Copyright Kaplan AEC Education, 2008 PROBLEM #1 Carbonate ion reacts with hydronium ion to produce...

Copyright Kaplan AEC Education, 2008

PROBLEM #1

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


PROBLEM #1 (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)


PROBLEM #2

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



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)


PROBLEM #3Start 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.


PROBLEM #3 (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).


PROBLEM #4

From Charles’ Law:

V1/T1 = V2/T2

From the information given: V1 = 1.50 litersT1 = 25C + 273 = 298 KT2 = 50 C + 273 = 323 KV2 = unknown


PROBLEM #4 (CONTINUED)

Solving for V2 and substituting: V2 = V1/T1 x T2 = 1.50 liters/298 K x 323 K = 1.63 liters

(answer a)


PROBLEM #5

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)

PROBLEM #1

PROBLEM #2

PROBLEM #3

PROBLEM #4

PROBLEM #5

PROBLEM #1

PROBLEM #2

PROBLEM #3

PROBLEM #4


PROBLEM #5






PROBLEM #1P = A (P/A, i, 24)900 = 40 (P/A, i, 24)(P/A, i, n) = 22.50 Looking at the tables this is true for i approximately 0.5% per month

iyear = (1 + imonth)12 – 1 = (1 + 0.005)12 – 1 = 0.0617 = 6.17%

Answer: (c)


PROBLEM #2

S = 650,000 (F/P, 15%, 5) – 150,000 (F/A, 10%, 5) = 650,000(2.0114) – 150,000(6.7424) = $296,050

Answer: (b)


PROBLEM #3EUACA = 75,000(A/P, 15%, 4) - 16,000 - 9,000(A/F, 15%, 4) = 75,000(.35027) - 16,000 - 9,000(0.20027) = $8,468

EUACB = 8,468 = 105,000(A/P, 15%, n) - 24,000(A/P, 15%, n) = (8,468 + 24,000)/105,000 = 0.3092From table look-up, the value of n that most nearly makes the

above relation true is 5.

Answer: (b)


PROBLEM #4EUACA = 25,000 (A/P, 10%, n) + 4,350EUACB = 32,000 (A/P, 10%, n) + 2,500

EUACA = EUACB

25,000 (A/P, 10%, n) + 4,350 = 32,000 (A/P, 10%, n) + 2,500(A/P, 10%, n) = (4,350 – 2,500) / 7,000 = 0.2643From table look-up, the value of n that most nearly makes the above relation true is 15.

Answer: (a)


PROBLEM #5

21,000 = 15,000 (1 + f)10

f = (21,000/15,000)0.1 – 1 = 0.034 = 3.4%

Answer: (c)

PROBLEM #1

PROBLEM #2

PROBLEM #3

Example 4 Solution


PROBLEM #1




PROBLEM #2


PROBLEM #3


PROBLEM #4


PROBLEM #5


PROBLEM #1




PROBLEM #2






PROBLEM #3


PROBLEM #4


PROBLEM #5





PROBLEM #1


PROBLEM #2

PROBLEM #3

PROBLEM #4


PROBLEM #5


PROBLEM #1


PROBLEM #2






PROBLEM #3


PROBLEM #4






PROBLEM #5




PROBLEM #1


PROBLEM #2


PROBLEM #2(continued)


PROBLEM #2(continued)


PROBLEM #3




PROBLEM #4


PROBLEM #5


PROBLEM #1




PROBLEM #2


PROBLEM #3


PROBLEM #4


PROBLEM #5


PROBLEM #1


PROBLEM #2


PROBLEM #3


PROBLEM #4


PROBLEM #5





Date post:	31-Dec-2015
Category:	Documents
Upload:	madeleine-gregory
View:	212 times
Download:	0 times

Copyright Kaplan AEC Education, 2008 PROBLEM #1 Carbonate ion reacts with hydronium ion to produce...

Documents