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Andru [333]
3 years ago
5

How many moles are 2.54 x 10^29 molecues of H2O

Chemistry
1 answer:
shtirl [24]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 421,786.7818 \ mol \ H_2O}}

Explanation:

We are asked to convert from molecules of water to moles.

<h3>1. Avogadro's Number </h3>

1 mole of any substance contains the same number of particles (atoms, molecules, formula units, etc). This is Avogadro's Number: 6.022*10²³. In this problem, the particles are molecules of water.

<h3>2. Convert Molecules to Moles</h3>

Use Avogadro's Number to make a ratio.

\frac {6.022 *10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Multiply by the number of molecules given: 2.54 *10²⁹

2.54 *10^{29}  \ molecules \ H_2O \times \frac {6.022 *10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Flip the ratio so the units of molecules of water cancel.

2.54 *10^{29}  \ molecules \ H_2O \times \frac {1 \ mol \ H_2O}{6.022 *10^{23} \ molecules \ H_2O}

2.54 *10^{29}   \times \frac {1 \ mol \ H_2O}{6.022 *10^{23} }

Condense the problem into 1 fraction.

\frac {2.54 *10^{29} }{6.022 *10^{23} }\ mol \ H_2O

421786.7818 \ mol \ H_2O

2.54 *10²⁹ molecules of water are 421, 786.7818 moles of water.

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
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Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

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Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

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