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erastovalidia [21]
4 years ago
12

Unknown element has two isotopes. Isotope A has a mass of 34 amu and abundance of 52%, isotope B has a mass of 33 amu and abunda

nce of 48%. Find average atomic mass of this element and express your
Chemistry
1 answer:
ZanzabumX [31]4 years ago
4 0

Answer:

x = 33.52 amu

Explanation:

It is given that,

Isotope A has a mass of 34 amu and an abundance of 52%, isotope B has a mass of 33 amu and an abundance of 48%.

Let x is the average atomic mass of this element. It can be calculated as follows :

x=52\%\ \text{of}\ 34+48\%\ \text{of}\ 33\\\\x=\dfrac{52}{100}\times 34+\dfrac{48}{100}\times 33\\\\x=0.52\times 34+0.48\times 33\\\\x=33.52\ \text{amu}

So, the average atomic mass of this element is 33.52 amu.

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Answer:

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Explanation:

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2CO_{(g)}      +      O_{2(g)}        ⇄       2CO_{2(g)}

By Applying the ICE Table; we have

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K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}

Given that K_c = 1.4*10^2 ; Then:

1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}

1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}

3.5 =( \frac{(0.001)}{(x)})^2

\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}

1.87=\frac{(0.001)}{(x)}

(x)= \frac{(0.001)}{1.87 }

x = 5.35 *10^{-4}M

∴ The equilibrium concentration of CO = x = 5.35 *10^{-4}M

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