Aballoon contains 12.88 g of nnitrogene gas (n2). how many nitrogene molecules are in the balloon?

Molar mass N₂ = 14 x 2 = 28.0 g/mol

number of moles: 12.88 / 28.0 => 0.46 moles of N₂

number molecules N₂ :

1 mole 6.02 x 10²³ molecules
0.46 moles ? ( molecules )

molecules = ( 0.46 ) x ( 6.02 x 10²³ ) / 1

= 2.7692 x 10²³ molecules of N₂

hope this helps!

Multiply by avagadro numberr
1.2× 6.022×10²³ = 7.2264×10²³

We can first find the number of moles of nitrogen gas in the container using th ideal gas law equation 
PV = nRT
where P - pressure - 112 100 Pa
V - volume - 8.86 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 30.8 °C + 273.15 = 303.95 K
substituting these values in the equation 
112 100 Pa x 8.86 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 303.95 K
n = 0.393 mol 
number of N₂ moles are 0.393 mol

In 1 mol of N₂ there are 6.022 x 10²³ molecules of N₂
therefore in 0.393 mol - 6.022 x 10²³ mol⁻¹ x 0.393 mol = 2.37 x 10²³ 
there are 2.37 x 10²³ molecules of N₂

We will use this law PV = nRT
we have to convert pressure from kPa to atm
1 atm = 101.325 kPa
? atm = 112.1 kPa
P = 1.11 atm
T = 30.8 °C + 273.15 = 303.95 K
V = 8.86 L
R = 0.088205 atm. L / mol . Kelvin
n = [tex]\frac{PV}{RT} = \frac{1.11 * 8.86}{0.08205 * 303.95}[/tex] = 0.39 mol
number of N₂ molecules = number of moles * Avogadro's number
                                        = 0.39 * (6.022 * 10²³) = 2.37 x 10²³ molecules