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aniked [119]
2 years ago
11

A hypothetical element consists of two isotopes of masses 86.95amu and 88.95amu with abundances of 35.5% and 64.5% respectively.

What iis the average atomic mass of this element
Chemistry
1 answer:
Art [367]2 years ago
5 0

Answer: The average atomic mass of the element = 88.242amu

Explanation:

The abundance of the first isotope is =35.5%

 Atomic mass of first isotope = 68.9257

The average atomic mass of the first isotope =86.95amu X 35.5%  =86.95amu X 0.355 =30.8725 amu

The abundance of the second isotope =64.5%

Atomic mass of the second isotope =88.95amu

The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu

Now the average atomic mass =30.8725 +57.37275 = 88.242amu

OR using the formulae

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100

8,824/100

=88.24amu

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Answer:

<h2>6.64 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{4 \times  {10}^{24} }{6.02 \times  {10}^{23} }   \\  = 6.644518...

We have the final answer as

<h3>6.64 moles</h3>

Hope this helps you

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