Application of Gas Laws: Show all calculations! How many grams of NO_2, nitrogen

Question

Application of Gas Laws: Show all calculations! How many grams of NO_2, nitrogen dioxide, are contained in 0.500 L of the gas at STP? Calculate the density of NO_2 gas. in grams per liter, at 110degree C and 12 atm. Calculate the molar mass of a gas if 0.125 g of the gas occupies 93.3 mL at STP. On a spring morning (15degree C) you fill your tires to a pressure of 2.25 atmospheres. As you ride along, the tire heats up to 45degree C from the friction on the road. What is the pressure in your tires now? What volume of H_2 is formed at STP when 6.0 g of Al is treated with excess NaOH? 2NaOH + 2Al + 6H_2O rightarrow 2NaAl(OH)_4 + 3H_2(g) Application of Gas Law What is the effect on the volume of an ideal gas when the temperature is decreased from 700K to 350K at constant pressure? What is the effect on the volume of an ideal gas when both the pressure and temperature arc reduced by half? What is the effect on the volume of an ideal gas when the pressure is changcd from 722 torr to 0.950 atm. and the temperature is changcd from 5degree C to 273 K? How many grams of phosphorus react with 35.5 L of O: at STP to form tetraphosphorus oxide, according to the following reaction equation? Pa (s) + 5 O_2 (g) rightarrow P_4O_10 (s) Ammonia decomposes to nitrogen and hydrogen when in the presence of heated steel wool. What is the final volume of the gas when 45 g of ammonia decomposes at 150degree C and 1 atm pressure. Write a balanced equation before you start your calculations! Earth's atmosphere consist of 78% N_2 and 20% O_2, and other gases. What is the partial pressure of N_2 and O_2 in torr, when the atmospheric pressure is 1 atm? Ammonia is completely decomposed to nitrogen and hydrogen, as in question 9b). At the end of the experiment the pressure of the gas mixture is 866 mmHg. Calculate the partial pressures of N_2 and H_2.

Explanation / Answer

a)

PV = nRT

n = PV/(RT) = 1*0.5/0.082*273

n = 0.0223353 mol

mass = mol*MW = 0.0223353*46.0055 = 1.0274 g

b)

T = 110 = 383K

P = 12 atm

MW = 46.0055

PV = nRT

PV = mass/MW*RT

D = mass/V then

D = P*MW/(RT)

D = (12)(46.0055)/(0.082*383) = 17.5783 g/L

D = 17.5783 g/L

c)

From:

PV = mass/MW*RT

MW = mass*RT/(PV)

MW = 0.125*0.082*273/(1*0.0933) = 29.991 g/mol

d)

T1 = 15°C = 288 K

P1 = 2.25 atm

T2 = 45°C = 318K

P2 = solve

P1/T1 = P2/T2

P2 = P1*(t2/T1) = 2.25 *318/288 = 2.484375 atm

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