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4 years ago

Hydrogen bonds:__________

Chemistry

1 answer:

Elina [12.6K]4 years ago

5 0

Answer:

Option e.

Explanation:

The option 'e' fact about Hydrogen bonds are correct, that is; ''form weak interactions but can provide structural stability when many are found in a single molecule''.

So, what is this Hydrogen bond?

Hydrogen bonds is a kind of strong dipole- dipole attractions. Hydrogen bonding occurs as a result of the bonding between hydrogen and strongly electronegative atoms for example oxygen, Fluorine and so on.

Since we now know what Hydrogen bond is, let us take a look at the options.

(a) for the first option, we can see from above meaning of Hydrogen bonds that Hydrogen bonds occur between Hydrogen and a strong Electronegative atom. So, this option is wrong.

(b). Option 'b' is also wrong Because Hydrogen bonds does not occur between hydrogen and oxygen atoms ONLY but also with other strong Electronegative atoms such as oxygen.

(c). Option 'c' is wrong because between a strong and not only between a weak electronegative atom and hydrogen.

(d). Option 'd' is also not correct.

(e). Option 'e' is correct. Hydrogen bonds contribute to the effect of boiling and melting point of substance, solubilty, dimerization, strength of bonds, shape of molecules and so on

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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 half-reactions:

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

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

With this in mind we can add the electrons:

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

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

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 flip the reduction potential value:

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

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

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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