2007 Solutions

 

1.               B

2.               C 2100g/(44g/mole) = 47.7 moles; 47.7 moles(32g/mole) = 1527 g

3.               D P = nRT/V = (5/32)(8.31(-25+273)/0.3 = 1073 kPa

4.               B

5.               B (snow= liquid water to solid is exothermic; because the reverse(melting) needs to absorb heat )

6.               C (really -396 kJ/mole)

7.               A

8.               C

9.               D

10.            A

11.            D

12.            A

13.            C

14.            A

 

 

15. PV = 90 (choose any two values of P and V)

If P = 12, then V= 90/12 = 7.5 mL

16. 1.27 g KClO3 (mol/ 122.5 g) = 0.010367 moles KClO3

but it's not a gas, so we need to to look at equation:

0.010367 moles KClO3(3 O2/2KClO3) = 0.01555 moles O2

 

R = PV/(nT) = 94.7 kPa(0.468 L)/(0.01555 moles)/(293 K) = 9.72 kPaL/(Kmole) : not an IDEAL gas

(question is not based on real data or experiment went awry becuase in reality oxygen does behave like an ideal gas!)

 

17. n = PV/(RT) = 0.0134 moles

 

Molar mass = m/n = (51.96 -51.02)g/ 0.0134 moles = 70.2 g/mole = CHF3

 

18. n = PV/(RT) = 0.00691 moles of H2

 

0.00691 moles of H2 (1 Mg/ 1 H2) = 0.00691 moles of Mg

 

0.00691 moles of Mg (24.3 g/mole) = 0.168 g

 

19. Q = mcDT

 

= (225 + 375)g (4.19 J/gC)(30.0 C)

= 75420 J

 

n = CV = 1mole/L (0.225L) = 0.225 moles

 

DH = -Q = -75420 J

 

DH/n =-75420 J/0.225 moles = -335 kJ/mole

 

20.            Multiply equation(2) by 3 and double equation (1). Reverse equation(3).

Note equation (4) is no good because it does not have liquid water.

Add the three equations: +278 kJ + 3(-285 kJ) + -2(394kJ) = -1365 kJ

 

DH = -1365 kJ which is for 1 mole of C2H5OH

 

-5509 kJ (1mole/-1365 kJ)(46g/mole) = 184 g.

 

21. Expt 1

Q = mcDT = 566 J

DH = -Q = -566 J

n = 0.200/65.3 = 0.00306 moles

DH/n = -566J / 0.00306 moles = -185 000 J = -185 k J/mole

 

Do the same for expt 2: DH/n = -91 .9 kJ/mole

 

Zn + 2HCl ZnCl2 + H2 DH = -185 kJ

ZnCl2 + H2O ZnO + 2 HCl DH = 91.9 kJ (notice: we reversed equation and sign to cancel ZnCl2)

H2 + 0.5 O2 H2O DH = -286 kJ

________________________________________________________________________________

Zn + 0.5 O2 ZnO DH = -379 kJ

 

 

 

22.                      use _Qlost = Q gained

c = 0.623 J/goC

 

23. A CH4: fewest bonds to break

B CH3OH gaseous states react faster

 

 

24. after 7 minutes, 0.244 moles HCl remain. There were originally 0.250.

So 0.006 moles of HCl reacted

 

Ratio is 2:1, so only 0.006 moles/2 = 0.003 moles of carbon dioxide formed in 7 minutes

 

Average rate = Dn/Dt = 0.003 moles/7 minutes = 4.29 X10-4 moles CO2/min

 

25. A curve 1 has no catalyst. Less is being made in the same given time.

B Curve 2 has a catalyst. Ae is lower.

 

 

26.

 

2 H2

2 NO

N2 (dont use liquid water)

Initial

4/2 = 2M

2 M

0

Changing

0.40 *2 =0.80

0.80

0.40

Equilibrium

2 0.80 = 1.2

1.2

0.40

 

K = [N2]/([H2]2[NO]2

= 0.19

 

27.

 

HCOOH

H2O

H3O+

HCOO-

Initial

0.10

-

0

0

Changing

x

-

x

x

Equilibrium

0.10 - x

-

x

x

 

 

 

 

 

 

KA = x(x)/(0.10-x) = 1.77 X 10-4

x = 0.004119567520

pH = -log(H+) = -log(0.004119567520) = 2.39

 

28. A) aluminum is oxidized

B) 2 Cu+2 + 6e- 3 Cu 0.34 V

2 Al 2 Al+3 + 6e 1.66 V

_________________________________________

2 Cu+2 + 2 Al 2 Al+3 + 3 Cu 2.00 V

c) 2.00 V

d) Cu+2 at the reduction half cell (cathode)

 

29. a) 2 Ag+ + 2e 2 Ag 0.80 V

Cu Cu+2 + 2e- -0.34 V

_______________________________________

2 Ag+ + Cu Cu+2 + 2 Ag 0.46 V