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4 years ago

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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

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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Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

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According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

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$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

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