CHEM 1411 Exam # 2
(Chapters 4,5,6, and 7)
Name:________________________________
Score:
Part I- ( 3 points each) - Please write your correct answer next to each question number, DO NOT CIRCLE.
____ 1. Which of the following are nonelectrolytes in water?
(i) HF
A.
(ii) ethanol, C2H5OH
(iii) CH3 OCH3
B. i, ii, and iii
C. iii only
ii and iii
(iv) KClO3
D. ii only
____ 2. How many milliliters of 1.50 M KOH solution are needed to supply 0.125 mole of KOH?
A. 0.0833 ml
B. 0.188 ml
C. 12.0 ml
D. 83.3 ml
____ 3. Which of the following is/are soluble in water?
(i) NiCl2
A. iv only
(ii) Ag2S
(iii) Cs3PO4
B. i, ii, and iii
C.
(iv) (NH4)2SO4
i, ii, and iv
D. i, iii and iv
____ 4. Which of the following reactions will occur?
(i)Ni(s) + Zn2+(aq) 
(ii)Pb(s) + Ag+(aq) 
(iii)Zn(s) + Ca2+(aq) 
(iv)Al(s) + Fe2+(aq) 
A.
i only
B.
ii only
C. ii and iv only
D. i and iii only
C. NH4OH
D. RbOH
____ 5. Which of the following is a weak base?
A. NaOH
B. Ca(OH)2
____6. How many kJ of heat must be removed from 1.0x103 g of water (heat capacity of 4.184 J /g.K) to
lower the temperature from 18.0°C to 12.0°C?
A. 2.5x10-2 kJ
B. 1.4 kJ
C. 4.2 kJ
D. 25 kJ
____ 7. From the following heats of reaction,
2C (graphite) + H2 (g)  C2H2 (g)
∆ H = 227 kJ/mole
6C (graphite) + 3H2 (g)  C6H6 (l)
∆ H = 49 kJ/mole
calculate the heat for the reaction
3C2H2 (g)  C6H6 (l)
A. 632 kJ/mole
B. -632 kJ/mole
C. -178 kJ/mole
D. 178 kJ/mole
____8. The heat of combustion of fructose, C6H12O6, is -2812 kJ. Using the following information,
1
f
for fructose.
C6H12O6 (s) + 6 O2(g)  6 CO2(g) + 6 H2O (l)
∆ H°f ( CO2 ) = -393.5 KJ/mole
A -210.3 kJ/mol
∆ H°f (H2O) = - 285.83 KJ/mole
B. 210.3 kJ/mol
C. -1264 kJ/mol
D. 1264 kJ/mol
____9. What is the kinetic energy in J and cal of a 45-g golf ball moving at 61 m/s?
A. 168 J, 40 cal
B. 84 J, 20 cal
C. 84 J, 350 cal
D. 84 kJ, 20 kcal
____10. Consider the combustion reaction of ethane gas, C2H6(g):
C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(g) , ∆ H = -1430 kJ
What is the enthalpy change for the reverse reaction if whole number of coefficients are used?
A. +1430 kJ
B. -1430 kJ
C. -2860 kJ
D. +2860 kJ
____11. What is the frequency of radiation that has a wavelength of 0.589 pm?
A. 1.96 x 10-21 s-1
B. 5.09 s-1
C. 5.09 x 108 s-1
D. 5.09 x 1020 s-1
____ 12. Statement, electrons fill the orbital singlet, then double up is called………….
A. Aufba principle
C. Puali's exclusion principle
B. Hund's rule
D. none of these
____ 13. For n = 4, what are the possible values of l?
A. 3, 2, 1
B. 4, 3, 2, 1
C.
3, 2, 1, 0
D. 4, 3, 2, 1, 0
____ 14. What is the maximum number of electrons that can occupy the subshell 3d?
A. 1
B. 3
C. 5
D. 10
____15. Write the electron configuration for the atom Cu, using the appropriate noble-gas inner core for
abbreviation.
A. [Ar]4s23d10
B. [Ar]4s24d9
C. [Ar]4s13d10
D. [Kr]4s13d10
____ 16. Which of the following gas has higher density at STP condition?
A. CH4
B. Cl2
C. CO2
D. O2
____ 17. How many moles of N2 gas occupy 11.2 liters volume a STP condition?
A. 1.0 mol
B. 2.0 mol
C. 0.50 mol
____ 18. Which of the following is/are not characteristic of gases?
2
D. cannot be determine
I. high density
A. I only
II. formation of homogeneous mixtures
B. I and III
III. low intermolecular forces
C. I, II, and III
D. II and III
____19. An unknown gas "X" effuses two times faster than a sample of SO3(g) through a porous container.
Which of the following is the unknown gas ?
A. H2
B. CH4
C. Ne
D. Ar
____20. Which of the following is not a statement of Boyle's law? All statements assume constant
temperature and amount of gas.
A. P = constant/V
B. V α 1/P
C. P/V = constant
D. PV = constant
____21.Which one has the largest radius ?
A. K+
B. K
C. Al 3+
D. Na +
____22. Arrange the following atoms in order of increasing atomic radius:
A. N < K < As < Fr
D. Fr < K < As < N
B. N < As < K < Fr
E. K < Fr < N < As
E. Mg
N, K, As, Fr
C. As < K < N < Fr
PART II- ( 5 points each) Please show all your work.
23. Write a complete ionic and net-ionic equation for the following reaction.
K2CO3 (aq) + H2SO4 (aq) 
24. a) What volume(ml) of 0.115 M HClO4 solution is required to neutralize 50.00 ml of 0.0875 M
Ca(OH)2 ?
b) A solution contains 3.2 g NaOH ( MW = 40) in 20.0 ml of solution. What is the molarity of the
solution?
25. A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter
of negligible heat capacity containing 80.0 mL water at 24.0 oC. The final temperature of the system
was found to be 28.4 oC. Calculate the Specific heat of the metal if density of water is 1.00 g/ml.
3
26. From the following heats of reaction,
I) N2(g) + 2O2(g)  2NO2(g)
∆H = +67.6 kJ
II) 2NO(g) + O2(g)  2NO2(g)
∆H = 113.2 kJ
calculate the heat of the reaction ,
N2(g) + O2(g)  2NO(g)
27. For the electronic transition from n = 3 to n = 5 in the hydrogen atom, calculate the energy,
and wavelength (in nm).
28. Write electron and core configuration for Bromine atom and determine the total number of unpaired
electrons.
29. 2.50 g CO2 gas occupies 5.60 liters at 789 torr.
a) Calculate the temperature of gas in 0C.
b) What is the density of CO2 gas at STP condition?
30. What volume of O2(g), measured at 22 °C and 763 torr, is consumed in the combustion of 7.50 L
of C2H6(g), measured at STP?
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)
Bonus Question ( 10 points) - Please show all your work.
Phosphorous pentachloride is used in the industrial preparation of many organic phosphorous compounds.
Equation I shows its preparation from PCl3 and Cl2:
(I)
PCl3 (l) + Cl2(g)  PCl5(s)
Use equation II and III to calculate ∆Hrxs of equation I:
P4 (s) + 6 Cl2 (g)  4 PCl3 (l)
∆H = 1280 KJ
(III) P4 (s) + 10 Cl2 (g)  4 PCl5 (s)
∆H = 1774 KJ
(II)
4
1411 SAMPLE EXAM # 2 (Key)
PART - I
1. A
2. D
n = M.VL  VL = (n /M) = (0.125 mol/ 1.50 mol/L) = 0.0833 L  x 1000 = 83.3 mL
3. D
4. C
5. C
6. q = m .c . ∆T = (1.0x103 g)(4.184 J/g.K)( 18.0 – 12.0 oC) = 25104 J = 25100 J /1000 = 25 kJ
7. B
3C2H2 (g)  6C (graphite) + 3H2 (g)
Reaction I (3 x rev.)
∆ H = 3(-227) kJ/mol
Reaction II (same)
6C (graphite) + 3H2 (g)  C6H6 (l)
∆ H = 49 kJ/mole
__________________________________________________________________________
Overall reaction
3C2H2 (g)  C6H6 (l)
∆ H = -632 kJ/mol
8. C
∆ H = ∆ H°f (reactants0 - ∆ H°f (products)
-2812 = [6(-285.83) + 6(-393.5)] – [ 6 (0) + ∆ H°f (fructose)] ∆ H°f (fructose)= -1264 kJ/mol
9. B
K.E. = (mV2)/2 = (.045 kg )(61 m/s)2 / 2 = 84 J / 4.184 = 20 cal
10. D
For reverse reaction with whole number
11. D
ν = (c / λ ) = ( 3.00x108 m/s )/ 0.589x10-12 m ) = 5.09 x 1020 s-1 or Hz
12. B
16. B
17. C
-2x (∆ H) = -2( -1430 kJ) = + 2860 kJ
13. C
14. D
15. C
Cl2 has highest molecular weight = 71.0 gr/mol
(11.2 L) ( 1 mol / 22.4 L) = 0.50 mol N2
18. A
19. C rate(X) / rate (SO3) = ( MSO3/Mx)½  2 = ( 80 / Mx)½  4 = 80 / Mx  Mx = 80/4 = 20 Neon
20. C
21. B
22. B
PART - II
23. molecular equation: K2CO3 (aq) + H2SO4 (aq)  K2SO4(aq) + H2CO3(aq)
ioninc equation : 2K+(aq) +CO32-(aq)+ 2H+(aq) +SO42- (aq)  2K+(aq) + SO42-(aq) + H2O(l) +CO2(g)
net-ionic equation : 2H+(aq) + CO32-(aq)  H2O(l) +CO2(g)
spectator ions : 2 K+ and SO42-
5
nbMbVb
(2)(50.00)(0.0875)
24. a) Va = ----------------- = --------------------------------- = 76.1 ml
naMa
(0.115)
(3.2/40 mole)
M = ----------------------- = 4 mol/ L ( mol.L -1 ) or molar
(20.0/1000 L)
b)
25. mc ∆T (metal) = mc ∆T (water)
(44.0)(Cmetal) (99.0-28.4) = (80.0)(4.184)(28.4-24)  Cmetal = 0.474 J/g.oC
H=
26. N2 + 2O2  2NO2
+ 67.6 kJ
 2NO + O2  H = + 113.2 kJ
2NO2
 H = + 180.8 kJ
N2 + O2  2NO
27.  E = 2.18 x 10 18 ( 1/32 - 1/52 ) = 1.55 x 10 19 J
 E = hc/
( 6.63 x 10 34 )( 3.00 x 108 )
 = ------------------------------------------ = 1.28 x 106 m = 1280 nm
( 1.55 x 10 19 )
28.
35Br
= 1ss 2s2 2p6 3s2 3p6 4ss 3d10 4p5 = [ Ar ] 4ss 3d10 4p5
One unpaired electron



29. a) PV = nR T  PV = (m/M) RT  T = ( MPV/m R)
= (44)(789)(5.60)/(760)(2.50)(0.0821) =1246 K = 973 oC 102 K
b) D(CO2) STP = (M /22.4) = 44/22.4 = 1.96 g/L
30. (7.50 L C2H6)( 1mol C2H6 / 22.4 L) ( 7 mol O2 / 2 mol C2H6 ) = 1.17 mol. O2
V = ( nRT/P) = (1.17)(0.0821)(273+22)(760)/(763) = 28.2 L
Bonus
Rev x 1/4 (ii)
1/4 (iii)
PCl3
 1/4 P4 + 6/4 Cl2
;H=
;  H =  443.5 kJ
1/4 P4 + 10/4 Cl2  PCl5
PCl3
+ Cl2
+ 320 kJ
;  H =  123.5 kJ
 PCl5
6