Magnesium occurs naturally in only three isotopes. ^24 Mg has an isotopic mass of 23.9850 amu and an abundance of 78.99%. ^25 Mg has an isotopic mass of 24.9858 amu and an abundance of 10.00%. Find the mass number of the last isotope of magnesium.

Can someone please explain how to solve this? *done mine the notes*

Magnesium Occurs Naturally In Only Three Isotopes. ^24 Mg Has An Isotopic Mass Of 23.9850 Amu And An

Answers

Answer 1

Answer:

25,98 amu and 11.01% abundance

Explanation:

1. Find the missing abundance

100 - 78.99 - 10 = 11.01 %

2. Using each mass and abundance and the periodic table for the Mg amu

And let Isotope = Mg

23.9850(78.99%) + 24.9858(10%) + Mg(11.01%) = 24.305 (100%)

Solve for Mg

Mg = 25,98


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Answers

Answer:

Explanation:

Hello there, I think this is a limiting agent question.

When 3 moles of Mg is reacted with 2 moles of Br2, because of ratio, we just need 1.5 moles of Br2, so Mg is a limiting agent.

So 3 moles of Mg will make 3 moles of MgBr

Answer:

Use the given functions to set up and simplify

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3=

2=

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Explanation:

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Answers

Answer:

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[tex].[/tex]

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[tex]{{\boxed{\mathcal{\red{It \: will \: be \: in \: your\: \: text \: book \: see}}}}}[/tex]

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