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storchak [24]
3 years ago
15

Rhenium has two naturally occurring isotopes, Re-185 and Re-187. The relative atomic mass of rhenium is 186.2. What are the natu

ral abundances of these isotopes
Chemistry
1 answer:
Ahat [919]3 years ago
8 0

Answer:

The natural abundance of Re-185 is 40% and of Re-187 is 60%

Explanation:

Relative atomic mass is defined as the sum of the mass of each isotope times its relative abundances.

Is Rhenium has only 2 isotopes and the abundances are X and Y we can write:

X + Y = 1 <em>(1)</em>

And:

185X + 187Y = 186.2 <em>(2)</em>

<em>Where X is abundance of Re-185 and Y abundance of Re-187</em>

<em />

Replacing (1) in (2):

185X + 187(1-X) = 186.2

185X + 187-187X = 186.2

-2X = -0.8

X = 0.4 = 40%

And Y = 100% - 40% = 60%

<h3>The natural abundance of Re-185 is 40% and of Re-187 is 60%</h3>

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(a)Determine the number of KNO3 molecules in 0.750 mol KNO3.
svp [43]

A. The number of molecules in 0.750 mole of KNO₃ is 4.515×10²³ molecules

B. The mass (in milligrams) of 2.39×10²⁰ molecules of Ag₂SO₄ is 124 mg

C. The number of molecules in 3.429 g of NaHCO₂ is 3.04×10²² molecules

<h3>Avogadro's hypothesis </h3>

1 mole of substance = 6.02×10²³ molecules

<h3>A. How to determine the number of molecules </h3>

1 mole of KNO₃ = 6.02×10²³ molecules

Therefore,

0.750 mole of KNO₃ = 0.75 × 6.02×10²³

0.750 mole of KNO₃ = 4.515×10²³ molecules

<h3>B. How to determine the mass of Ag₂SO₄</h3>

6.02×10²³ molecules = 312 g of Ag₂SO₄

Therefore,

2.39×10²⁰ molecules = (2.39×10²⁰ × 312) / 6.02×10²³

2.39×10²⁰ molecules = 0.124 g

Multiply by 1000 to express in mg

2.39×10²⁰ molecules = 0.124 g × 1000

2.39×10²⁰ molecules = 124 mg of Ag₂SO₄

<h3>C. How to determine the number of molecules </h3>

68 g of NaHCO₂ = 6.02×10²³ molecules

Therefore,

3.429 g of NaHCO₂ = (3.429 × 6.02×10²³) / 68

3.429 g of NaHCO₂ = 3.04×10²² molecules

Learn more about Avogadro's number:

brainly.com/question/26141731

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