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skad [1K]
3 years ago
10

How many atoms are present in 179.0 g of iridium?

Chemistry
2 answers:
patriot [66]3 years ago
6 0
I dont know the answer


posledela3 years ago
4 0
160 atoms are present
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A compound is 70.0% iron and 30.0% oxygen. Its molecular mass is 479.1 g/mol. Determine the
Strike441 [17]

Answer:

Fe_6O_9

Explanation:

Hello there!

In this case, since these problems about formulas, firstly require the determination of the empirical formula, assuming that the given percentages are masses, we can calculate the moles and mole ratio of oxygen to iron as shown below:

n_{Fe}=70/55.85=1.25\\\\n_O=30/16=1.875

In such a way, by rounding to the first whole number we multiply by 8 and divide by 5 to obtain:

Fe_{2}O_{3}

Whose molar mass is 159.69 g/mol and the mole ratio of the molecular to the empirical formula is:

479.1/159.69=3

Therefore, the molecular formulais:

Fe_6O_9

Regards!

4 0
3 years ago
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the answer is conduction

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bazaltina [42]

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160.3g

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We need to find the mass, so make mass the subject of the formula.

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Calculate the formula/molecular mass of following
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