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Gnoma [55]
3 years ago
12

What is the average atomic mass of the following isotopic mixture - 42.10 g/mole- 93.50%; 48.30 g/mole 6.200%; 50.40 g/mole - 0.

3000%? Show your work!!
Chemistry
1 answer:
bazaltina [42]3 years ago
8 0

Answer:

40.03g/mole

Explanation:

Given parameters:

Mass =  42.10 g/mole and abundance  = 93.50%

Mass = 8.30 g/mole and abundance =  6.200%

Mass  = 50.40 g/mole and abundace  = 0.3000%

Find;

Average atomic mass = ?

Solution:

Average atomic mass = sum (mass x abundance)

 Average atomic mass = (\frac{93.50}{100}  x 42.1) + (\frac{6.2}{100} x 8.3) + (\frac{0.3}{100} x 50.4)

                                      = 39.36g/mole + 0.52g/mole + 0.15g/mole

                                      = 40.03g/mole

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<u>Answer:</u>

<u>For a:</u> The isotopic representation of iodine is _{53}^{131}\textrm{I}

<u>For b:</u> The isotopic representation of cesium is _{55}^{137}\textrm{Cs}

<u>For c:</u> The isotopic representation of strontium is _{38}^{52}\textrm{Sr}

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The isotopic representation of an atom is: _Z^A\textrm{X}

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A = Mass number of the atom

X = Symbol of the atom

  • <u>For a:</u>

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Mass number = 53 + 78 = 131

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Thus, the isotopic representation of cesium is _{55}^{137}\textrm{Cs}

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Number of neutrons = 52

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