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2 years ago

Which statement describes the bonds in iron sulfate, FeSO4?

Chemistry

1 answer:

Feliz [49]2 years ago

5 0

___

Regarding the bonds in FesO₄, Fe and S have an ionic bond, while S and O have covalent bonds.

Elements form bonds to increase their stability. The main types of bonds are:

Metallic bonds: they are formed between metals and the electrons are in a delocalized cloud.
Ionic bonds: they are formed between metals (lose electrons) and nonmetals (gain electrons)
Covalent bonds: they are formed between nonmetals, which share electrons.

Regarding the bonds in FesO₄:

Fe is a metal and S a nonmetal, thus they will form ionic bonds.
S and O are both nonmetals, thus they will form covalent bonds.

Regarding the bonds in FesO₄, Fe and S have an ionic bond, while S and O have covalent bonds.

Learn more: brainly.com/question/23882847

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2 years ago

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1. 0.97 V

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

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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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