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2 years ago

2. Nitrogen has two naturally occurring

Chemistry

1 answer:

kvv77 [185]2 years ago

8 0

Answer:

Suppose that:

Mass of N-14= 14

and Mass of N-15= 15

fraction N-14 x 14 + fraction N-15 x 15 = 14.007

now.. if the fractions were =, that is if fraction N-14 = fraction N-15 = 0.5, then...

0.5 x 14 + 0.5 x 15 = 14.5

more N-15 would make that number larger...

0.3 x 14 + 0.7 x 15 = 14.7

let x = abundance of N-14, then (1-x) = abundance of N-15, and we could solve for x.

(X)x14 + (1-X)x 15 = 14.007

14X + 15 - 15X = 14.007

X = 0.993

the relative abundance of N-14 = 99.3%

the relative abundance of N-15 = 0.7%

So, according to the above explanation N-14 is abundant

Explanation:

Suppose that:

Mass of N-14= 14

and Mass of N-15= 15

fraction N-14 x 14 + fraction N-15 x 15 = 14.007

now.. if the fractions were =, that is if fraction N-14 = fraction N-15 = 0.5, then...

0.5 x 14 + 0.5 x 15 = 14.5

more N-15 would make that number larger...

0.3 x 14 + 0.7 x 15 = 14.7

let x = abundance of N-14, then (1-x) = abundance of N-15, and we could solve for x.

(X)x14 + (1-X)x 15 = 14.007

14X + 15 - 15X = 14.007

X = 0.993

the relative abundance of N-14 = 99.3%

the relative abundance of N-15 = 0.7%

So, according to the above explanation N-14 is abundant

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Answer:

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In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

$atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu$

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