Answer : The moles of methane gas could be, 
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
![[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%28%5Cfrac%7Bn_1%7D%7Bt_1%7D%29%7D%7B%28%5Cfrac%7Bn_2%7D%7Bt_2%7D%29%7D%5D%3D%5Csqrt%7B%5Cfrac%7BM_2%7D%7BM_1%7D%7D)
where,
= rate of effusion of fluorine gas
= rate of effusion of methane gas
= moles of fluorine gas = 
= moles of methane gas = ?
= time = 12.3 min (as per question)
= molar mass of fluorine gas = 38 g/mole
= molar mass of methane gas = 16 g/mole
Now put all the given values in the above formula 1, we get:
![[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%28%5Cfrac%7B5.13%5Ctimes%2010%5E%7B-3%7Dmol%7D%7B12.3min%7D%29%7D%7B%28%5Cfrac%7Bn_2%7D%7B12.3min%7D%29%7D%5D%3D%5Csqrt%7B%5Cfrac%7B16g%2Fmole%7D%7B38g%2Fmole%7D%7D)
Therefore, the moles of methane gas could be,
Here we have to compare the state of helium gas at STP and high temperature and low pressure.
At STP (standard condition of temperature and pressure) i.e. 273K temperature and 1 bar pressure. At STP helium gas will behave as a real gas.
At higher temperature and low pressure Helium will behave as an ideal gas.
The ideal gas conditions are developed on taking into account two factors: (i) the gas molecules are point of mass and having no volume. (ii) there is no existence of force of attraction between the molecules.
The deviation from ideal gas to the real gas depends upon the van der waals' interaction between the gas molecules. Now in low pressure and high temperature, we can ignore the volume and also the inter-molecular force of attraction. Thus the gas sample can behaved as ideal gas.
But at elevated pressure and low temperature i.e. STP the assumptions are not valid and it will behave as real gas.
Answer:
441.28 g Oxygen
Explanation:
- The combustion of hydrogen gives water as the product.
- The equation for the reaction is;
2H₂(g) + O₂(g) → 2H₂O(l)
Mass of hydrogen = 55.6 g
Number of moles of hydrogen
Moles = Mass/Molar mass
= 55.6 g ÷ 2.016 g/mol
= 27.8 moles
The mole ratio of Hydrogen to Oxygen is 2:1
Therefore;
Number of moles of oxygen = 27.5794 moles ÷ 2
= 13.790 moles
Mass of oxygen gas will therefore be;
Mass = Number of moles × Molar mass
Molar mass of oxygen gas is 32 g/mol
Mass = 13.790 moles × 32 g/mol
<h3> = 441.28 g</h3><h3>Alternatively:</h3>
Mass of hydrogen + mass of oxygen = Mass of water
Therefore;
Mass of oxygen = Mass of water - mass of hydrogen
= 497 g - 55.6 g
<h3> = 441.4 g </h3>
Molarmass of <span>Zn(C2H3O2)4 is 301.5561 g/mol
moles of </span>
<span>Zn(C2H3O2)4
= 62 g * 1 mol/(</span><span>301.5561 g) = 0.2056 mol
concentration = moles / volume
concentration * volume = moles
volume = moles / concentration
volume = 0.2056 mol / 1.5 M
volume = 0.13706 L
volume = 137 mL
volume = 140 mL
</span>
