The answer is (3) Cu2O. Copper (I) has an oxidation state of +1 (that's what the "I" indicates). You can also think of this as copper (I) having a charge of +1. Oxygen has an oxidation state of -2 (that's just a rule you have to know), and you can think of it as oxygen having a charge of -2. You need oxidation numbers in a neutral compound to add up to 0 (or charges in a neutral compond to add up to 0), so you need two Cu to balance the O, which is Cu2O.

3 0

3 years ago

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Solid sodium hydroxide <br> NaOH<br> decomposes into gaseous water and solid sodium oxide .

kari74 [83]

What about it do you want answered

8 0

3 years ago

In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?

Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In $H_2O$ , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in $Cl^-$ is -1. On product side, the oxidation number of hydrogen in $H_2$ is zero and in $OH^-$ the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in $Cl_2$ is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

8 0

2 years ago