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Amiraneli [1.4K]
3 years ago
11

The element copper exists in nature as two isotopes: has a mass of 62.9296 u and has a mass of 64.9278 u.The average atomic mass

of copper is 63.55 u. Calculate the relative abundance of the two copper isotopes.
Chemistry
1 answer:
Reil [10]3 years ago
5 0

Answer:

%isotope 1 = 68.92%

%isotope 2 = 31.08 %

Explanation:

we have the following data:

misotope 1 = 62.9296 u

misotope 2 = 64.9278 u

maverage = 63.55 u

%isotope 1 = X

%isotope 2 = 100-X

the average atomic mass is equal to:

maverage = [(misotope 1x%isotope 1)+(misotope 2x%isotope 2)/100]

substituting values:

63.55 = [(62.9296 x X)+(64.9272 x (100-X))]/100

clearing the X:

X = 68.92

therefore we have to:

%isotope 1 = 68.92%

%isotope 2 = 100-68.92 = 31.08 %

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Answer:

Cu~+~PtCl_2->Pt~+~CuCl_2

Explanation:

In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case Pt^+^2) and the anion is the atom at the right of the name (in this case Cl^-). With this in mind, the <u>formula would be</u> PtCl_2.

Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u> Cu. So, we have as reagents:

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The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:

A~+~BC->AB~+~C

With this in mind, the reaction would be:

Cu~+~PtCl_2->Pt~+~CuCl_2

I hope it helps!

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