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Leviafan [203]
3 years ago
7

A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume

both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank
Chemistry
1 answer:
pshichka [43]3 years ago
6 0

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure   = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol

amount of SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol

mole fraction of N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330

mole of fractionSF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669

PV = nRT

P of N_2 O = \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa

mole of SF_4=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa

Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

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Which of the following is true for balancing equations?
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3 years ago
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The equilibrium shifts to produce more reactants.

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If 3.53 g of CuNO, is dissolved in water to make a 0.330 M solution, what is the volume of the solution in milliliters?
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Answer:

84.8 mL

Explanation:

From the question given above, the following data were obtained:

Mass of CuNO₃ = 3.53 g

Molarity of CuNO₃ = 0.330 M

Volume of solution =?

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Therefore, the volume of the solution is 84.8 mL.

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3 years ago
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