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2 years ago

Give the o.N. Of each of the elements magnesium and oxygen in the reactants and in the products 2Mg + O2=2MgO

Chemistry

1 answer:

vlada-n [284]2 years ago

4 0

Answer:

$2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}$

Explanation:

Hello there!

In this case, according to the rules for the oxidation states in chemical reactions, it is possible to realize that lone elements have 0 and since magnesium is in group 2A, it forms the cation Mg⁺² as it loses electrons and oxygen is in group 6A so it forms the anion O⁻²; therefore resulting oxidation numbers are:

$2Mg^0 + O_2^0\rightarrow2Mg^{2+}O^{2-}$

Best regards!

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Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$

But this is just a supposition.

Regards.

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2 years ago

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3 years ago

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hope this was helpful :)

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3 years ago

1. Oxygen<br> full electron configuration

Arada [10]

Answer:

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