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Chemistry

2 answers:

Sonbull [250]3 years ago

8 0

Hey How are you?

Explanation:

Nutka1998 [239]3 years ago

3 0

Answer:

The volume is 8 cm³

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0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reache

HACTEHA [7]

Answer:

12.531 grams will be the final mass of calcium carbonate.

Explanation:

$CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$

Partial pressure of carbon dioxide gas at equilibrium = $P=0.220 atm$

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

$PV=nRT$ ( ideal gas equation)

$n=\frac{PV}{RT}=\frac{0.220 atm\times 10.0 L}{0.0821 atm L/mol K\times 385 K}=0.06960 mol$

According to reaction , 1 mole of carbon dioxide gas is formed from 1 mole of calcium carbonate,0.06960 mole of carbon dioxide gas will be obtained from :

$\frac{1}{1}\times 0.06960 mol=0.06960 mol$ calcium carbonate

Moles of calcium carbonate at equilibrium = 0.100 mol - 0.06960 mol = 0.03040 mol

After addition of 0.300 atm of carbon dioxide gas, more amount of calcium carbonate will be be formed.

Here, at the same temperature, the equilibrium pressure of the carbon dioxide gas is 0.220 atm so, entire 0.300 atm of carbon dioxide will get convert to calcium carbonate.

So amount moles of carbon dioxide gas added = moles of calcium carbonate formed after re-establishment of an equilibrium :

Partial pressure of carbon dioxide gas added at equilibrium = $P=0.300 atm$