Copyright Kaplan AEC Education, 2008 Chemistry Outline Overview STATES OF MATTER, p. 546...

Copyright Kaplan AEC Education, 2008

Chemistry Outline Overview

STATES OF MATTER, p. 546

PERIODICITY, p. 550

NOMENCLATURE, p. 551

EQUATIONS, p. 554


Chemistry Outline Overview Continued

STOICHIOMETRY, p. 556

OXIDATION AND REDUCTION, p. 557

ELECTROCHEMISTRY, p. 559

ACIDS AND BASES, p. 561


Chemistry Outline Overview Continued

METALS AND NONMETALS, p. 563

EQUILIBRIUM, p. 566

SOLUTIONS, p. 567

KINETICS, p. 570


Acid/Base

A 0.4157 g sample containing sodium carbonate (Na2CO3) and inert impurities requires 37.46 ml of

0.1023 molar HCl for titration. The percentage by weight of sodium carbonate in the sample is most nearly

(a) 48.86% (c) 97.72%(b) 100% (d) 24.43%


Answer

Carbonate ion reacts with hydronium ion to produce carbonic acid according to the equation

CO32– + 2H3O

+ – H2CO3 + 2H2O

from which it is evident that 2 moles of HCl (source of the H3O

+) is required for the complete conversion of 1 mole of

carbonate ion to carbonic acid. The number of moles of HCl used is 0.03746 liter x 0.1023 mole/liter = 3.832 x 10 -3 mole


Answer (continued)

Since 2 moles of HCl is required to react with 1 mole of CO32 –, the

number of moles of sodium carbonate present is half this amount, or

3.832 x 10-3 mole/2 = 1.916 x 10-3 mole The formula weight of Na2CO3 is 106.0, and the weight of Na2CO3 in the

sample is therefore 1.916 x 10-3 mole x 106.0 g/mole = 0.2031 g

The percentage by weight of Na2CO3 in the sample is thus

0.2031 g/0.4157 g = 0.4886 = 48.86% (answer a)


Acid/Base

When 1.00 x 10-2 mole of p-chlorobenzoic acid is dissolved in enough water to make 1 liter of solution, the pH of the solution is 3.02. The pH of a 1.00 molar solution of this acid is most nearly

(a) 0 (c) 2(b) 1 (d) 3


Answer

We can represent the acid as simply ROOH, and thus the reaction it undergoes when it dissociates is

ROOH + H2O ROO– + H3O+

From the pH of the 0.0100 molar solution, we can determine equilibrium concentrations of the various species and then the Ka of the acid. With this, we can then calculate the hydronium

ion concentration and thus the pH for any other starting concentration of the acid.


Answer (continued)

For pH = 3.02 = log 1/[H3O+], [H3O

+] = 10-3.02 = 9.55 x

10-4 molar = [ROO–]. The concentration of the remaining undissociated acid is thus 0.0100 – 9.55 x 10-4 = 9.05 x 10-3. The concentration of water is assumed constant in aqueous solution, and thus it does not appear in the Ka expression:

Ka = [ROO–][H3O

+]/[ROOH]

= (9.55 x 10–4)(9.55 x 10–4)/(9.05 x 10–3) = 1.01 x 10–4


Answer (continued)

Now, with 1.00 molar ROOH, set the equilibrium concentrations of the anion and hydronium ion equal to X and the undissociated acid concentration equal to 1.00 – X. Put the terms into the Ka expression and solve for X.

Ka = 1.01 x 10-4 = (X)(X)/(1.00 –X)

X = 0.01 molar = [H3O

+]

Therefore, the expected pH = –log [H3O

+] = –log (0.01) = 2

(answer c)


Electrochemistry

In the compound potassium bromate (KBrO3), the

oxidation number of bromine (Br) is (a) +1(b) +3(c) +5(d) -1


Answer

We begin by assigning oxidation numbers to the potassium and the oxygen. Potassium, being an alkali metal, will always have an oxidation number of +1, while oxygen, unless it is a part of a peroxide, will almost always have an oxidation number of -2. Thus, with three oxygen atoms present, the total oxidation number of the oxygens is -6, while that of the potassium is +1. To make the molecule neutral as a whole, bromine must be +5. Answer (c).


Electrochemistry

The weight, in grams, of chromium that will be plated from a solution of Cr(NO3)3 by a

current of 6.0 amp in 2.0 hours is most nearly (a) 3.9(b) 7.8(c) 15.6(d) 23.4


Answer

The equivalent weight of chromium is its atomic weight (52.0) divided by the number of chromium ions produced per dissolved chromium nitrate molecule, which is 3. equivalent weight = 52.0/3 = 17.3 g/eq


Answer (continued)

From Faraday’s law, each equivalent requires 96,500 coulombs to produce. In this problem, we used 6.0 amps (coul/sec) in 2 hours (= 7200 sec). Thus, the number of coulombs used is 6 coul/sec x 7200 sec = 43,200 coulombs 43,200 coul/96,500 coul/eq = 0.448 equivalents 0.448 eq x 17.3 g/eq = 7.75 g (answer b)


Stoichiometry

The weight, in grams, of a single atom of silicon (atomic mass = 28.1) is most nearly

(a)2.33 x 10-23

(b)4.67 x 10-23

(c)9.33 x 10-23 (d) can’t be determined from the data given


Answer

The atomic mass of silicon is 28.1, which is the mass, in grams, of 1 mole of silicon atoms. One mole of silicon is Avogadro’s number of silicon atoms. Thus (28.1 g/mole)/(6.023 x 1023 atoms/mol) = 4.67 x 10-23 g/atom (answer b)


Stoichiometry

The percentage by weight of C in chloroform, CHCl3 is most nearly

(a) 0.84%(b) 10.0%(c) 89.1%(d) 20.0%


Answer

One gram-formula weight of the compound is 12.0 + 1.0 + 3(35.5) = 119.5 g. Thus, there are 12 g of carbon in every 119.5 g of the compound, and therefore the percentage of carbon in the compound is 12.0/119.5 x 100 = 10.0% C (answer b) By the same reasoning, the percentages of H and of Cl in CHCl3 are

1.0/119.5 x 100 = 0.84% H

and (3 x 35.5)/119.5 x 100 = 89.1% Cl


Stoichiometry

When the equation below is balanced, what will be the coefficient of H2O?

C6H6 + O2 CO2 + H2O

(a) 1 (c) 6(b) 2 (d) 15


AnswerStart with the carbon atoms. Since there six on the left side, and since carbon appears only in CO2 on the right side, put a 6

before the CO2:

C6H6 + O2 6CO2 + H2O

Putting a 3 before the H2O will balance the hydrogens:

C6H6 + O2 6CO2 + 3H2O

All that remains is to balance the oxygen.


Answer (continued)With 12 + 3 = 15 oxygens on the right side, a coefficient of 7.5 is needed for the O2 on the left side:

C6H6 + 7.5O2 6CO2 + 3H2O

We could consider the equation balanced. However, we wish to think of the equation as representing a reaction involving whole molecules and avoid fractional coefficients. We can accomplish this by multiplying each coefficient by 2, yielding:

2C6H6 + 15O2 12CO2 + 6H2O

Thus, the coefficient of H2O is 6 (answer c).


Electronic Structure

What is the electronic configuration for Fe? (a) 1s22s22p63s23p64s23d6

(b) 1s22s22p63s23p64s24p6

(c) 1s22s22p63s23p64s23d44p4

(d) none of the above


Answer

The answer is (a) based on the principle of filling orbitals from lowest to highest energy.


Chemical Bonding/PeriodicityArrange the following compounds in order of

increasing ionic character in its bonds. LiBr, FeBr2, BeBr2

(a) LiBr, FeBr2, BeBr2

(b) LiBr, BeBr2, FeBr2

(c) FeBr2, BeBr2, LiBr

(d) FeBr2, LiBr, BeBr2


AnswerWe obtain this answer by examining electronegativity differences. Electronegativity increases as one moves from left to right and from bottom to top across the periodic table. Also, the greater the electronegativity difference between two atoms in a bond, the greater the ionic character. With the anion the same in all of these compounds, we need examine only the relative electronegativities of the cations. Lithium is in Group 1, beryllium is in Group 2, and iron is among the transition metals, to the right of both Li and Br. Thus, Li is the most electronegative, followed by Be and then by Fe. Iron-bromine bonds will be the least ionic followed by beryllium-bromium bonds and then lithium-bromine bonds. The answer is (c).


Stoichiometry

The percentage composition of a compound containing only carbon and hydrogen is 85.6% C and 14.4% H. The molecular weight is 56.0. What is the molecular formula for this compound?

(a) CH2

(b)C2H4

(c) C3H6

(d) C4H8


AnswerChoose a basis of 100 grams of the compound. Thus, we have 85.6 grams of C and 14.4 grams of H. Convert these to moles and form a molar ratio of the elements. 85.6 g C/(12.0 g/mol) = 7.13 moles C 14.4 g H/(1.0 g/mol) = 14.4 moles H


Answer (continued)

Thus, the empirical formula is C7.13H14.4 CH2,

which has a formula weight of 12.0 + 2 x 1.0 = 14.0. Divide the formula weight into the molecular weight to determine the number of formulas (and thus the molecular formula) in the compound.

56.0/14.0 = 4.0 Thus, the molecular formula is the empirical formula multiplied by 4, or C4H8 (answer d).


States of Matter

A closed vessel is partially filled with a liquid at a certain temperature. As the temperature is increased, what will happen to the density of the liquid and the density of the vapor above the liquid?

(a) both will decrease(b) both will increase(c) liquid density decreases and vapor density increases(d) no change in liquid density while vapor density increases


Answer

As the temperature increases, the vapor pressure of the liquid increases and more liquid evaporates into the gas phase. Because the vessel is closed, the vapor volume cannot expand to accommodate the additional vapor molecules, and thus the density of the vapor phase increases. The liquid phase expands very slightly with an increase in temperature. Coupled with this is the slight loss of liquid to the gas phase as the liquid evaporates. Both of these phenomena act to decrease the number of liquid phase molecules while at the same time increasing the volume the liquid phase occupies. Thus, the liquid phase density decreases. The correct answer is (c).


States of Matter

A sample of methane gas has a volume of 1.50 liters at 25C and 1 atm pressure. What volume will it occupy at 50C and 1 atm pressure?

(a)1.63 liters (c) 0.75 liters(b) 3.0 liters (d) 1.0 liter


Answer

From Charles’ Law:

V1/T1 = V2/T2

From the information given: V1 = 1.50 liters

T1 = 25C + 273 = 298 K

T2 = 50 C + 273 = 323 K

V2 = unknown


Answer (continued)

Solving for V2 and substituting:

V2 = V1/T1 x T2 = 1.50 liters/298 K x 323 K

= 1.63 liters

(answer a)


States of Matter

What weight of N2 will occupy a volume of 2.5

liters at a pressure of 1 atm and a temperature of 200C?

(a)0.9 g(b)1.8 g(c)3.6 g(d) can’t be determined from the data given


Answer The ideal gas law is:

PV = nRT = (g/M)RTwhereP = pressure = 1 atmV = volume = 2.5 litersT = temperature = 200 C + 273 = 473 KR = ideal gas constant = 0.0821 liter-atm/mole-KM = molecular weight = 28.0 g/mole for N2

g = mass = unknown


Answer (continued)

Solving for the mass and substituting: g = (PVM)/(RT) = (1 x 2.5 x 28.0)/(0.0821 x 473) = 1.8 g

(answer b)


Solution Chemistry

The molarity of a 20 weight percent NaCl solution which has a density of 1.15 g/ml is most nearly (a) 0.34 (b) 0.39 (c) 7.86 (d) 3.93


Answer

One milliliter of the solution weighs 1.15 g, so 1 liter weighs 1.15 g x 1000 = 1,150 g. Since the solution is 20 weight percent NaCl, the weight of NaCl present is 20% of the total weight, or 0.20 x 1,150 = 230 g. To get the molarity, we need to convert the mass of NaCl to moles and divide by 1 liter. The molecular weight of NaCl is 23.0 + 35.5 = 58.5: (230 g NaCl/58.5 g/mole NaCl)/1 liter = 3.93 molar (answer d)


Solutions

The vapor pressure of pure water at 29C is 30.0 torr. The vapor pressure in torr of a 20 weight percent glucose (C6H12O6) solution in water at

29 C is most nearly (a) 20.3 (b) 29.4 (c) 0.60 (d) 9.7


Answer

From Raoult’s Law, p = p0 Xwherep0 = vapor pressure of pure water at 29 C = 30.0 torrX = mole fraction of water in the solutionp = vapor pressure of the solution, unknown


Answer (continued)

The molecular weight of glucose is 6 x 12 + 12 x 1 + 6 x 16 = 180 and that of water is 2 x 1 + 1 x16 = 18. Thus, assuming a basis of 100 grams of solution, there are 20/180 = 0.11 moles of glucose and 80/18 = 4.4 moles water. Thus, the mole fraction of water is given by:

X = 4.4/(4.4 + 0.11) = 0.98.


Answer (continued)

Thus, the vapor pressure of the solution is p = p0 X = 30.0 torr x 0.98 = 29.4 torr. (answer b)


Solutions

A mixture of 40.0 g of chloroform and 80 g of carbon tetrachloride is prepared. The mole fraction of chloroform in the resulting solution is most nearly (a) 0.196 (b) 0.304 (c) 0.392 (d) 0.608


Answer

The molecular weight of chloroform (CHCl3) is

12.0 + 1.0 + 3 x 35.5 = 119.5.The molecular weight of carbon tetrachloride (CCl4) is 12.0 + 4 x 35.5 = 154.0

Thus, the number of moles of chloroform is 40.0 g/(119.5 g/mol) = 0.335 moles


Answer (continued)

and the number of moles of carbon tetrachloride is

80.0 g/(154.0 g/mol) = 0.519 moles and the mole fraction of chloroform is thus 0.335 moles/(0.335 moles + 0.519 moles) = 0.392 (answer c)


Solutions

The volume, in milliliters, of a 6.0 molar HCl solution required to prepare 250 ml of Cl2 at STP according to

the following equation MnO2 + 4HCl MnCl2 + Cl2 + 2H2O

is most nearly (a) 3.75 (b) 6.00 (c) 7.50 (d) 15.0


Answer

The number of moles of Cl2 (MW = 71.0) contained in 250 ml

at STP can be gotten from the ideal gas equation:

PV = nRT Solving for the number of moles:

n = PV/RT n = (1 atm)(0.250 lit)/(0.0821 lit-atm/mol-K)(273 K) = 0.0112 moles Cl2


Answer (continued)

Thus, from the balanced chemical reaction, we require four times this much HCl, or 4 x 0.0112 = 0.0448 moles HCl. We have a 6 molar solution from which to get it: 0.0448 mole/(6.0 mol/lit) = 0.00750 lit = 7.50 ml (answer c)


KineticsFor the reaction CO + Cl2 COCl2, the rate of formation

of COCl2 at a given temperature, T, is found to be

6.7 x 10-3 moles/liter-min when the concentration of CO and of Cl2 are each 0.10 molar. The rate law for this

reaction is

r = k [CO][Cl2]3/2

The rate of formation, in the same units, of COCl2 at this same

temperature if the concentrations of CO and Cl2 are each

0.02 molar is most nearly (a) 2.71 x 10-5 (b) 3.79 x 10-5

(c) 1.18 x 10-4 (d) 4.57 x 10-4


Answer

First we must calculate the rate constant from the given information: k = r/[CO][Cl2]

3/2 = (6.7 x 10-3)/(0.10)(0.10)3/2

= 2.09 Therefore, our rate law is

r = 2.09 [CO][Cl2]3/2


Answer (continued)

For our new set of conditions, with [CO] = [Cl2] = 0.02

molar, the rate of reaction is r = (2.09)(0.02)(0.02)3/2 = 1.18 x 10-4 mole/liter-min (answer c) This is a smaller rate than in the first case, which is what we would expect, since the temperature is unchanged and the concentrations of the reactants are smaller.


Kinetics/EquilibriumEthyl acetate can be prepared from acetic acid and ethanol according to the equation CH3COOH + C2H5OH – CH3COOC2H5 + H2O

At some temperature, 0.10 mole of acetic acid is mixed with 0.10 mole of ethyl alcohol in 1.0 liter of solution (a non-reactive solvent is used). When equilibrium is established, 0.060 mole of ethyl acetate is present. The equilibrium constant, K, for this reaction is most nearly: (a) 2.2 (b) 0.44 (c) 1.0 (d) can’t be determined from the data given


Answer

From the stoichiometry of the balanced reaction (the reaction as given is balanced), one mole of each reactant reacts to produce one mole of each product. Thus, if 0.060 mole of ethyl acetate is produced, 0.060 mole of water must also be produced. This also means that 0.060 mole of each of acetic acid and ethanol must have reacted, leaving 0.10 – 0.060 = 0.040 mole of each reactant remaining.


Answer (continued)To summarize, at equilibrium: [CH3COOC2H5] = [H2O] = 0.060 mole/liter

[CH3COOH] = [C2H5OH] = 0.040 mole/liter

From the definition of the equilibrium constant K = ([[CH3COOC2H5][H2O])/([CH3COOH][C2H5OH]) =

(0.060)(0.060)/(0.040)(0.040)

K = 2.2 (answer a)


Kinetics/Extent of Reaction

Solid CuO reacts with gaseous CO to give solid Cu2O

and gaseous CO2.

2CuO(s) + CO(g) – Cu2O(s) + CO2(g)

If the system is at equilibrium, how will a decrease in the volume of the system affect the number of moles of CO2?

(a) decreases (b) increases (c) remains unchanged d) cannot tell from the data given


Answer

By Le Chatelier’s Principle, any system at equilibrium will act to return to an equilibrium condition if anything is done to it to perturb that equilibrium. In this case, two moles of solid CuO react with one mole of gaseous CO to yield one mole of solid Cu2O and one mole of gaseous CO2. Decreasing the volume

of the container will cause the system to respond in such a way as to reduce its own volume. The system can do this by decreasing the number of moles of gas present. (The volume of solids is neglected because a change in pressure, which will occur when the volume is changed, has a negligible effect on the volume of a solid.)


Answer (continued)

Examining the above equation, we find that the number of moles of gas on each side is the same. Thus, the system cannot respond to a change in volume because driving the reaction in either direction will have no effect on the total moles of gas present. Thus, a decrease in volume will not affect the number of moles of CO2, and the correct answer is (c).

Copyright Kaplan AEC Education, 2008 Chemistry Outline Overview STATES OF MATTER, p. 546...

Documents

Transcript of Copyright Kaplan AEC Education, 2008 Chemistry Outline Overview STATES OF MATTER, p. 546...