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marusya05 [52]
3 years ago
14

How does the abundance of various isotopes affect the atomic mass?

Chemistry
1 answer:
eduard3 years ago
6 0

solution:

The quoted atomic mass on the Periodic Table is the WEIGHTED average of the individual isotopic masses. The higher the isotopic percentage, the MORE that isotope will contribute to the isotopic mass. For this reason, most masses that are quoted on the Table are non-integral.

By way of example we could look to the hydrogen atom. The VAST majority of hydrogen atoms (in this universe) are the protium isotope. i.e. 1H, whose nuclei contain JUST the defining proton. There is a smaller percentage (>1%) of hydrogen atoms WITH one NEUTRON in their nuclei to give the deuterium isotope. i.e. 2H, and because this is relatively cheap, and easily incorporated into a molecule, deuterium labelling is routinely used in analysis.

And there is even a smaller percentage of hydrogen atoms with TWO NEUTRONS in their nuclei, to give the tritium isotope. i.e. 3H. The weighted average of the isotopic percentages gives 1.00794⋅g⋅mol^{−1}


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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
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Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

and,

The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

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