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3 years ago

Given the equation representing a system at equilibrium in a sealed, rigid container:

Chemistry

2 answers:

pantera1 [17]3 years ago

8 0

Answer: 1) HI to increase

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)+energy$

As the given reaction is exothermic , on increasing the tempearture, the equilibrium will shift in the direction where decrease in tempearture is taking place.So, the equilibrium will shift in the left direction. i.e. towards reactants. Thus amount of $HI$ will increase.

LenaWriter [7]3 years ago

4 0

Answer:

Choice 1. "HI to increase".

Explanation:

I found out the hard way.

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marusya05 [52]

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A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions

Mars2501 [29]

Answer:

Qm = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

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Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

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3 years ago

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Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

tensa zangetsu [6.8K]

Answer:

$[SO_2Cl_2] = 0.09983 M$

Explanation:

Write the balance chemical equation ,

$SO_2Cl_2((g) = SO_2(g) + Cl_2(g)$

initial concenration of $SO_2Cl_2((g) =0.1M$

lets assume that degree of dissociation= $\alpha$

concenration of each component at equilibrium:

$[SO_2Cl_2] = 0.1-0.1\alpha$

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$[Cl_2] = 0.1\alpha$

$Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}$

$Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}$

as $\alpha$ is very small then we can neglect $1-\alpha$

therefore ,

$Kc ={0.1\alpha \times \alpha}$

$\alpha =\sqrt{\frac{Kc}{0.1}}$

$\alpha = 1.73 \times 10^{-3}$

Eqilibrium concenration of $[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173$

$[SO_2Cl_2] = 0.09983 M$

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Answer:

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