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timurjin [86]
2 years ago
14

Will a double replacement reaction take place if a solution of sodium hydrogen carbonate is added to a solution of sodium chlori

de? Explain.
Chemistry
1 answer:
VARVARA [1.3K]2 years ago
7 0

Answer:

A double replacement reaction <u> do not</u> take place if a solution of sodium hydrogen carbonate is added to a solution of sodium chloride.

NO reaction occur.

Explanation:

Double replacement Reaction  : A reaction in which positive and negative ions of two different compound exchange to form new compounds.

This is also called metathesis reaction.

For Example :

Na_{2}SO_{4} + SrCl_{2}\rightarrow 2NaCl + SrSO_{4}

Here Na+ and Sr2+ has exchanged

Cl- and SO4(2-) has exchanged their places.

So , Requirement of Double displacement reaction is :

Their should be salts of <u>different cation and anion</u>

NaHCO_{3} =  sodium hydrogen carbonate =Na+ cation

and HCO3 - anion

NaCl = Sodium Chloride = Na+ cation and Cl- anion

In both the reacts the cation is : Na+

So No-exchange of ions occur .

No- double displacement reaction.

The reality is , this reaction can't occur at all. It can neither be acid-base nor Redox Reaction

So ,

NaCl + NaHCO_{3}\rightarrow No\ Reaction

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Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H2O).
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Answer:

5.9x10^-2 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Concentration of CO, [CO] = 0.30 M

Concentration of H2, [H2] = 0.10 M

Concentration of H2O, [H2O] = 0.020 M

Equilibrium constant, K = 3.90

Concentration of CH4, [CH4] =..?

Step 2:

The balanced equation for the reaction. This is given below:

CO(g) + 3H2(g) <=> CH4(g) + H2O(g)

Step 3:

Determination of the concentration of CH4.

The expression for equilibrium constant of the above equation is given below:

K = [CH4] [H2O] / [CO] [H2]^3

3.9 = [CH4] x 0.02/ 0.3 x (0.1)^3

Cross multiply to express in linear form

[CH4] x 0.02= 3.9 x 0.3 x (0.1)^3

Divide both side by 0.02

[CH4] = 3.9 x 0.3 x (0.1)^3 /0.02

[CH4] = 5.9x10^-2 M

Therefore, the equilibrium concentration of CH4 is 5.9x10^-2 M

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Answer:

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Explanation:

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