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

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

c) should be the answer.