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3 years ago

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Charra [1.4K]3 years ago

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A 143.1 g sample of a compound contains 53.4 g of carbon, 16.9 g of hydrogen, 43 g of nitrogen, and some amount of oxygen. what

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A non-stoichiometric compound is a compound that cannot be represented by a small whole-number ratio of atoms, usually because o

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<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38

Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
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</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

7 0

3 years ago

If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at 127oC in an evacuated

zhuklara [117]

Answer : The the partial pressure of the nitrogen gas is 0.981 atm.

The total pressure in the tank is 2.94 atm.

Explanation :

The balanced chemical reaction will be:

$(CH_3)_2N_2H_2(l)+2N_4O_4(l)\rightarrow 3N_2(g)+4H_2O(g)+2CO_2(g)$

First we have to calculate the moles of dimethylhydrazine.

Mass of dimethylhydrazine = 150 g

Molar mass of dimethylhydrazine =60.104 g/mole

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

$\text{Moles of dimethylhydrazine}=\frac{150g}{60.104g/mole}=2.49mole$

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.

8 0

3 years ago