Bromine has two naturally-occurring isotopes. 79Br has a mass of 78.9 amu and accounts for 50.3% of bromine atoms. If the atomic mass of bromine is 79.9 amu, what is the mass of an atom of the second bromine isotope?
A) 77.9 amu.
B) 80.0 amu.
C) 80.1 amu.
D) 80.9 amu.
E) 88.9 amu.

Answers

Answer 1

The mass of an atom of the second bromine isotope is 80.9amu

Calculation of Mass of an atomic isotope

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Now given that

Atomic mass of bromine = 79.9 amuPercent abundance of 1st isotope = 50.3%Atomic mass of 1st isotope = 78.9 amuPercent abundance of 2nd isotope = 100-50.3%=49.7%Atomic mass of 2nd isotope =?

Plugging in values, we have that

79.9= (50.3×78.9)+(49.7× ?) /100

79.9x 100= (50.3×78.9)+(49.7× ?) /100

7990= 3968.7+(49.7x)

7990- 3968.7=49.7x

4021.3= 49.7x

x=4021.3/49.7

x=80.9amu

Therefore, the  mass of an atom of the second bromine isotope is 80.9amu.

See related question on finding mass of an isotope here: https://brainly.com/question/19533439


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Give the picture please

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Answers

Answer:

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What is one reason why transpiration is important in plants?

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Give Brainliest.

A compound has a percent composition of 15.24% sodium (molar mass
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Answers

This problem is providing us with the percent composition of a compound containing sodium, bromine and oxygen. Thus, the empirical formula is required and found to be NaBrO₃.

Empirical formulas

In chemistry, empirical formulas are used in order to figure out the minimum whole-number representation of a molecular formula, which contains the actual number of atoms.

Thus, we start determining it by assuming the given percentages as masses and calculating the moles with the atomic mass of the involved elements:

[tex]n_{Na}=\frac{15.24}{22.99}=0.700 \\\\n_{Br}=\frac{52.95}{79.90} =0.663\\\\n_O=\frac{31.81}{16.00}=1.99[/tex]

Next, we divide by the 0.633 as the smallest number of moles in order to determine their subscripts in the chemical formula:

[tex]Na:\frac{0.700}{0.663}=1\\ \\Br:\frac{0.663}{0.663}=1\\ \\O=\frac{1.99}{0.663} =3[/tex]

Hence, the empirical formula turns out to be NaBrO₃.

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Answer:

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Explanation: Hope this helps

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Answer:

Explanation:

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Answer:

Will it make a difference if we swap the anode and cathode in electroplating?

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[tex]TextFromYourMrEx...[/tex]

#CarryOnLearning

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Layla~

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Based on the data provided;

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