MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
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Chapter 22, Problem 22.17P

(a)

Interpretation Introduction

Interpretation:

The volume of fluorine gas at given pressure and temperature has to b calculated.

Concept introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

   nTPV = RnTPPV = nRT

Where,

    n = molesofgasP = pressureT = temperatureR = gas constant

(a)

Expert Solution
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Explanation of Solution

The number of moles of fluorine and SiF4 is calculated as,

Moles of F =(100. kg Ca5 (PO4)3F)(103gCa5 (PO4)3F1kgCa5 (PO4)3F)(1molCa5 (PO4)3F504.31 gCa5 (PO4)3F)(1molF1molCa5 (PO4)3F)(15%100%) =29.74361 mol FMoles of SiF4 =(29.74361 mol F)(1molSiF44molF)=7.43590 mol SiF4

The volume of fluorine gas is determined using ideal gas equation.

    PV = nRTV=nRTP=(7.43590 mol SiF4)(0.0821L.atmmol.atm)((273 1450)K)1atm =1.1×103L

(b)

Interpretation Introduction

Interpretation:

The volume of drinking water can be fluorinated to given level has to be calculated.

Concept introduction:

1 m3 = 1000L

Density of water = 1.00 g/mL .

(b)

Expert Solution
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Explanation of Solution

The number of moles of Na2SiF6 and mass of fluorine are determined as,

    Moles of Na2SiF6 =(7.43590 mol SiF4)(1molNa2SiF62molSiF4)=3.71795 molNa2SiF6Mass (g) of F-=(3.71795 molNa2SiF6)(6 mol F-1molNa2SiF6)(19gF-1molF-)=4.2x102 g F-

If one gram F– will fluoridate 106 g water to a level of one ppm, then it is to determine how many grams (converted to mL using density, then converted from mL to L to m3) of water can 423.8463 g F fluoridate to the one ppm level.

Necessary conversion factors: 1 m3 = 1000L, density of water = 1.00 g/mL .

Volume=(423.8463 gF-)(106gwater1gF-)(mLwater1gwater)(1cm31mL)(10-2m1cm)3=4.2x102 m3

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Chapter 22 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 22 - Prob. 22.11PCh. 22 - Prob. 22.12PCh. 22 - Prob. 22.13PCh. 22 - Prob. 22.14PCh. 22 - Prob. 22.15PCh. 22 - Prob. 22.16PCh. 22 - Prob. 22.17PCh. 22 - Prob. 22.18PCh. 22 - Define: (a) ore; (b) mineral; (c) gangue; (d)...Ch. 22 - Define: (a) roasting; (b) smelting; (c) flotation;...Ch. 22 - What factors determine which reducing agent is...Ch. 22 - Prob. 22.22PCh. 22 - Prob. 22.23PCh. 22 - Prob. 22.24PCh. 22 - Prob. 22.25PCh. 22 - Prob. 22.26PCh. 22 - Prob. 22.27PCh. 22 - Why is cryolite used in the electrolysis of...Ch. 22 - Prob. 22.29PCh. 22 - How is Le Châtelier’s principle involved in the...Ch. 22 - Elemental Li and Na are prepared by electrolysis...Ch. 22 - A Downs cell operating at 77.0 A produces 31.0 kg...Ch. 22 - Prob. 22.33PCh. 22 - Prob. 22.34PCh. 22 - The last step in the Dow process for the...Ch. 22 - Prob. 22.36PCh. 22 - Prob. 22.37PCh. 22 - Prob. 22.38PCh. 22 - Prob. 22.39PCh. 22 - Prob. 22.40PCh. 22 - Prob. 22.41PCh. 22 - Prob. 22.42PCh. 22 - Prob. 22.43PCh. 22 - Prob. 22.44PCh. 22 - Prob. 22.45PCh. 22 - Prob. 22.46PCh. 22 - Prob. 22.47PCh. 22 - Prob. 22.48PCh. 22 - Prob. 22.49PCh. 22 - Prob. 22.50PCh. 22 - Prob. 22.51PCh. 22 - Prob. 22.52PCh. 22 - Prob. 22.53PCh. 22 - Prob. 22.54PCh. 22 - Prob. 22.55PCh. 22 - Prob. 22.56PCh. 22 - Prob. 22.57PCh. 22 - Prob. 22.58PCh. 22 - Prob. 22.59PCh. 22 - In the production of magnesium, Mg(OH)2 is...Ch. 22 - Prob. 22.61PCh. 22 - Prob. 22.62PCh. 22 - The production of S8 from the H2S(g) found in...Ch. 22 - Prob. 22.64PCh. 22 - Prob. 22.65PCh. 22 - Prob. 22.66PCh. 22 - Prob. 22.67PCh. 22 - Prob. 22.68PCh. 22 - Prob. 22.69PCh. 22 - Prob. 22.70PCh. 22 - Prob. 22.71PCh. 22 - Prob. 22.72PCh. 22 - Prob. 22.73PCh. 22 - Prob. 22.74PCh. 22 - Prob. 22.75PCh. 22 - Prob. 22.76PCh. 22 - Prob. 22.77PCh. 22 - Prob. 22.78PCh. 22 - Prob. 22.79PCh. 22 - Prob. 22.80PCh. 22 - Prob. 22.81PCh. 22 - Prob. 22.82PCh. 22 - Prob. 22.83PCh. 22 - Prob. 22.84PCh. 22 - Prob. 22.85PCh. 22 - Prob. 22.86P
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