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soldier1979 [14.2K]

4 years ago

13

Significant figures 5.3316 + 6.87 + 37.48

2 answers:

Andre45 [30]4 years ago

6 0

Explanation:

5.3316

37.48

6.87

my answer is 49.6816

Aleksandr-060686 [28]4 years ago

5 0

Answer:

Explanation:

01: 5.3316+6.87+37.48

02: 12.2016+37.48

03: 49.6816

04: 49.68

Answer: 49.68 (Decimals: 2; Significant Figures: 4)

HOPE THIS HELPS..

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The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63x10^-3 at 527 C. Calculate the equilibrium partial

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<u>Answer:</u> The partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

<u>Explanation:</u>

We are given:

$K_c=4.63\times 10^{-3}$

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Relation of $K_p$ with $K_c$ is given by the formula:

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Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $4.63\times 10^{-3}$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

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$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304$

The chemical reaction for the decomposition of phosgene follows the equation:

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

At t = 0 0.760 0 0

At $t=t_{eq}$ 0.760-x x x

The expression for $K_p$ for the given reaction follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

We are given:

$K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x$

Putting values in above equation, we get:

$0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656$

Negative value of 'x' is neglected because partial pressure cannot be negative.

So, the partial pressure for the components at equilibrium are:

$p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm$

Hence, the partial pressure of the $CO,Cl_2\text{ and }COCl_2$ are 0.352 atm, 0.352 atm and 0.408 atm respectively.

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