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aivan3 [116]
3 years ago
8

Can someone PLEASE help me! I have missed a few days of school due to sickness and my teacher is making me make up a lab! She ga

ve me some things to study, but Chemistry is NOT one of my strong suits by any means! I only have 3 questions! I will post them separately to save room. Here is the first.
1. List at least two chemical reactions that result in a yellow, orange, or red precipitate. For these reactions list the possible chemical name of the precipitate. What do these reactions (and the others similar to them) have in common?
Chemistry
1 answer:
OLEGan [10]3 years ago
3 0
<span>2,4-dinitrophenylhydrazine (Brady’s reagent) is an important reagent related to hydrazine. Most aldehydes and ketones very readily with this reagent to give the yellow orange and red precipitates of 2,4-dinitrophenylhydrazones.
Unconjugated aldehydes and ketones give precipitates toward the yellow while conjugated compound tend to be deeper colour of red.
The conversion of aldehydes and ketones into hydrazone is an example of the addition-elimination reaction occurring at the unsaturated carbonyl group.</span>
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Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

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i could give you a scientist just try albert Einstein

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