chem woot diagnostic so ln

7/29/2019 Chem Woot Diagnostic So Ln

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Art of Problem Solvingwww.aops.com

ChemWOOTDiagnostic Exam Solutions

Below are the solutions for the Diagnostic Exam for Art of Problem Solvings ChemWOOT program. Hereshow to interpret your results:

1620 questions correct: You are very well-prepared for ChemWOOT. We hope to see you in class!

1215 questions correct: Youll need to work extra diligently to keep up if you enroll in Chem-WOOT. Be sure to review the questions you got wrong, and read the appropriate sections of yourtextbook.

011 questions correct: ChemWOOT is probably not for you at this stage.

Question 1

(c) 12C10H12O4S(s) + 12 O2(g) 10CO2(g) + SO2(g) + 6 H2O(g)

Question 2

(a) 66 H+(aq) + 2 MnO4 (aq) + 5 SO

23 (aq) 2 Mn

2+(aq) + 5 SO24 (aq) + 3 H2O(g)

Manganese goes from oxidation state (VII) to (II) - a 5 electron difference, whereas sulfur goes from oxidationstate (IV) to (VI) - a 2 electron difference. This determines the ratio of manganese species to sulfur species,and the rest is bookkeeping.

Question 3

(d) 3.01 1023

No mass is lost during polymerization, so 14 grams corresponds to 0 .50 moles of ethylene. Multiplying by

Avogadros number ( 6.021023 molecules

1 mole ), we get 3.01 1023 molecules.

Question 4

(b) 6.04 L

6 g Mg 1 mol Mg

24.3 g Mg

1 mol H21 mol Mg

= 0.247 mol H2

Using the ideal gas law, convert moles of gas into volume:

V =nRT

P=

(0.247 mol)(0.0821 LatmmolK )(298 K)

1 atm= 6.04 L


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Question 5

(c) 2.06 L

Volume is proportional to temperature and inversely proportional to pressure. Therefore, the new volumeof the balloon is

2.00 L 283 K

293 K

1.00 atm

0.94 atm= 2.06L

Question 6

(b) 0.80(0.0821)(393)7.0 atm

Hydrogen gas is the limiting reagent in this reaction. 0.40 mol of H2 will react with 0.20 mol of O2 to form0.40 mol of H2O. After the reaction goes to completion, there are 0.40 mol of O 2 remaining and 0.40 mol ofH2O produced, for a total of 0.80 mol of gas. Remembering to convert 120

C to 393 K, (b) is the correctideal gas law expression.

Question 7

(a) H2S

PH3 has one electron pair, and the other three choices all have zero electron pairs on the central atom.

Question 8

(b) III only

Xenon trioxide has three ligands and a lone pair, and is thus trigonal pyramidal. Carbon tetrachloride hasfour ligands and is tetrahedral. Boron trichloride has three ligands, and is trigonal planar.

Question 9

(c) (5,1,1, 12 )

Indium has three valence electrons: two 5s electrons and one 5p electron. A p-orbital has quantum numberl = 1, and there is the additional restriction |ml| l.

Question 10

(d) Si

The first four ionization energies (IE) steadily increase, but the fifth IE is much larger than the fourth. Thishints that the fifth electron must have come from a different energy level; therefore there must have beenfour valence electrons. Silicon is the only choice with four valence electrons.

Question 11

(a) The melting point of NaCl (808C) is higher than the melting point of sugar (186C).

The melting point is a direct measurement of how much thermal energy is required to break a lattice heldtogether by ionic bonds or hydrogen bonds. The higher melting point of NaCl as compared to sugar providesevidence that ionic bonds are stronger than hydrogen bonds.


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Question 12

(d) CaCl2

The goal of adding salt to ice is to lower the freezing point of ice so that it will melt at a lower tempera-ture. Colligative effects such as freezing point depression and boiling point elevation occur proportionally tothe number of ions generated in solution. Each substance has a different molar mass, and additionally, cangenerate different numbers of ions in solution. 3111 >

2101

2103

> 1180 , so calcium chloride is the clear winner.

Question 13

(b) 8 days

If only 6.25 percent of the original material remains, this corresponds to four half lives. ( 124 = 0.0625). Thus,each half life must be 8 days.

Question 14

(c) Rate = k[NO][O2]2

Comparing the first and third experiments, we see that a fourfold increase in [NO] corresponds to a fourfoldincrease in rate. Therefore, the reaction is first order in [NO]. Comparing the first and second experiments,we see that twofold increase of [NO] and fourfold increase of [O 2] corresponds to a 32-fold increase in rate.Compensating for NO, we see that the reaction must be second order in [O2].

Question 15

(c) 80.0 mL

To neutralize exactly, the quantity of base must balance the quantity of acid. There are (50.0 mL)(0

.240 M) =12 mmol of calcium hydroxide, or 24 mmol hydroxide. We would need 24 mmol0.300 M = 80 mL of monoprotic acid

to match the hydroxide.

Question 16

(a)(0.30)2

(0.45)(0.10)2

According to stoichiometry, 0.60 moles of CO and 0.30 moles of O 2 must have been consumed as the systemreached equilibrium. This leaves us with 0.20 moles of CO, 0.90 moles of O2, and 0.60 moles of CO2. Convertto concentrations to get the final expression:

Kc =[CO2]

2

[O2][CO]2

=(0.30)2

(0.45)(0.10)2


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Question 17

(d) 2 10

9

Assuming that the majority of HCN stays undissociated, let x = [H+] = [CN] be the amount of dissociation.We have

Ka =[H+][CN]

[HCN]=

x2

0.05 x

x2

0.05

Solving for x gives:x =

Ka 0.05 = 5 10

6 = [H+]

so

[OH] =Kw

[H+]= 2 109

Question 18

(e) Let x be the molar solubility. Then, dissolving x M of Ag2CO3 will produce 2x M of Ag+ and x M of

CO23 . The solubility product is

Ksp = [Ag+]2[CO23 ] = (2x)

2 x = 4x3

sox =

3

2 1012

Question 19

(b) G, H, S are all negative

By definition, spontaneous means that G is negative. Since G = H TS, if the reaction becomes

nonspontaneous at higher T, then it must be true that S is negative. But since the reaction is spontaneousat room temperature, H must be negative as well.

Question 20

(c) I and II only

We know that S, G < 0. As in question 19, this implies that H < 0. Another way of saying that areaction is spontaneous is that the products are favored, so the protein is predominantly folded. Finally,thermodynamics says nothing about kinetics, so we cannot conclude anything about rate of the reaction.

chem woot diagnostic so ln

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