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lesya [120]
3 years ago
15

When 8.70 kJ of thermal energy is added to 2.50 mol of liquid methanol, it vaporizes. Determine the heat of vaporization in kJ/m

ol of methanol.
Please answer quickly, its urgent!
Chemistry
1 answer:
Anni [7]3 years ago
5 0

Answer:

The heat of vaporisation of methanol is "3.48 KJ/Mol"

Explanation:

The amount of heat energy required to convert or transform  1 gram of liquid to vapour is called heat of vaporisation

When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.

Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is =   \frac{8.7}{2.5}KJ

 = 3.48 KJ

So, the heat of vaporization \delta H_{vap} = 3.48KJ/Mol

Therefore, the heat of vaporization of methanol is 3.48KJ/Mol

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If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at 127oC in an evacuated
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The total pressure in the tank is 2.94 atm.

Explanation :

The balanced chemical reaction will be:

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First we have to calculate the moles of dimethylhydrazine.

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\text{Moles of dimethylhydrazine}=\frac{\text{Mass of dimethylhydrazine}}{\text{Molar mass of dimethylhydrazine}}

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Now we have to calculate the moles of N_2 gas.

From the balanced chemical reaction we conclude that,

As, 1 mole of (CH_3)_2N_2H_2 react to give 3 moles of N_2 gas

So, 2.49 mole of (CH_3)_2N_2H_2 react to give 2.49\times 3=7.47 moles of N_2 gas

Now we have to calculate the partial pressure of nitrogen gas.

Using ideal gas equation :

PV=nRT\\\\P_{N_2}=\frac{nRT}{V}

where,

P = Pressure of N_2 gas = ?

V = Volume of N_2 gas = 250 L

n = number of moles  N_2 gas = 7.47 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 127^oC=273+127=400K

Putting values in above equation, we get:

P_{N_2}=\frac{(7.47mole)\times (0.0821L.atm/mol.K)\times 400K}{250L}=0.981atm

Thus, the partial pressure of the nitrogen gas is 0.981 atm.

Now we have to calculate the total pressure in the tank.

Formula used :

P_{N_2}=X_{N_2}\times P_T

P_T=\frac{1}{X_{N_2}}\times P_{N_2}

P_T=\frac{1}{(\frac{n_{N_2}}{n_T})}\times P_{N_2}

P_T=\frac{n_{T}}{n_{N_2}}\times P_{N_2}

where,

P_T = total pressure = ?

P_{N_2} = partial pressure of nitrogen gas = 0.981 atm

n_{N_2} = moles of nitrogen gas = 3 mole  (from the reaction)

n_{T} = total moles of gas = (3+4+2) = 9 mole  (from the reaction)

Now put all the given values in the above formula, we get:

P_T=\frac{9mole}{3mole}\times 0.981atm=2.94atm

Thus, the total pressure in the tank is 2.94 atm.

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