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vladimir2022 [97]
4 years ago
10

Lithium has 2 isotopos. One has a mass of

Chemistry
1 answer:
Rina8888 [55]4 years ago
4 0

Answer:

The isotope with mass 6.015amu =7.5% abundance

The isotope with mass 7.016amu = 92.5%

Explanation:

<u>Step 1:</u> Given data

Lithium has an average atomic mass of 6.941 amu

Lithium has 2 isotopos with a mass, respectively 6.015amu and 7.016amu

<u>Step 2: </u>Calculate the abundances

Isotope with mass 6.015 amu has an abundance of X

Isotope with mass 7.016 amu has an abundance of Y

The average atomic mass is the average mass of the isotopes with its respectively abundance.

X+Y = 1 or X = 1 - Y

6.941 = 6.015X +7.016Y

This gives us:

6.941 = 6.015(1-Y) +7.016Y

6.941 = 6.015 - 6.015Y + 7.016Y

6.941 - 6.015 = -6.015Y +7.016Y

0.926 = 1.001Y

Y = 0.925

Y = 92.5%

X = 1- 92.5 = 7.5%  

To control we can do the following equation:

6.015*0.075 + 7.016*0.925 = 6.941

This means the abundances of the 2 isotopes are:

The isotope with mass 6.015amu =7.5% abundance

The isotope with mass 7.016amu = 92.5%

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A solution of iron(III) chlorate was poured into a lithium phosphate solution. Would you expect a precipitate to form if 354.0 m
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Answer:

K. Yes, because Qsp > Ksp for iron(III) phosphate

Explanation:

Lithium chlorate is very soluble in water, that means has a big ksp.

Iron (III) phosphate has a ksp of 1.3x10⁻²²

Ksp formula of iron (III) phospate is:

FePO₄(s) ⇄ Fe³⁺(aq) + PO³⁻(aq)

Ksp = [Fe³⁺] [PO₄³⁻] = 1.3x10⁻²²

Molar concentration of Fe³⁺ and PO₄³⁻ is:

[Fe³⁺] = 0.340L ₓ (2.52x10⁻⁹mol / L) ÷ (0.354L + 0.520L) = <em>9.8x10⁻¹⁰M</em>

[PO₄³⁻] = 0.520L ₓ (4.65x10⁻¹¹mol / L) ÷ (0.354L + 0.520L) = 2.78<em>x10⁻¹¹M</em>

Replacing in Ksp formula:

Qsp = [9.8x10⁻¹⁰] [2.78x10⁻¹¹M] = 2.72x10⁻²⁰

As Qsp > Kps, the equlibrium will shift to the right decreasing the ions concentrations producing FePO₄(s), <em>a precipitate</em>

<em>The ions decreases its concentration until Q = Kps</em>

<em />

Thus, right answer is:

<em>K. Yes, because Qsp > Ksp for iron(III) phosphate </em>

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What is the principal difference between a strong and weak acid?
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Substrate: N-benzoyl-L-tyrosine ethyl ester (BTEE) 0.001M
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Based on the dilution formula, 0.1 mL of the stock solution of the enzyme is required to prepare a 50-fold diluted enzyme in 0.01 M HCl.

<h3>How can 50-fold dilution of the enzyme be done?</h3>

The 50-fold dilution of the stock enzyme solution can be done by using the dilution formula to determine the given volume of the stock solution required.

The dilution formula is given below:

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From the given data for the enzyme dilution;

C1 = 1

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Making V1 subject of formula in the dilution formula:

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Therefore, 0.1 mL of the stock solution of the enzyme is required to prepare a 50-fold diluted enzyme in 0.01 M HCl.

Learn more about dilution at: brainly.com/question/24709069

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