Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chemistry 107

Chem 107

Hour Exam I
  Name Print KEY
Fall 1996

(100 pts)
  Lab Section (circle): Tu Th

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(10) 1. Complete the following table:

Isotope A Z p n e
205Pb4+  205 82 82 123 78
Cm 239 96 96 143 96
Te2- 132 52 52 80 54

(10) 2. Answer the following questions regarding the periodic table and the elements:

a. What is the charge on Group II-A ions? +2

b. What is the formula for elemental iodine? I2

c. Name an element occurring as a liquid at room temperature. Hg or Br

d. What charge would you expect for an Se ion? -2

e. What is the formula for elemental iron? Fe

f. What is the general name for the Group I-A elements? alkali metals

g. The elements with the lowest average ratios of neutrons to protons are located where on the Periodic Table? near the top

h. Which element has the greatest attraction for electrons? F

i. Which elements are called the halogens? Group VIIA

j. Which element would you expect to give up electrons most readily? Cs or Fr

(8) 3. For each of the following compounds determine whether it is ionic or covalent. Use hi, lo, intermediate rules

 a. TiCl4 Hi (lo, hi)   b. SiO2 Covalent (intermediate, hi)
 c. NH3 Covalent (hi, intermediate)  d. ZnS Ionic (lo, hi)

(8) 4 a. Imagine that you are given a sapphire (Al2O3) ring with a 1.50 carat stone in it. Having just finished chemistry, your immediate reaction of course is to figure out how many atoms of oxygen are in your new possession! 1.000 carat = 200.0 mg. Show your work in discovering this answer.

First want to find moles Al2O3; FW = 2(26.98) + 3(16.00) = 101.96 g/mol

(1.50 carat)(0.2000 g/1.000 carat)/(101.96 g/mol) = 2.942 x 10-3 mol Al2O3

but have 3 atoms O/Al2O3, therefore 3(2.942 x 10-3 mol) = 8.827 x 10-3 mol O

and (8.827 x 10-3 mol O)(6.022 x 1023 atom/mol) = 5.316 x 1021 atom

Checking sig figs = 5.32 x 1021 atom

(2) b. How many formula units of aluminum oxide are there in this stone? Show work for credit.

from (a) above have 2.942 x 10-3 mol Al2O3, therefore

(2.942 x 10-3 mol)(6.022 x 1023 FU/mol) = 1.772 x 1021 FU

checking sig figs   = 1.77 x 1021 FU

(12) 5. Name the compounds below.


a. KOH
Potassium hydroxide
b. Fe
2(SO4)3 Iron(III) sulfate
c. Na
2O Sodium oxide
d. Hg(C
2H3O2)2 Mercury(II) acetate
e. NH
3 Ammonia
f. H
3PO4 Phosphoric acid

(12) 6. Write formulae for each of the following compounds.

a. Hydrogen cyanide HCN

b. Ammonium hydroxide NH4OH

c. Iron(III) chromate Fe2(CrO4)3

d. Silicon tetrachloride SiCl4

e. Plumbic oxide PbO2

f. Hydrochloric acid HCl

(12) 7. Consider the reaction below:

  2 C4H10(g) + 13 02(g) Æ
 8 CO2(g)

 +
10  H2O

a. Balance the equation by placing the proper coefficients and compounds in the spaces provided.

b. How many moles of carbon dioxide may be produced by burning 8.50 grams of butane (C4H10) in air? Show work for credit

MWbutane= 4(12.01) + 10(1.008) = 58.10 g/mol

(8.50 g)/(58.10 g/mol) = 0.1463 mol

from equation have 4 moles CO2/mol butane

therefore 4(0.1463) = 5.850 x10-1 mol

Checking sig figs = 5.85 x10-1 mol CO2

(6) 8. Define each of the following terms:

a. Orbital: a particular electronic energy level; spatial distribution of electron probability density

b. Actinide: any of the elements in the second f group (elements 89 - 102)

c. Mass number: the number of nucleons (protons + neutrons) in an atom

(8) 9. Make each of the conversions below to the proper number of significant figures. Use exponential notation where appropriate!

a. 225.86 nm = 2.2586 x 10-7 m

b. 1.078 x 103 ° C = 1351 K

c. 54678.9 micrograms + 2.3 milligrams = 5.7 x 10-5 kilograms

d. 345.0 cm x 3400 cm x 54.3786 cm = 6.4 x 104liters

(6) 10. a. Write a spectroscopic convention electronic configuration for:

Al: 1s22s22p63s23p1

b. Write a Noble gas core convention electronic configuration for:

Pt: [Xe] 6s24f145d8

c. Write an orbital diagram convention electronic configuration for:

P:

(6) 11. A student attempting to determine the density of ring to see if it was gold first weighed it in air on a triple beam balance, determining its weight to be 12.42 g. She then placed it in a cylinder filled to the 15.00 ml mark and then read a new volume of 16.2 ml. What is the density of this piece of jewelry? Is it gold? (The density of gold is 19.3 g/cm3. Show work!

D = g/mL = g/cm3 = m/V

m = 12.42 g

V = (16.2 - 15.00)mL = 1.20 mL

D = 12.42g/1.2mL = 10.35g/mL

checking sig figs = 1.0 x 101g/cm3

 

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Last modified 15 August 2005