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6a. 0.898 mol sample of hydrogen gas at a temperature of 20.0 °C is found to occupy a volume of 24.5 liters. The pressure of this gas sample is
mm Hg.

6b. A sample of helium gas collected at a pressure of 991 mm Hg and a temperature of 300 K has a mass of 2.82 grams. The volume of the sample is
L.

8a. What volume is occupied by a 0.814 mol sample of xenon gas at a temperature of 0 °C and a pressure of 1 atm?

8b. A 1.21 mol sample of oxygen gas occupies a volume of
L at STP.

9b.What volume of carbon dioxide is produced when 7.95 liters of butane (C4H10) react according to the following reaction? (All gases are at the same temperature and pressure.)

butane (C4H10)(g) + oxygen(g) carbon dioxide(g) + water(g)

liters carbon dioxide

10b. What volume of carbon dioxide is produced when 0.504 mol of calcium carbonate reacts completely according to the following reaction at 0°C and 1 atm?

calcium carbonate ( s ) calcium oxide ( s ) + carbon dioxide ( g )

liters carbon dioxide

Respuesta :

pstnonsonjoku

Gas law refers to the relationship between the temperature, volume and pressure of a gas.

What is gas laws?

The term gas law refers to the relationship between the temperature, volume and pressure of a gas.

6a) PV =nRT

P = nRT/V

P =  0.898 * 0.082 * 293/24.5

P = 0.88 atm

6b)  Number of  moles of helium = 2.82 g/4 g/mol = 0.705 moles

Pressure =  991 mm Hg or 1.3 atm

V = nRT/P

V = 0.705 * 0.082 * 300/1.3

V = 13.3 L

8a) V = nRT/P

V = 0.814 * 0.082 * 273/1

V = 18.22 L

8b) 1 mole of oxygen occupies 22.4 L

1.21 mole of oxygen occupies   1.21 mole *  22.4 L/ 1 mole

= 27.1 l

9b) C4H10(g) + 13/2O2 -----> 4CO2 + 5H20

1 moles of butane produces 4 moles of CO2, It then follows that the volume of carbon dioxide produced is 4( 7.95 liters) = 31.8 litres

10b) CaCO3 ------> CO2 + CaO

Since the reaction is 1:1,  0.504 mol of CO2 is produced

PV =nRT

V = 0.504 * 0.082 * 273/1 atm

V = 11.3 L

Learn more about the ideal gas equation:https://brainly.com/question/79835

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