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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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If I have 4.5 liters of gas at a temperature of 33 0C and a pressure of 6.54 atm, what will be the pressure of the gas if I rais
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This is an exercise in the general or combined gas law.

To start solving this exercise, we must obtain the following data:

<h3>Data:</h3>
  • V₁ = 4.5 l
  • T₁ = 33 °C + 273 = 306 k
  • P₁ = 6.54 atm
  • T₂ = 94 °C + 273 = 367 k
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  • P₂ = ¿?

We use the following formula:

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Where

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  • V₁ = Initial volume
  • T₂ = Initial temperature
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  • V₂ = Final volume
  • T₁ = Initial temperature

We clear the general formula for the final pressure.

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2} }{V_{2}T_{1}}  \ \to \ Clear \ formula \end{gathered}$}

We solve by substituting our data in the formula:

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(6.54 \ atm)(4.5 \not{l})(367 \not{K}) }{(2.3 \not{l})(306 \not{k})}  \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{10800.81}{ 703.8 } \ atm  \end{gathered}$}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=15.346 \ atm \end{gathered}$} }

If I raise the temperature to 94°C and decrease the volume to 2.3 liters, the pressure of the gas will be 15,346 atm.

8 0
2 years ago
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