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2 years ago

If the oxidation state of H is +1 and O is -2, what is the oxidation state of C in C2H4O?

Chemistry

1 answer:

Sedbober [7]2 years ago

8 0

Answer:

-1

Explanation:

According to this question, the oxidation state/number of H and O in C2H4O is +1 and -2 respectively.

The oxidation state of carbon in the compound can be calculated thus:

Where;

x represents the oxidation number of C

C2H4O = 0 (net charge)

x(2) + 1(4) - 2 = 0

2x + 4 - 2 = 0

2x + 2 = 0

2x = -2

Divide both sides by 2

x = -1

The oxidation number of C in C2H4O is -1.

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Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb ( NO 3 ) 2

nikklg [1K]

Answer : The volume of $NH_4I$ solution required is, 2.93 L

The number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

Explanation :

First we have to calculate the initial moles of $Pb(NO_3)_2$ .

$\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}$

$\text{Moles of }Pb(NO_3)_2=0.700M\times 0.945L=0.662mol$

Now we have to calculate the moles of $NH_4I$

The balanced chemical reaction is:

$Pb(NO_3)_2(aq)+2NH_4I(aq)\rightarrow PbI_2(s)+2NH_4NO_3(aq)$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react with 2 moles of $NH_4I$

So, 0.662 mole of $Pb(NO_3)_2$ react with $0.662\times 2=1.32$ moles of $NH_4I$

Now we have to calculate the volume of $NH_4I$

$\text{Volume of }NH_4I=\frac{\text{Moles of }NH_4I}{\text{Concentration of }NH_4I}$

$\text{Volume of }NH_4I=\frac{1.32mol}{0.450mol/L}=2.93L$

Now we have to calculate the moles of $PbI_2$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react to give 1 moles of $PbI_2$

So, 0.662 mole of $Pb(NO_3)_2$ react to give 0.662 moles of $PbI_2$

Thus, the number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

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