To find the number of moles of gas we can use the ideal gas law equation, we dont need to use the mass of gas given as we only have to find the number of moles
PV = nRT
P - pressure - 300.0 kPa
V - volume - 25.0 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 27 °C + 273 = 300 K
substituting these values in the equation
300.0 kPa x 25.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 300 K
n = 3.01 mol
number of mols of gas - 3.01 mol
Answer;
= 18.24
Explanation;
The ratio of N and O in the formula NO2 IS 1:2
Mass of nitrogen gas is 0.500 g
Moles of nitrogen will be;
= 0.500/16 = 0.03125 moles
Therefore;
The moles of Oxygen from the ratio will be;
= 0.03125 × 2 = 0.0625 moles
But; 0.0625 moles is equal to 1.140 g of Oxygen
The atomic number (mass in 1 mole) will be;
= 1.140 /0.0625
= 18.24
Thus the atomic number of Oxygen from the data is 18.24
As number of gaseous moles in reactant and prodict are same that is 4
So No change will occur
<span>Answer:
Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass.
râšM = constant
Therefore for two gases the ratio rates is given by:
r1 / r2 = âš(M2 / M1)
For Cl2 and F2:
r(Cl2) / r(F2) = âš{(37.9968)/(70.906)}
= 0.732 (to 3.s.f.)</span>
The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
