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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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30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

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Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

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$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

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