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Debora [2.8K]
3 years ago
15

There are two naturally occurring isotopes of boron. 10 B has a mass of 10.0129 u. 11 B has a mass of 11.0093 u. Determine the a

bundance of each isotope
Chemistry
1 answer:
Vanyuwa [196]3 years ago
5 0
<h2>Natural Abundance for 10B is 19.60%</h2>

Explanation:

  • The natural isotopic abundance of 10B is 19.60%.
  • The natural isotopic abundance of 11B is 80.40%.
  • The isotopic masses of boron are 10.0129 u and 11.009 u respectively.

For calculation of abundance of both the isotopes -

Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.

Determining it as an equation -

10x + 11y= 10.8

x+y=1 (ratio)

10x + 10y = 10

By taking the denominator away from the numerator

we get;

y = 0.8

x + y = 1

∴ x = 0.2

To get percentages  we need to multiply it by 100

So, the calculated abundance is 80% for 11 B and 20% 10  B.

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What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm
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Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}

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n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = 5\times 196.96 g

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Mass of {3.01\times 10^{24}} atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = 19.32 g/cm^{3}

Volume of the cuboid = length\times breadth\times height

Volume of the gold bar =6.00\times 4.25\times 2.00

Volume of the gold bar = 51cm^{3}

Using formula,

Density = \frac{mass}{Volume}

Mass = Density\times Volume

Mass = 19.32 \times 51

Mass = 985.32 g

So, A  gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

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Answer: The salt produced will be K_{2}SO_{4}

Explanation:

During a neutralization reaction, an acid reacts with a base for producing the correspondent salt, and water.

The strong acids release all the protons avalaible when are dissolved, such as sulfuric acid. As you can see, sulfuric acid have 2 protons ready for being released (H_{2}SO_{4}); and those places have to be occcupied for other ions equivalents to the H+: K+ from KOH in this case.

Therefore the answer will be K_{2}SO_{4}.

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