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Wittaler [7]
4 years ago
8

Acetylating agents such as acetic anhydride react preferentially with primary amines, iodoacetate reacts preferentially with sul

fhydryl groups, and atp-dependent kinases preferentially add a phosphoryl group to side-chain hydroxyl or phenolic −oh groups. which amino acid side chains, or main chain groups, in a polypeptide are most likely to be modified by treatment with: part a acetic anhydride check all that apply. check all that apply. glutamine arginine lysine asparagine proline n-terminal amine
Chemistry
1 answer:
nydimaria [60]4 years ago
8 0

II think thihe answer is to be part of the acid to be honest

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Which of the following is not an example of a molecule?
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A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
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<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

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3 years ago
These are elements that
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