PartA: PV = nRT n = PV/RT = 0.920 x 2.12 / (0.082 x 312.15) = 0.0762 moles Mass

Question

PartA:

PV = nRT

n = PV/RT = 0.920 x 2.12 / (0.082 x 312.15) = 0.0762 moles

Mass of NO2 = n x molar mass = 0.0762 x 46.01 = 3.506 g

Mass of CaCl2 = 49.6 g

Molar mass of CaCl2 = 110.98 g/mol

Number of moles of CaCl2 = mass / molar mass = 0.4469

Volume of solution = 600.0 mL = 0.600 L

Conentration of CaCl2 in mole/L = 0.4469/0.600 = 0.7448 mole/L

Conentration of Ca2+ = 0.7448 mole/L

Conentration of Cl2 = 2 x 0.7448 mole/L = 1.4897 mole/L

P = 735 mmH = 735/760 = 0.967 atm

V = 0.115 L

T = 20 + 273.15 K = 293.15 K

n = PV/RT = 0.967 x 0.115/(0.082 x 293.15) = 4.63 x 10-3 mole

Mass = n x molar mass = 4.63 x 10-3 x 44.01 = 0.304 g

Explanation / Answer

Part A Calculate the mass of nitrogen dioxide gas occupying a volume of 2.12 L at 39 Cand 0 920 atm pressure. Express your answer with the appropriate units. IA mNo2 = Value Units Submit My Answers Give Up

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