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"; Chemistry Atomic Mass Chemistry Test - Chem Homework Help
Chemistry Atomic Mass Chemistry Test

 

Last Name: _____________________________ First Name: _______________

Lab Sec. # ____ TA: ____________________ Lab day/time: _____________

 

 

 

CHEMISTRY 2A (D)

 

Instructions:

Final Exam 

 

 

Multiple Choice Circle one

CLOSED BOOK EXAM! DO NOT OPEN the exam until instructed to do so.

1. a b c d e

2. a b c d e

3. a b c d e

4. a b c d e

5. a b c d e

6. a b c d e

7. a b c d e

8. a b c d e

9. a b c d e

10. a b c d e

11. a b c d e

12. a b c d e

13. a b c d e

1-13 total points:

14. a b c d e

15. a b c d e

16. a b c d e

17. a b c d e

18. a b c d e

19. a b c d e

14-19 total points:

Possible Points

Points

# 1-13 (3 points each)

/ 39

# 14-19 (5 points each)

/ 30

# 20 (6 points)

/ 06

# 21 (42 points)

/ 42

# 22 (10 points)

/ 10

# 23 (10 points)

/ 10

# 24 (06 points)

/ 06

# 25 (30 points)

/ 30

# 26 (06 points)

/ 06

# 27 (16 points)

/ 16

Total Score (195)

/ 195

No books, notes, or additional scrap paper are permitted. All information required is contained on the exam. Place all work in the space provided. If you require additional space, use the back of the exam. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). Sharing of calculators is not allowed.

 

  1. Read each question carefully (27 problems, 15 pages).
  2. ANSWER ALL THE QUESTIONS YOU ARE SURE OF FIRST, AND THEN USE THE REMAINING TIME FOR THE REST.
  3. Please show all relevant work where it is indicated for full credit.
  4. The last 2 pages contain a periodic table and some useful information.

You may remove them for easy access and for scratch.

  1. If you finish early, RECHECK YOUR ANSWERS!

 

Concepts: Multiple Choice

Questions 1-13: 3 points each (no partial credit)

 

  1. Write the symbol of the species that has 28 protons, 26 electrons, and 32 neutrons.

 

    1. 58Fe2+
    2. 60Ni2-
    3. 60Ni2+
    4. 60Fe2+
    5. 60Ni3+

 

  1. In the reaction, H2O2  H2O + 1/2O2 oxygen is:

 

    1. Reduced
    2. Oxidized
    3. Both a. and b.
    4. Neither a. nor b.
    5. Annihilated

 

  1. How many σ bonds does the molecule of guanine have?

 

a)

17

O

NH

C

C

C

C

N

    1. 2
    2. 3
    3. 4 HC N N NH2
    4. 0 H

 

  1. Choose the INCORRECT formula-name combination.

 

    1. Na2CO3 sodium carbonate
    2. P2O5 diphosphorus pentoxide
    3. FeBr2 iron dibromide
    4. N2O3 dinitrogen trioxide
    5. K3N potassium nitride

 

  1. In which of the following compounds does chlorine have the highest oxidation number?

 

    1. ClO2
    2. Cl2
    3. FeCl3
    4. KClO3
    5. Mg(ClO)2
  1. Which of the following reactions is NOT a redox reaction?

 

    1. 5 I (aq) + IO3 (aq) + 6 H+ (aq)  3 I2 (s) + 3 H2O (l)
    2. Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
    3. 2 H2O2 (aq)  2 H2O (l) + O2 (g)

(d)

BaCl2 (aq) + Na2SO4 (aq)  BaSO4 (s) + 2NaCl (aq)

(e) CH3COOH (aq) + 2 O2 (g)  2 CO2 (g) + 2 H2O (l)

 

  1. Which statement is FALSE about the 5dxy orbital?

 

    1. It has a nodal plane in the xz plane.
    2. It has a nodal plane in the yz plane.
    3. For an electron in the 5dxy orbital there is a higher probability of finding the electron farther from the nucleus than if it is in the 5s orbital.
    4. It can hold a total of 10 electrons.
    5. None of the above.

 

  1. Choose the species from which one electron could most easily be removed.

 

    1. Cl
    2. Na+
    3. Na
    4. K
    5. Ar

 

  1. Which of the following has the largest radius?

 

    1. As3–
    2. Br
    3. Sr2+
    4. Cl
    5. They all have approximately the same size

 

  1. Using the VSEPR theory, which one of the following chemical species would you expect to have a dipole moment (polar)? a. I3

b.

O3

    1. CO2
    2. CH4
    3. None of the above
  1. What is the bond order of the nitrogen-oxygen bonds in the nitrate ion? (a) 2, 1, and 1
    1. 1, 2, and 2
    2. 1.33
    3. 1
    4. 2

 

  1. Choose the INCORRECT statement about the molecule of HCN.

 

    1. There are a total of two σ bonds in the molecule.
    2. There are a total of two π bonds in the molecule.
    3. There is a lone pair on nitrogen.

d.

The molecule is bent.

e. The carbon atom has sp hybridization.

 

  1. Which of the following is NOT TRUE about a mixture of gases?

 

    1. The sum of the partial pressures of each gas is equal to the total pressure.
    2. The sum of the moles of each gas is equal to the total number of moles.
    3. The sum of the temperature of each gas is equal to the total temperature.
    4. The sum of the partial volumes of each gas is equal to the total volume.
    5. All of the above statements are true.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Calculations: Multiple Choice

Questions 14-19: 5 points each (no partial credit)

 

  1. A hypothetical element, E, has two stable isotopes:

E-38 = 38.012 u 75.68%

E-46 = 45.981 u 24.32%

The element’s atomic mass would be closest to which element?

A.M. = 39.10

A.M. = 39.95

A.M. = 40.08

A.M. = 44.96

A.M. = 87.62

(b) Ar

(a) K

        1. Ca
        2. Sc
        3. Sr

 

(38.012)(0.7568) + (45.981)(0.2432) = 39.95 u

 

 

 

 

  1. Find the empirical formula for a compound with the following mass composition: K 42.4%, Fe 15.2%, C 19.5% and N 22.8%.
        1. KFeCN
        2. K4FeC6N6
        3. KFeC6N6
        4. K4FeCN6
        5. K4FeC6N

 

 

 

 

 

K42.4 Fe15.2 C19.5 N22.8

K42.4/39.1 Fe15.2/55.8 C19.5/12.0 N22.8/14.00

K1.08 Fe0.272 C1.63 N1.63

K1.08/0.272 Fe0.272/0.272 C1.63/0.272 N1.63/0.272

K4FeC6N6  (K4FeC6N6)

 

  1. You have a 1.212 M stock aqueous solution of Sr(OH)2. What volume of the stock solution (in mL) must be diluted with water to prepare 50.00 mL of a solution that has a hydroxide concentration [OH] = 0.2000 M?
      1. 6.455 mL
      2. 1.294 mL
      3. 2.528 mL
      4. 4.125 mL
      5. 8.173 mL

 

[OH]= 2 [Sr(OH)2] 

 

[Sr(OH)2] after dilution = [OH] after dilution /2 = 0.2000 M/2 = 0.1000 M

 

 

+ H2O

M1 of Sr(OH)2 = 1.212 M M2 of Sr(OH)2 = 0.1000 M

V1 = ? mL V2 = 50.00 mL

 

M1 V1 = M2 V2  V1 = (M2V2)/M1  V1 = 4.125 mL

 

 

  1. When chlorine is added to acetylene, 1,1,2,2-tetrachloroethane is formed:

 

2 Cl2(g) + C2H2(g)  C2H2Cl4(l)

 

How many liters of chlorine gas at STP will be needed to make 75.0 grams of

C2H2Cl4?

a. 10.0 L

b.

20.0 L

      1. 30.0 L
      2. 40.0 L
      3. 50.0 L

 

 

Molar mass of C2H2Cl4 = 167.8 g/mol

(75.0 g C2H2Cl4) (mol C2H2Cl4 /167.8 g C2H2Cl4) = 0.447 mol C2H2Cl4

Moles of Cl2 gas needed = 2 x 0.447 = 0.894

PV = nRT 

V = (nRT)/P = [(0.894 mol Cl2) x (0. 0.0821 L atm mol-1 K-1) x (273 K)] /1 atm =

= 20.0 L

 

 

  1. Consider an electron moving at 1 x 106 m/s. Find its wavelength in angstroms. a. 1 Å
      1. 3 Å
      2. 5 Å

d.

7 Å

e. 9 Å

 

 

h

λ=

mυ

 

 

word image 698

 

 

 

  1. The energy of a photon is used to completely remove the electron of a hydrogen atom. What is the wavelength of the photon in nanometers if the electron is removed from its second orbit?

 

a)

365 nm

      1. 410 nm
      2. 434 nm
      3. 486 nm
      4. 656 nm

 

 

En = -RH (Z2/n2) RH = 2.179 x 10 -18 J

 

Z = 1 and n =2

 

E2 = -(2.179 x 10 -18 J) (12/22)

 

E2 = – 5.448 x 10 -19

J

E = hν = hc/λ  λ = hc/E

 

λ = (6.626×10-34 Js)( 3.00 x 108 m s-1)/(5.448 x 10 -19 J) = 3.65 x 10 -7 m = 365 nm

 

 

 

 

  1. (6 points) Fill in the blanks with the correct ground state electron configuration (noble gas configuration) for the given atom or the atom for the given ground state electron configuration of the neutral atom.

 

Ato

m

Electron Configuration

Sn

 

[Kr] 5s24d10 5p2

 

Cu

 

[Ar] 3d10 4s1

 

 

  1. (42 points) Fill in the following table (see the ClF3 example).

For the determination of formal charge(s) consider the Lewis formula with the smallest formal charges (lowest energy).

 

 

Molecule

Electron-Gr

Geometr

(central at

oup

y om)

Molecular Geometry

Ideal

Bond

Angles

Polar or

Nonpolar

Molecule

Hybridization

 

(central

atom)

Formal

Charge(s)

of the

underlined element

Oxidation

Number of the

underlined element

ClF3

TrigonalBipyrami

dal

T-shape

90º,180º

polar

sp3d

0

-1

CCl2Br2

Tetrahedr

al

Tetrahedral

109.5º

Polar

sp3

0

+4

PCl3

Tetrahedr

al

Trigonal Pyramidal

109.5º

Polar

sp3

0

+3

CO2

Linear

 

Linear

180º

 

sp

0

+4

Nonpolar

 

H2O

Tetrahedr

al

Bent

109.5º

Polar

sp3

0

-2

XeF4

Octahedr

al

Square-planar

 

90º, 180º

 

 

sp3d2

0

+4

Nonpolar

 

 

SF4

 

Trigonal-

 

 

Seesaw

 

90º, 180º

 

Polar

sp3d

0

+4

 

Bipyramid

al

 

 

120º

 

 

 

 

 

 

 

 

 

Partial Credit. Show ALL Your Work. Explain.

 

22. (10 points)

If 10.0 g of sodium hydrogen carbonate is added to 450 mL of a 0.155 M aqueous solution of copper(II) nitrate and left to react according to the reaction:

 

2NaHCO3(aq) + Cu(NO3)2(aq)  CuCO3(s) + 2NaNO3(aq) + H2O(l) + CO2(g)

 

How many liters of carbon dioxide are produced at STP?

 

Moles of Cu(NO3)2 = (0.450 L) x (0.155 mol/L) = 0.06975 mol

Moles of NaHCO3 = (10.0 g)/(84.01 g/mol) = 0.119 mol

Moles of NaHCO3 < 2 times moles of Cu(NO3)2 so NaHCO3 is the limiting reactant

Moles of CO2 = ½ moles of NaHCO3 = 0.119 mol /2 = 0.0595 mol

V = nRT/P = (0.0595 mol x 0.0821 L atm mol-1 K-1 x 273 K)/1 atm = 1.33 L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V =

1.33 L

 

 

 

 

23. (10 points)

A civil engineer wants to reduce odors at a wastewater treatment plant by adding hydrogen peroxide to the sewage. The hydrogen peroxide is delivered as 50% by mass solution, but for maintenance and safety issues, the H2O2 is diluted to a 3% by mass solution. If the engineer needs 20.0 gallons of the 3% by mass aqueous solution of H2O2, how many gallons of water does the engineer need to add to the 50% solution?

(Density of 50% solution = 1.197 g/mL; density of 3% solution = 1.015 g/mL)

 

50% by mass = 50 g H2O2 per 100 g of aqueous solution

 

Change both mass percents to molarity (mol/L) using density and molar mass.

 

(50.0 g pure H2O2 /100.0 g aq. solution) x (1.197 g solution/1 mL solution)

x

(1000 mL/L)x (1 mol H2O2 / 34.01 g H2O2) = 17.59 mol H2O2 / L solution = 17.6 M

 

(3.0 g pure H2O2 / 100.0 g aq. solution) x (1.015 g solution / mL solution) x

(1000 mL / L) x (1 mol H2O2 / 34.01 g H2O2) = 0.895 mol / L = 0.90 M

 

 

Using the dilution formula

 

:

MiVi = MfVf  (17.6 M) Vi = (0.90 M)(20.0 gal)  Vi = 1.0 ga

l

Volume of water to add = Vf – Vi = 20.0 gal – 1.0 gal = 19 gal of water.

Volume of water =

19 gal

 

 

  1. (6 points) Consider the molecule of ozone, O3.
    1. Draw its two resonance Lewis electron dot diagrams.
    2. What is the electron group geometry of its central atom?
    3. What is its molecular geometry?
    4. What is the hybridization of its central atom?
    5. What is the ideal bond angle?
    6. Is it polar or nonpolar molecule?

 

word image 2258

A)

 

 

 

 

 

B)

Trigonal planar

C)

Ben

t

D)

sp

2

 

E)

120

o

F)

Pola

r

  1. (30 points) Complete the MO energy level diagram for the cation B2+.
    1. Fill in the electrons using arrows (↑ and/or ↓) for the atomic and molecular orbitals.
    2. Designate all the energy levels (i.e., σ2s*, π2p, σ2s, etc.)
    3. Calculate the bond order for B2+.
    4. Is B2+ a paramagnetic or diamagnetic chemical species?

 

 

a. and b.

 

 

E

σ

2

p

*

 

π

2

p

*

 

 

2

s

 

2

s

 

2

p

 

2

p

 

σ

2

s

 

σ

2

s

*

 

π

2

p

 

σ

2

p

 

↑↓

 

↑↓

 

↑↓

 

↑↓

 

 

 

B

2

+

 

B

+

 

 

B

 

 

 

 
 

 

 

c. Bond order for B2+ = 0.5

 

 

 

Paramagnetic or Diamagnetic?

Answer:

Paramagnetic

 

 

d

.

 

 

 

 

26. (6 points)

Write the possible four quantum numbers n, l, ml, and ms for each of the following electrons.

 

 

 

2s 4d 5f

 

n = 2 n = 4 n = 5

l = 0 l = 2 l = 3

ml = 0 ml = -2, -1, 0, +1, +2 ml = -3, -2, -1, 0, +1, +2, +3

ms = -1/2 ms = -1/2 ms = +1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

27. (16 points)

Balance the following chemical equation by the half-reaction method and show your work. Next, answer the following questions.

 

CH4O + O2 CO2 + H2O

 

Note! The above chemical equation is very easy to balance just by inspection. You are NOT asked to balance it by the inspection method. Balancing it by the inspection method will result in zero points for credit.

 

CH4O + H2O  CO2 + 6 H+ + 6 e

 

4 H+ + 4 e- + O2  2 H2O

 

Or

 

2 CH4O + 2 H2O  2 CO2 + 12 H+ + 12 e

 

12 H+ + 12 e+ 3 O2  6 H2O

 

2 CH4O + 3 O2  2 CO2 + 4 H2O

 

 

 

  1. The oxidizing chemical species is: O2
  2. The reducing chemical species is: CH4O
  3. The name of the oxidized element is: carbon
  4. The name of the reduced element is: oxygen
  5. The oxidant’s molecular formula is: O2
  6. The reductant’s molecular formula is: CH4O
  7. The name of the element that gains electrons is: oxygen
  8. The name of the element that loses electrons is: carbon
  9. The chemical species that causes oxidation is: O2
  10. The chemical species that causes reduction is: CH4O
  11. The oxidation number of _C__ (give the chemical symbol of the element) increases from _-2__ to _+4_
  12. The oxidation number of _O__ (give the chemical symbol of the element) decreases from _0__ to _-2__

Solubility rules:

 

Compounds which are soluble or mostly soluble:

  • Group 1, NH4+, chlorates, acetates, nitrates
  • Halides (except Pb2+, Ag+, and Hg22+)
  • Sulfates (except Sr2+, Ba2+, Pb2+, and Hg22+) Compounds which are insoluble:
  • Hydroxides, sulfides (except above rule, and sulfides of group 2)
  • Carbonates, phosphates, chromates (except above rules)

Some useful equations and data:

PLEASE NOTE: Important values and equations required for calculations are given with the respective problem. The following may or may not be of any use.

 

 

 

h

λ= mυ

d = PM

RT

Molarity = Moles

Liter

mass (g)

n (moles) =

molar mass (g/mol)

PV = nRT

NA = 6.022 x 1023

h = 6.626×10-34 Js

1 J = 1 kg m2 s-2 = 1 N m

mL∙M = mmol

Ptotal = P1 + P2 + …

TK = ToC + 273

R = 0.0821 L atm mol-1 K-1

1 nm = 10-9 m

c = 3.00 x 108 m s-1

c=νλ

xA = nA / ntot = PA / Ptot

Ephoton = hν

Rate2/Rate1=(M1/M2)1/2

1 atm = 760 mmHg

∆E = RH(1/ni2 – 1/nf2)

 

x⋅∆p = h

1 g = 6.022×1023 amu

1 Å = 10-10 m

d = m V

me = 9.109387×10-28 g

p=m∆υ

Actual Yield

% Yield = x100

Theoretical Yield

En =−RH Zn22

RH = 2.179 x 10 -18 J

 

 

 

 

 

 

 

 

 

 

 

 

 

Potentially Useful Information

(You may remove this page for ease of access)

 

 

 

Key

word image 699

 

 

 

57 La

138.9

1.10

58 Ce

140.1

1.12

59 Pr

140.9

1.13

60

Nd

144.2

1.14

61

Pm

(145)

1.13

62

Sm

150.4

1.17

63 Eu

152.0

1.2

64 Gd

157.3

1.20

65 Tb

158.9

1.2

66 Dy

162.5

1.22

67 Ho

164.9

1.23

68 Er

167.3

1.24

69

Tm

168.9

1.25

70 Yb

173.0

1.1

 

89 Ac

(227)

1.1

90 Th

232.0

1.3

91 Pa

(231)

1.5

92 U

238.0

1.38

93

Np

(237)

1.36

94 Pu

(244)

1.28

95

Am

(243)

1.3

96

Cm

(247)

1.3

97 Bk

(247)

1.3

98 Cf

(251)

1.3

99 Es

(252)

1.3

100

Fm

(257)

1.3

101

Md

(258)

1.3

102

No

(259)

1.3

 

 

 

 

 

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