
In face-to-face interactions, nonverbal communication is continuous, ongoing, and in constant motion. While we can stop talking to end verbal communication, we can’t turn off nonverbal communication. Our posture, eye contact (or lack of eye contact), facial expressions, and physical appearance are always communicating something about us, whether intentional or unintentional.
A nonverbal message is rarely sent through just one channel in isolation, rather it accompanies multiple channels and occurs in clusters. For example, squinted eyes might be accompanied with a furrowed brow and/of a pursing of the lips. Because of this, we need to be aware of the entire cluster when decoding a particular message and look for nonverbal congruence. Nonverbal congruence is the consistency among the cluster of nonverbals we are observing, which influences our understanding of what message is being conveyed. For example, someone crossing their arms usually has a negative interpretation. However, since nonverbal communication is multichanneled, this particular body language should not be read in isolation; instead, other nonverbal cues should be considered.  However, if someone has their arms crossed but they are smiling, then they likely are not angry.
Nonverbal communication is both conscious and unconscious, which means we use it both intentionally and unintentionally. Conscious communication means that we think about our communication before we communicate. Unconscious communication means that we do not think about every message we communicate.   Nonverbal communication can occur as unconscious reactions to situations. However, at times, we certainly make conscious choices to use or withhold nonverbal communication to share meaning. Angry drivers use many conscious nonverbal expressions to communicate to other drivers, and in a job interview you are making conscious decisions about your wardrobe, posture, and eye contact.
While the symbolic and abstract nature of verbal language can often lead to misunderstandings, nonverbal communication is even more ambiguous. As with verbal communication, most of our nonverbal signals can be linked to multiple meanings, but unlike words, many nonverbal signals do not have any one specific meaning. If you’ve ever had someone wink at you and didn’t know why, you’ve probably experienced this uncertainty. Did they wink to express their affection for you, their pleasure with something you just did, or because you share some inside knowledge or joke? Just as we look at context clues in a sentence or paragraph to derive meaning from a particular word, we can look for context clues in various sources of information like the physical environment, other nonverbal signals, or verbal communication to make sense of a particular nonverbal cue. Unlike verbal communication, however, nonverbal communication doesn’t have explicit rules of grammar that bring structure, order, and agreed on patterns of usage. Instead, we implicitly learn norms of nonverbal communication, which leads to greater variance. In general, we exhibit more idiosyncrasies in our usage of nonverbal communication than we do with verbal communication, which also increases the ambiguity of nonverbal communication.
When most people think of nonverbal communication, they think of behaviors and actions directly connected to the body, such as gestures and facial expressions. And when most people think of the type of communication that takes place through mediated platforms, like text messaging and social media posts, they usually think of it as verbal communication. However, technology has changed and reshaped our conceptualization and definitions of what communication is and looks like in a variety of contexts. Scholars have equated communication such as text messages to ‘fingered speech’ (McWhorter, 2013) and these types of communication have transcended past our traditional definitions as they exhibit characteristics of both verbal and nonverbal communication simultaneously.
Due to the dynamic and ambiguous nature of this type of communication, we may use a variety of strategies to emphasize, substitute for, modify, and/or contradict the accompanying verbal message, which aligns with the definition of nonverbal communication. For example, facial expressions, gestures, and movement can be relayed through emojis, emoticons, and gifs. We can change, modify, and contradict the meaning of verbal messages through punctuation, the spelling of a word, and word elongation. Consider how the following message sent through a text message might be interpreted a variety of different ways based on these strategies:
On the decoding end, mediated messages can present more opportunities for misunderstanding than do face-to-face messages. This form of communication is not only light on nonverbal information, but it lacks opportunity for immediate feedback that may help to clarify intentions in order to achieve shared meaning. Have you ever heard someone talk about the need for a “sarcasm font?” Visuals can help, but the pitch and tone of intended sarcasm is difficult to convey through technology. Additionally, consider how lack of response to a text message can easily be misunderstood. We all have different norms and expectations for reply time. Consider how someone who checks their phone every few minutes might unintentionally misunderstand a less timely reply (a lengthy silence) from another person who only checks their phone every few hours.


11.9: Effects of Temperature and Pressure on Equilibrium Position
The advancement \(\xi\) of a chemical reaction in a closed system describes the changes in the amounts of the reactants and products from specified initial values of these amounts. We have seen that if the system is maintained at constant temperature and pressure, \(\xi\) changes spontaneously in the direction that decreases the Gibbs energy. The change continues until the system reaches a state of reaction equilibrium at the minimum of \(G\). The value of the advancement in this equilibrium state will be denoted \(\xi\eq\), as shown in Fig. 11.15. The value of \(\xi\eq\) depends in general on the values of \(T\) and \(p\). Thus when we change the temperature or pressure of a closed system that is at equilibrium, \(\xi\eq\) usually changes also and the reaction spontaneously shifts to a new equilibrium position.
To investigate this effect, we write the total differential of \(G\) with \(T\), \(p\), and \(\xi\) as independent variables \begin{equation} \dif G = -S\dif T + V\difp + \Delsub{r}G\dif\xi \tag{11.9.1} \end{equation} and obtain the reciprocity relations \begin{equation} \Pd{\Delsub{r}G}{T}{p, \xi} = -\Pd{S}{\xi}{T,p} \qquad \Pd{\Delsub{r}G}{p}{T, \xi} = \Pd{V}{\xi}{T,p} \tag{11.9.2} \end{equation} We recognize the partial derivative on the right side of each of these relations as a molar differential reaction quantity: \begin{equation} \Pd{\Delsub{r}G}{T}{p, \xi} = -\Delsub{r}S \qquad \Pd{\Delsub{r}G}{p}{T, \xi} = \Delsub{r}V \tag{11.9.3} \end{equation} We use these expressions for two of the coefficients in an expression for the total differential of \(\Delsub{r}G\): \begin{gather} \s{ \dif\Delsub{r}G = -\Delsub{r}S\dif T + \Delsub{r}V\difp + \Pd{\Delsub{r}G}{\xi}{T,p}\dif\xi } \tag{11.9.4} \cond{(closed system)} \end{gather} Since \(\Delsub{r}G\) is the partial derivative of \(G\) with respect to \(\xi\) at constant \(T\) and \(p\), the coefficient \(\pd{\Delsub{r}G}{\xi}{T,p}\) is the partial second derivative of \(G\) with respect to \(\xi\): \begin{equation} \Pd{\Delsub{r}G}{\xi}{T,p} = \Pd{^2 G}{\xi^2}{T,p} \tag{11.9.5} \end{equation} We know that at a fixed \(T\) and \(p\), a plot of \(G\) versus \(\xi\) has a slope at each point equal to \(\Delsub{r}G\) and a minimum at the position of reaction equilibrium where \(\xi\) is \(\xi\eq\). At the minimum of the plotted curve, the slope \(\Delsub{r}G\) is zero and the second derivative is positive (see Fig. 11.15). By setting \(\Delsub{r}G\) equal to zero in the general relation \(\Delsub{r}G = \Delsub{r}H - T\Delsub{r}S\), we obtain the equation \(\Delsub{r}S = \Delsub{r}H/T\) which is valid only at reaction equilibrium where \(\xi\) equals \(\xi\eq\). Making this substitution in Eq. 11.9.4, and setting \(\dif\Delsub{r}G\) equal to zero and \(\dif\xi\) equal to \(\dif\xi\eq\), we obtain \begin{gather} \s{ 0 = -\frac{\Delsub{r}H}{T}\dif T + \Delsub{r}V\difp + \Pd{^2 G}{\xi^2}{T,p}\dif\xi\eq } \tag{11.9.6} \cond{(closed system)} \end{gather} which shows how infinitesimal changes in \(T\), \(p\), and \(\xi\eq\) are related.
Now we are ready to see how \(\xi\eq\) is affected by changes in \(T\) or \(p\). Solving Eq. 11.9.6 for \(\dif\xi\eq\) gives \begin{gather} \s{ \dif\xi\eq = \frac{\displaystyle \frac{\Delsub{r}H}{T}\dif T - \Delsub{r}V\difp} { \Pd{^2 G}{\xi^2}{T,p}} } \tag{11.9.7} \cond{(closed system)} \nextcond{} \end{gather}
The right side of Eq. 11.9.7 is the expression for the total differential of \(\xi\) in a closed system at reaction equilibrium, with \(T\) and \(p\) as the independent variables. Thus, at constant pressure the equilibrium shifts with temperature according to \begin{gather} \s{ \Pd{\xi\eq}{T}{p} = \frac{\Delsub{r}H} { T\Pd{^2 G}{\xi^2}{T,p}} } \tag{11.9.8} \cond{(closed system)} \nextcond{} \end{gather}
and at constant temperature the equilibrium shifts with pressure according to \begin{gather} \s{ \Pd{\xi\eq}{p}{T} = - \frac{\Delsub{r}V} { \Pd{^2 G}{\xi^2}{T,p}} } \tag{11.9.9} \cond{(closed system)} \nextcond{} \end{gather}
Because the partial second derivative \(\pd{^2G}{\xi^2}{T,p}\) is positive, Eqs. 11.9.8 and 11.9.9 show that \(\pd{\xi\eq}{T}{p}\) and \(\Delsub{r}H\) have the same sign, whereas \(\pd{\xi\eq}{p}{T}\) and \(\Delsub{r}V\) have opposite signs.
These statements express the application to temperature and pressure changes of what is known as Le Chatelier’s principle : When a change is made to a closed system at equilibrium, the equilibrium shifts in the direction that tends to oppose the change. Here are two examples.
-
It is easy to misuse or to be misled by Le Chatelier’s principle. Consider the solution process B\(^*\)(s)\(\arrow\)B(sln) for which \(\pd{\xi\eq}{T}{p}\), the rate of change of solubility with \(T\), has the same sign as the molar differential enthalpy of solution \(\Delsub{sol}H\) at saturation. The sign of \(\Delsub{sol}H\) at saturation may be different from the sign of the molar
integral
enthalpy of solution, \(\Del H\m\sol\). This is the situation for the dissolution of sodium acetate shown in Fig. 11.9. The equilibrium position (saturation) with one kilogram of water is at \(\xi\subs{sol} \approx 15\mol\), indicated in the figure by an open circle. At this position, \(\Delsub{sol}H\) is positive and \(\Del H\m\sol\) is negative. So, despite the fact that the dissolution of 15 moles of sodium acetate in one kilogram of water to form a saturated solution is an exothermic process, the solubility of sodium acetate actually
increases
with increasing temperature, contrary to what one might predict from Le Chatelier’s principle (L. K. Brice,
J. Chem. Educ.
,
60
, 387–389, 1983).
Another kind of change for which Le Chatelier’s principle gives an incorrect prediction is the addition of an inert gas to a gas mixture of constant volume. Adding the inert gas at constant \(V\) increases the pressure, but has little effect on the equilibrium position of a gas-phase reaction regardless of the value of \(\Delsub{r}V\). This is because the inert gas affects the activities of the reactants and products only slightly, and not at all if the gas mixture is ideal, so there is little or no effect on the value of \(Q\subs{rxn}\). (Note that the dependence of \(\xi\eq\) on \(p\) expressed by Eq. 11.9.9 does not apply to an open system.)


1.23: Intellect, Cognition, and Emotion
-
- Last updated
- Save as PDF
Humans have emotions - feelings are tangible while emotions are - or could be considered to be deep and complicated. The idea that feelings are tangible basically means that they could be more sensory or less intellectual and deep. Emotions are more powerful than feelings; however, they could also trigger the human intellect.
What would it mean for emotion to be powerful? Would that involve physical feelings? Physical stimulation can also be deep or shallow, emotional or intellectual. If the feeling (physical, emotional) is intellectual then it could be emotional or it could also be tied in with sensory feelings (say when you touch something).
What would it mean for something to be intellectual? Would that mean that it is different from the persons emotions? Emotions can be tied in with feelings - however that means that the emotion could be shallow and thought provoking or deep, or a strong emotion that is also deep.
It is important to distinguish deep feelings from sensory feelings. Deep feelings are probably intellectual - they are tied in with complicated cognitions which include memory processes, executive functioning (control of thoughts, ideas and images) and understanding concepts.
Concepts can also be emotional since they are intellectual or intelligent. A concept is like an idea only it is general or generic. An idea is something that occurs to someone while a concept could be the definition of an idea or the idea that the person refers to or already understood. Those deeper concepts can trigger emotions that are related to the idea or concept. A single concept could be powerful or significant to the person.
A humans emotions could influence their thoughts - and their physical feelings can also influence either their thoughts or their emotions (or both at the same time). Thoughts could be complicated - they are a mix of goals and motivations with the persons environment and experience. Furthermore, a motivation could have complicated emotions, and their present situation could be causing complicated emotions.
The difference between feelings and thoughts is simple and complex - a thought could be complex because it could involve the persons motivations mixed in with the objects in their environment and their experience. They could have a thought for each object or each objective reality in their situation.
The difference between their feelings and thoughts then is that their feelings cause feeling, or stimulation and could be complex and intellectual while their thoughts could be unconscious or complex.
title
Humans have feelings. Humans can also think about their feelings. Other factors in reality help the thinking process - such as what is in the persons environment, and what they are paying attention to all assist the persons thinking process.
But what exactly is a thought process? Is it a sentence? Is it a single idea? Is it a few ideas that the person is trying to think about or understand?
The ideas someone is thinking about could be complicated and internal - or simple and related to their environment.
Humans have ideas - mulitple ideas can compose a thought process. The ideas can be about different things - stuff in the persons environement, other ideas or memories that they want to think about, and they can form thoughts or sentences about those ideas - they can also think about their feelings (with ideas or sentences).
For instance, a feeling could be an idea - or an idea could become a feeling
What does that mean an idea is? An idea is something that occurs to someone - it is a concept or intention, or an understanding of some sort.
Ideas can relate to a persons feelings - and to the persons thought process. That is, ideas can complete a thought process.
title
Ideas are thoughts that occur to people, 'that is an excellent idea' wold be the expression.
People have emotions. Their emotions are feelings that they feel. That means that they like to think about things
If humans think, then however it is more fun , however , it is not fun , however I now think that that makes sense.