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Butoxors [25]
3 years ago
8

What volume of 1.75 M hydrochloric acid (HCl aq) must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid?

Chemistry
1 answer:
Airida [17]3 years ago
7 0

Answer:

V_1=0.0714L

Explanation:

Hello there!

In this case, since we need to dilute the 1.75-M HCl, and the total number of moles remain unchanged, we can write:

n_1=n_2

And in terms of volume and concentration:

C_1V_1=C_2V_2

Thus, we can solve for the volume of the concentrated HCl as shown below:

V_1=\frac{C_2V_2}{C_1}

Therefore, we plug in the data to get:

V_1=\frac{0.250M*0.500L}{1.75 M}\\\\V_1=0.0714L

Best regards!

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Answer:

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Draw the major resonance contributors for the enamine shown. do not draw resonance and curved arrows. include formal charges and
Makovka662 [10]

<em>Answer:</em>

There are two major resonance contributors for the en-amine. One contributor have no formal charges, only have one lone pair at N while the other one has positive and negative formal charges.

<em>Explanation:</em>

An en-amine is formed by the condensation reactions of aldehydes or ketone with secondary amine.

The contributor<em> </em>I have no formal charges. It has only one lone pair at Nitrogen atom.

The contributor II has +1 formal charges at N, and -1 formal charges at α Carbon.

Please see attachment. The resonance contributor are given.


Download docx
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