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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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Name the following alkyne.<br> Please help me &lt;3
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Answer:

D. 7-methyl-3-octyne

Explanation:

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2) Now, identify the location of the functional group.

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Further Explanation:

A) This option is incorrect as there are only 8 carbons in the parent chain. Although there are 9 carbons in total, the 9th carbon is taken care of in '7-methyl'.

B) Location number of the functional group should be as low as possible, so start counting the number of carbons from the left!

C) Since the functional group is an alkyne, the word 'octane' should be 'octyne' instead.

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