1. How many air molecules are in a 11.5×12.0×10.0 ft room? Assume atmospheric pr
ID: 530042 • Letter: 1
Question
1. How many air molecules are in a 11.5×12.0×10.0 ft room? Assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 C, and ideal behavior.
Volume conversion:There are 28.2 liters in one cubic foot.
2. To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 6.3-L bulb, then filled it with the gas at 1.30 atm and 23.0 C and weighed it again. The difference in mass was 9.5 g . Identify the gas.
Express your answer as a chemical formula.
3.
When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction
CaCO3(s)CaO(s)+CO2(g)
What is the mass of calcium carbonate needed to produce 51.0 L of carbon dioxide at STP?
4. Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is
2C4H10(g)+13O2(g)8CO2(g)+10H2O(l)
At 1.00 atm and 23 C, what is the volume of carbon dioxide formed by the combustion of 4.00 g of butane?
5. A quantity of CO gas occupies a volume of 0.47 Lat 1.5 atm and 290 K . The pressure of the gas is lowered and its temperature is raised until its volume is 2.8 L .
Find the density of the CO under the new conditions.
6. A 48.5-mL sample of gas in a cylinder is warmed from 25 C to 89 C.
What is its volume at the final temperature? (Assume constant pressure.)
7. A balloon contains 0.128 mol of gas and has a volume of 2.84 L .
If an additional 0.164 mol of gas is added to the balloon (at the same temperature and pressure), what will its final volume be?
Explanation / Answer
1.We know that ideal gas equation is PV = nRT
Where
T = Temperature = 20 oC = 20+273 = 293 K
P = pressure = 1.0 atm
n = No . of moles = ?
R = gas constant = 0.0821 L atm / mol - K
V= Volume of the gas = 11.5x12.0x10.0 ft3
= 1380 ft3 x 28.2 L/ft3
= 38916 L
Plug the values we get n = (PV)/(RT)
= 1618 moles
We know that 1 mole of any compound contains avagadro number (6.023x1023) molecules
1618 moles of air contains 1618 x 6.023x1023 =9.74x1026 molecules
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