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Mumz [18]
3 years ago
7

A solution is created by dissolving 11.0 grams of ammonium chloride in enough water to make 305 mL of solution. How many moles o

f ammonium chloride are present in the resulting solution?
Chemistry
1 answer:
vredina [299]3 years ago
4 0
Reacting mass=11.0
molar mass =53.45
moles=reacting Mass /molar mass
11.0/53.45=0.21moles
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Hello!
blondinia [14]

Answer:

[CO]=[Cl_2]=0.01436M

[COCl_2]=0.00064M

Explanation:

Hello there!

In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:

K=\frac{[CO][Cl_2]}{[COCl_2]}

Which can be written in terms of x, according to the ICE table:

0.32=\frac{x^2}{0.015M-x}

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:

[CO]=[Cl_2]=0.01436M

[COCl_2]=0.015M-0.01436M=0.00064M

Regards!

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