A bicycle tire holds 1.50 L of air at 5atm and 20.0 °C. How many moles of air is this? If the average mass of air is 29.0 g/mol, what is the mass of air in the tire?​

9.05 g

Explanation:

PV=nRT

Use the ideal gas equation. Substitute values.

P = 5 atm

V = 1.50 L

n = ?

R (gas constant) = 0.08206 L-atm/mol-K

T = 20.0°C

*Always convert °C to K.

T = 20.0° + 273 = 293K

Substitute values.

(5 atm)(1.50 L) = n(0.08206 L-atm/mol-K)(293K)

n = (5 atm)(1.50 L) / (0.08206 L-atm/mol-K)(293K)

n = 0.3119335... mol

Convert to grams with the given average mass of air.

0.3119335... mol x (29.0 g/1 mol) = 9.05 g

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.