Chemistry, 17.07.2019 04:30 crowdedallys5385

Carbon exists in nature as three isotopes. given the atomic masses and relative abundance of the isotopes, what is the average atomic mass of carbon? a. 12.000 amu b.12.011 amu c.13.002 amu d.13.003 amu e.14.003 amu

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Answer from: ShexSmart1646

The correct answer is B. 12.011 amu

Answer from: barstr9146

Answer : The correct option is, (B) 12.011 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

We are assuming that:

Mass of isotope C-12 = 12.000 amu

% abundance of isotope C-12= 98.93% = 0.9893

Mass of isotope C-13 = 13.003 amu

% abundance of isotope C-13 = 1.07% = 0.0107

Mass of isotope C-14= 14.003 amu

% abundance of isotope C-14 = 0.0000000001% = 0.000000000001 = $1.0\times 10^{-12}$

In this case, the abundance of C-14 is very low. So, their influence is negligible.

Now put all the given values in the above formula, we get:

$\text{ Average atomic mass of an element}=\sum[(12.000\times0.9893)+(13.003\times 0.0107)+(14.003\times 1.0\times 10^{-12})]$

$\text{ Average atomic mass of an element}=12.011amu$

Therefore, the average atomic mass of carbon is, 12.011 amu

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

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Carbon exists in nature as three isotopes. given the atomic masses and relative abundance of the iso...

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