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7nadin3 [17]
4 years ago
14

Copper has two naturally occurring isotopes with atomic masses of 62.9296 u () and 64.9278 u (). The atomic mass of copper is 63

.546 u. What is the percent distribution of the isotopes
Chemistry
1 answer:
natta225 [31]4 years ago
4 0

Answer:

The abundance of first isotope is 69.15 %

The abundance of second isotope is 30.85 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope:

% = x %

Mass = 62.9296 u

For second isotope:

% = 100  - x  

Mass = 64.9278 u

Given, Average Mass = 63.546 u

Thus,  

63.546=\frac{x}{100}\times {62.9296}+\frac{100-x}{100}\times {64.9278}

Solving for x, we get that:

x = 69.15 %

<u>The abundance of first isotope is 69.15 %</u>

<u>The abundance of second isotope is 100 - 69.15 % = 30.85 %</u>

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[A]²

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8 0
3 years ago
Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the
Cloud [144]

Answer : The equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

Solution : Given,

Equilibrium constant, K_c=1.8\times 10^{-5}

Initial concentration of HC_2H_3O_2 = 0.260 m

Let, the 'x' mol/L of H_3O^+ are formed and at same time 'x' mol/L of HC_2H_3O_2 are also formed.

The equilibrium reaction is,

                  HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)

Initially                0.260 m                       0                 0

At equilibrium    (0.260 - x)                     x                 x

The expression for equilibrium constant for a given reaction is,

K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

Now put all the given values in this expression, we get

1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}

By rearranging the terms, we get the value of 'x'.

x=2.154\times 10^{-3}m

Therefore, the equilibrium concentration of H_3O^+ at 25^oC is, 2.154\times 10^{-3}m.

4 0
3 years ago
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