SOLUTIONS

Homework 1 Chem 1151 Spring 2008 DUE: Feb 12 in Discussion

10 points Chapters 1, 2 and the first part of 3

Before working on this assignment, I strongly suggest that you read the text, study the examples and try the suggested odd-numbered problems and the problems done in class. Show all work for credit. You are encouraged to work with other students, seek help on the material from the TAs, instructor and Tutoring Center although none of us will work these problems out for you. Finally, put your name and section number on your paper.

1. Write a balanced chemical equation for each of these reactions (include states of matter).

a.  Aqueous solids SOLUTIONS of nickel(II) chloride and sodium phosphate react to produce solid nickel(II) phosphate and aqueous sodium chloride.

3NiCl2(aq) + 2Na3PO4(aq) à Ni3(PO4) 2(s) + 6NaCl(aq)

b.  Dinitrogen pentoxide decomposes to form nitrogen dioxide and oxygen. All of the species are gases

2N2O5(g) à 4NO2 (g) + O2 (g) or N2O5(g) à 2NO2 (g) + ½ O2 (g)

c.  Solid calcium fluoride dissolves in liquid sulfuric acid to produce hydrogen fluoride gas and solid calcium sulfate.

CaF2(s) + H2SO4(l) à 2HF(g) + CaSO4(s)

d.  Aqueous solutions of ammonium carbonate and magnesium chloride react to form solid magnesium carbonate and aqueous ammonium chloride.

(NH4)2CO3(aq) + MgCl2(aq) à MgCO3(s) + 2NH4Cl(aq)

2.  Consider a solid platinum sphere of diameter 1.24 cm.

a.  Calculate the volume in cubic cm, in liters, in cubic ft. V = (4/3) πr3.

r = 1.24/2 = 0.62 cm

V = (4/3) π (0.62 cm)3 = 0.998 cm3

0.998 cm3 x (1 L/1000 cm3) = 9.98E-4 L

0.998 cm3 x (1 in/2.45 cm) 3 x (1 ft/12 in) 3 = 3.52E-5 ft3

b.  If the density of platinum is 21.4 g/cm3, how much does the sphere weigh?

Density = mass/volume or mass = density x volume = 21.4 g/cm3 x 0.998 cm3 = 21.4 g

c.  On Feb 1, 2008, the cost of Pt was $1736 per troy oz. How much is the Pt sphere worth? Look at Problem 1.41 for conversion factors.

1 grain = 0.0648 g 24 grains = 1 pw 20 pw = 1 troy oz

21.4 g x (1 grain/0.0648 g) x (1 pw/24 grains) x (1 troy oz/20 pw) x ($1736/troy oz) = $1194

3.  Complete this table

Symbol / # protons / # neutrons / # electrons / Net charge
5726Fe / 26 / 31 / 26 / 0
3517Cl / 17 / 18 / 17 / 0
8838Sr2+ / 38 / 50 / 36 / +2
12852Te2- / 52 / 76 / 54 / -2

4.  Text Problem 3.56

Mass / Mol / # molecules / # atoms
4.24 g / 5.43E-2 / 3.27E22 / 3.92E23
4.04 g / 0.224 mol water / 1.35E23 / 4.05E23
1.98 g / 4.50E-2 / 2.71E22 / 8.13E22
0.297 g / 9.27E-3 / 5.58E21 / 3.35E22 atoms

5.  Copper has two naturally occurring isotopes 63Cu with a mass of 62.930 amu and 65Cu with a mass of 64.928 amu. Calculate the fractional abundance of each isotope.

Atomic Weight = Σ (Ai x Mi)/100 = 63.55 g/mol from PT

63.55 = [A1 x 62.930 + A2 x 64.928]/100

Note: A1 + A2 = 100 or A2 = 100 - A1

Then 63.55 = [A1 x 62.930 + (100 - A1) x 64.928]/100

Solve for A1 to get 137.8 = A1 x 1.998 or A1 = 68.97% and A2 = 31.03%