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Chemistry: Atoms First

Chapter 6

Chemistry: Atoms FirstChapter 6

1.

(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu

3.

(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu

5.

(a) 56.107 amu;
(b) 54.091 amu;
(c) 199.9976 amu;
(d) 97.9950 amu

8.

(a) % N = 82.24%
% H = 17.76%;
(b) % Na = 29.08%
% S = 40.56%
% O = 30.36%;
(c) % Ca2+ = 38.76%

10.

% NH3 = 38.2%

12.

(a) CS2
(b) CH2O

14.

C6H6

16.

Mg3Si2H3O8 (empirical formula), Mg6Si4H6O16 (molecular formula)

18.

C15H15N3

20.

We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.

22.

(a) 0.679 M;
(b) 1.00 M;
(c) 0.06998 M;
(d) 1.75 M;
(e) 0.070 M;
(f) 6.6 M

24.

(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g

26.

(a) 37.0 mol H2SO4;
3.63 ×× 103 g H2SO4;
(b) 3.8 ×× 10−6 mol NaCN;
1.9 ×× 10−4 g NaCN;
(c) 73.2 mol H2CO;
2.20 kg H2CO;
(d) 5.9 ×× 10−7 mol FeSO4;
8.9 ×× 10−5 g FeSO4

28.

(a) Determine the molar mass of KMnO4; determine the number of moles of KMnO4 in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 ×× 10−3 M

30.

(a) 5.04 ×× 10−3 M;
(b) 0.499 M;
(c) 9.92 M;
(d) 1.1 ×× 10−3 M

32.

0.025 M

34.

0.5000 L

36.

1.9 mL

38.

(a) 0.125 M;
(b) 0.04888 M;
(c) 0.206 M;
(e) 0.0056 M

40.

11.9 M

42.

1.6 L

44.

(a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units:
%mass1×mass1=%mass2×mass2%mass1×mass1=%mass2×mass2
This equation can be rearranged to isolate mass1 and the given quantities substituted into this equation.
(b) 58.8 g

46.

114 g

48.

1.75 ×× 10−3 M

50.

95 mg/dL

52.

2.38 ×× 10−4 mol

54.

0.29 mol

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