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galben [10]
3 years ago
5

A 49.48-mL sample of an ammonia solution is analyzed by titration with HCl. It took 38.73 mL of 0.0952 M HCl to titrate the ammo

nia. What is the concentration of the original ammonia solution?
Chemistry
1 answer:
Kryger [21]3 years ago
8 0

Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂  

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

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When nitrogen gas reacts with hydrogen gas, ammonia gas is formed. how many grams of hydrogen gas are required to react complete
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The balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of N₂ to H₂ is 1:3
we need to find the moles of N₂, volume of N₂ has been given 
molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP.
if 22.4 L is occupied by 1 mol 
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Answer:

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Explanation:

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Answer:

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Explanation:

Hello there!

In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:

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Best regards!

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