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Sedbober [7]
3 years ago
13

A new compound was recently discovered and found to have an atomic weight of 342.38 amu. This element has two isotopes, the ligh

ter of which has a mass of 340.91 amu and an abundance of 68.322%. What is the mass of the heavier isotope?
Chemistry
1 answer:
Rufina [12.5K]3 years ago
5 0

Answer:

The mass of heavier isotope is 345.6 amu.

Explanation:

Given data:

atomic wight of compound = 342.38 amu

lighter isotope mass = 340.91 amu

abundance of lighter isotope = 68.322%

mass of heavier isotope = ?

Solution:

average atomic mass = ( % age abundance of lighter isotope × its atomic mass) + (% age abundance of heavier isotope × its atomic mass)  / 100

percentage of heavier isotope = 100- 68.322 = 31.678

Now we will put the values in formula.

342.38  = (68.322× 340.91) + (31.678 × X) / 100

342.38 = 23291.65302 +  (31.678 × X) / 100

342.38 × 100 = 23291.65302 +  (31.678 × X)

34238 -23291.65302 =  (31.678 × X)

10946.35 / 31.678 = X

345.6 = X

The mass of heavier isotope is 345.6 amu.

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Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338
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<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

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Putting values in equation 1, we get:

0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol

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Volume of solution = 664 mL

Putting values in equation 1, we get:

0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = (2\times 0.224)=0.448mol

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

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Looking at the Lewis structure of SO2 attached to this answer, we can see the twelve valence electrons in the molecule and how they are distributed around each atom as shown.

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