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Anit [1.1K]
4 years ago
12

Be sure to answer all parts. the average atomic mass of nitrogen is 14.0067. the atomic masses of the two stable isotopes of nit

rogen, 14n and 15n, are 14.003074002 and 15.00010897 amu, respectively. use this information to determine the percent abundance of 14n.
Chemistry
1 answer:
Sergio039 [100]4 years ago
8 0
To answe this question, we will assume that the percentage abundance of 14n is y. Since the percentage abundance of 14n + percentage abundance of 15n = 100% = 1
Therefore, percentage abundance of 15n = 1 - y.

Now we know that the average atomic mass of nitrogen is 14.0067
Therefore:
average atomic mass of nitrogen = atomic mass of 14n x its percentage abundance + atomic mass of 14n x its percentage abundance
14.0067 = 14.003074002 y + <span>15.00010897 (1 - y)
</span>14.0067 = 14.003074002 y + 15.00010897 - 15.00010897 y
14.0067 - 15.00010897 = 14.003074002 y  - 15.00010897 y 
<span>0.99703497 y =  0.99340897
</span>
Therefore y = <span> 0.99636

Based on this, the percentage abundance of 14n is </span> 0.99636x100= <span>99.636%
while percentage abundance of 15n = 100% - </span>99.636% = <span>0.364%</span>
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Answer:

^{32}_{14}Si\rightarrow ^{32}_{15}P+e^-+\bar{v}_e

Explanation:

Beta decay conserves the lepton number. In \beta^- decay, the atomic number of the element increases which is accompanied by the release of  an electron antineutrino, e^-+\bar{v}_e.

For example:-

^A_ZX\rightarrow ^A_{Z+1}X+e^-+\bar{v}_e

The \beta^- decay of silicon-32 is shown below as:-

^{32}_{14}Si\rightarrow ^{32}_{15}P+e^-+\bar{v}_e

8 0
4 years ago
Fireworks, KNO3<br> Give only the names of the elements alphabetically, separating them with commas.
Alborosie
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The answer will be K, N, O.

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Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

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=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

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