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Naily [24]
4 years ago
14

A solution is created by dissolving 11.5 grams of ammonium chloride in enough water to make 255 mL of solution. How many moles o

f ammonium chloride are present in the resulting solution
Chemistry
1 answer:
Nata [24]4 years ago
3 0

Answer:

<h2>The no. of moles present in one liter solution is 0.843</h2>

Explanation:

Mass of ammonium chloride  = 11.5 gm

Molar mass  of ammonium chloride  = 53.491 gm

No. of moles

N = \frac{mass}{molar \ mass}

N = \frac{11.5}{53.491}

N = 0.215 moles

Molarity=  \frac{N}{volume}

molarity = \frac{0.215}{0.255}

M = 0.843 M

Thus the no. of moles present in one liter solution is 0.843

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Free_Kalibri [48]

Answer:

3.2 \div 22.4 =  \frac{1}{7} mol \: no

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Explanation:

NO is limit reactant so:

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3 0
3 years ago
A+gas+at+250k+and+15atm+has+a+molar+volume+12%+less+than+the+perfect+gas+law+character. +calculate+the+compressibility+factor
shutvik [7]

The compression factor based on the information given is 0.083.

<h3>How to calculate the compression factor?</h3>

The way to calculate the compression factor goes this:

V(ideal) = RT/P

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V(real) will be:

= V(ideal)/12

= 1.368/12 = 0.114

The compression factor will be:

= 1.368/(12 × 1.368)

= 0.083

Therefore, the compression factor is 0.083.

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2 years ago
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