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mars1129 [50]
3 years ago
9

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

Chemistry
1 answer:
Oksanka [162]3 years ago
8 0

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

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How to make 100 ml of 0.001 mM solution with 0.0405mM solution?
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Measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water

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To make 100 mL of 0.001 mM solution from 0.0405mM solution, we need to determine the volume of 0.0405mM solution needed. This can be obtained as follow:

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Volume of stock solution needed (V₁) =?

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Divide both side by 0.0405

V₁ = 0.1 / 0.0405

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Therefore, to make 100 mL of 0.001 mM solution from 0.0405mM solution, measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water.

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3 years ago
How many molecules of SO2 are in 2.45 g of SO2? (molar mass=64.06 g/mol). (Hint: you need to use your entire stoichiometry flow
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To what pressure must a gas be compressed in order to get into a 3.00 cubic foot
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To solve this problem, we  apply Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".

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1 and 2 are initial and final states;

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