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Naddik [55]
3 years ago
13

Solid solution are called:

Chemistry
1 answer:
forsale [732]3 years ago
7 0

Answer:

Chemical solution in solid form; whose solvent's crystal structure is not altered by solute

Explanation:

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For the reaction below, Kp 5 1.16 at 800.8C. CaCO3(s) 34 CaO(s) 1 CO2(g) If a 20.0-g sample of CaCO3 is put into a 10.0-L contai
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Answer:

The mass percentage of calcium carbonated reacted is 2.5%.

Explanation:

The reaction is:

CaCO_{3}(s)--->CaO(s)+CO_{2}(g)

Thus the Kp of the equilibrium will be:

Kp = partial pressure of carbon dioxide [as the other are solid]

Moles of calcium carbonate initially present = \frac{mass}{molarmass}=\frac{20}{100}=0.2

Let us apply ICE table to the equilibrium given:

                        CaCO_{3}(s)--->CaO(s)+CO_{2}(g)

Initial                       0.2                       0          0

Change                 -x                            +x        +x

Equilibrium           0.2-x                         x          x

Kp = partial pressure of carbon dioxide

Kp = Kc(RT)ⁿ

where n = difference in the number of moles of gaseous products and reactants

for given reaction n = 1

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T = temperature = 800 ⁰C = 1073 K

Putting values

Kc =\frac{Kp}{RT}=\frac{1.16}{8.314X1073}=1.3X10^{-4}

Kc = \frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}

1.3X10^{-4}(0.2-x)=x^{2}

x^{2} = 0.26X10^{-4}-1.3X10^{-4}x

On calculating

x =  0.005

where x = the moles of calcium carbonate dissociated or reacted.

Percentage of the moles or mass reacted = \frac{molesreacted X100}{initialmoles}=\frac{0.005X100}{0.2}=2.5%

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