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sleet_krkn [62]
3 years ago
11

There are two naturally occurring isotopes of boron determine the abundance of each isotope

Chemistry
1 answer:
MAVERICK [17]3 years ago
6 0

Answer : The correct answer is:

The abundance of isotopes can be found out using following formula :

atomic mass of element = ( abundance of isotope 1 * mass of isotope 1 ) + ( abundance of isotope 2 * mass of isotope 2 ) +.....

Boron ( mass = 10.811 u ) has two isotopes : ¹⁰B ( mass = 10.0129 u ) and ¹¹ B ( mass = 11.0093 u )

Sum of abundance of all isotopes = 100%

Lets assume abundance of isotope ¹¹B = x % = 0.01 x , so abundance of isotope ¹⁰B = 100 % - x% = 100/100 - x/100 = 1 - 0.01 x

Plugging values in formula :

10.811 = ( 0.01 x * 11.0093) + [( 1 - 0.01 x) * 10.0129 ]

10.811 = (0.110093 x ) + ( 10.0129 - 0.100129 x )

10.811 - 10.0129 = 0.110093 x - 0.100129 x

0.7981 = 0.009964 x

Dividing both side by 0.00964

0.7981 / 0.009964 = 0.00964 /0.00964 x

x = 82.7 %

Abundance of of isotope ¹¹ B = 82.7 %

Abundance of isotope ¹⁰ B = 100% - x % = 100 %- 82.7 % = 17.3 %

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<u>Given information:</u>

Concentration of HCl = 0.035 M

<u>To determine:</u>

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<u>Explanation:</u>

Hydrochloric acid, HCl is a strong acid. It will completely dissociate to give H+ and Cl- ions

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A 13 g sample of P4010 contains how many
Alex777 [14]

Answer:

\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅,  then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}

2. Moles of P₂O₅

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The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

\text{Moles of P$_{2}$O}_{5} = \text{0.0458 mol P$_{4}$O}_{10} \times \dfrac{\text{2 mol P$_{2}$O}_{5}}{\text{1 mol P$_{4}$O}_{10}} = \text{0.0916 mol P$_{2}$O}_{5}

3. Molecules of P₂O₅

\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}

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Answer:

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Explanation:

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