a.) The formula for dinitrogen tetraoxide is N2O4. How many Grams of nitrogen ar

Question

a.) The formula for dinitrogen tetraoxide is N2O4. How many Grams of nitrogen are present in 1.40 moles of dinitrogen tetraoxide?

b.) How many Moles of oxygen are present in 2.98 grams of dinitrogen tetraoxide?

c.) The formula for silver sulfide is Ag2S. How many Grams of Ag are present in 1.48 moles of silver sulfide?

d.) How many Moles of S^2- ion are present in 1.59 grams of silver sulfide?

e.) What volume is occupied by a 1.13 mol sample of carbon dioxide gas at a temperature of 0 degrees Celsius and a pressure of 1 atm?

f.) The stopcock of connecting a 1.00 L bulb containing methane gas at a pressure of 558 mmHg, and a 1.00 L bulb containing nitrogen gas at a pressure of 674 mmHg, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, what is the final pressure in the system?

g.) It is observed that 18.0 kcal of energy are released when a 32.7 g sample of an unknown liquid condenses. What is the heat of vaporaization of the unknown liquid in cal/g?

Explanation / Answer

(a): From the molecular formulae N2O4 it is clear that 1 mole of N2O4 contains 2 mol N atom.

molar mass of N2O4 = 92.01 g/mol

atomic mass of N atom = 14 g/mol

Hence

1.4 mol N2O4

= 1.4 mol N2O4 x (2 mol N atom / 1 mol N2O4) x (14 g N atom / 1 mol N atom)

= 39.2 g N atom (answer)

(b): 1 mol of N2O4 contians 4 mol of O atoms.

atomic mass of O = 16.0 g/mol

Hence

2.98 g N2O4

= 2.98 g N2O4 x (1 mol N2O4 / 92.01 g N2O4) x ( 4 mol O atom / 1 mol N2O4)

= 0.130 mol oxygen (answer)

(c):

From the molecular formulae Ag2S it is clear that 1 mole of Ag2S contains 2 mol Ag atom.

molar mass of Ag2S = 247.80 g/mol

atomic mass of Ag atom = 107.868 g/mol

Hence

1.48 mol Ag2S

= 1.48 mol Ag2S x (2 mol Ag atom / 1 mol Ag2S) x ( 107.868 g Ag atom / 1 mol Ag atom)

= 319.3 g Ag atom (answer)

(d):

1 mol of Ag2S contians 1 mol of S2- ion

atomic mass of S = 32.06 g/mol

Hence

1.59 g Ag2S

= 1.59 g Ag2S x (1 mol Ag2S / 247.80 g Ag2S) x ( 1 mol S^2- / 1 mol Ag2S)

= 0.00642 mol S^2- (answer)

(e): Given P = 1 atm, n = 1.13 mol, T = 0 DegC = 273 K

We can calculate volume from ideal gas equation

PV = nRT

=> V = nRT/P = (1.13 mol x 0.0821 L.atm.mol-1K-1 x 273 K) / 1 atm

=> V = 25.33 L (answer)

(f): Here moles of gas doesn't change

=> total mol = n(methane) + n(nitrogen)

=> PVt/RT = (558 mmHg x 1.00 L / RT) + (674 mmHg x 1.00 L / RT)

=> P x 2.00 = 558 mmHg + 674 mmHg = 1232 mmHg

=> P = 1232 mmHg / 2.00 = 616 mm Hg (answer)

(g): dHv = 18.0 x 103 cal / 32.7 g = 550.46 cal/g

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