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emmasim [6.3K]
3 years ago
14

A pure sample contains only nitrogen and oxygen atoms. If the sample is 30.4% nitrogen, by weight, what is the empirical formula

of the molecule? 1. N2O3 2. NO2 3. N2O 4. N2O5 5. NO
Chemistry
1 answer:
andrew-mc [135]3 years ago
8 0

Answer: 2. NO_2

Explanation:

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

a) If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of N = 30.4 g

Mass of O= 69.6 g

Step 1 : convert given masses into moles

Moles of N=\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{30.4g}{14g/mole}=2.2moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{69.6g}{16g/mole}=4.4moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For N = \frac{2.2}{2.2}=1

For O =\frac{4.4}{2.2}=2

The ratio of N: O = 1; 2

Hence the empirical formula is NO_2

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Answer : The number of iron atoms present in each red blood cell are, 1.077\times 10^9

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First we have to calculate the moles of iron.

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Now we have to calculate the number of iron atoms.

As, 1 mole of iron contains 6.022\times 10^{23} number of iron atoms

So, 0.0519 mole of iron contains 0.0519\times 6.022\times 10^{23}=3.125\times 10^{22} number of iron atoms

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Number of iron atoms are present in each red blood cell = \frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}

Number of iron atoms are present in each red blood cell = \frac{3.125\times 10^{22}}{2.90\times 10^{13}}

Number of iron atoms are present in each red blood cell = 1.077\times 10^9

Therefore, the number of iron atoms present in each red blood cell are, 1.077\times 10^9

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