Answer : The final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.
Explanation :
As we know that:

At constant volume and temperature of gas, the pressure will be directly proportional to the number of moles of gas.
The relation between pressure and number of moles of gas will be:

where,
= initial pressure of gas = 24.5 atm
= final pressure of gas = 5.30 atm
= initial number of moles of gas = 1.40 moles
= final number of moles of gas = ?
Now put all the given values in the above expression, we get:


Therefore, the final number of moles of gas that withdrawn from the tank to lower the pressure of the gas must be, 0.301 mol.
1. Berkelium(Berkeley, CA) 2. Dubnium(Dubna, Russia) 3. Darmstaditum (Darmstadt, Germany) 4. Erbium(Ytterby, Sweden) 5. Strontium(Strontian, Scotland) 6. Terbium(Ytterby, Sweden) 7. Yttebium(Ytterby, Sweden) 8. Yttrium(Ytterby, Sweden)
The given formula contains two carbon along with six hydrogen and an oxygen.
The possible isomers from
are ethanol and dimethyl ether.
Ehtanol is an alcohol containing a hydroxyl group whereas dimethyl ether is an ether containing an oxygen between two methyl
The structure of two isomers are:
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol