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1 year ago

How many atoms of oxygen are present in 7.51 grams of glycine with formula C₂H5O2N?

1 answer:

3 0

1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N. Details about number of atoms can be found below.

How to calculate number of atoms?

The number of atoms of a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

However, the number of moles of oxygen in glycine can be calculated using the following expression:

Molar mass of C₂H5O2N = 75.07g/mol

Mass of oxygen in glycine = 32g/mol

Hence; 32/75.07 × 7.51 = 3.2grams of oxygen in glycine

Moles of oxygen = 3.2g ÷ 16g/mol = 0.2moles

Number of atoms of oxygen = 0.2 × 6.02 × 10²³ = 1.205 × 10²³ atoms

Therefore, 1.205 × 10²³ atoms of oxygen will be present in 7.51 grams of glycine with formula C₂H5O2N.

Learn more about number of atoms at: brainly.com/question/8834373

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What mass of water (in grams) is produced by the reaction of 23.0 g of SiO2?

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<em>"Your question is not complete, it seems to be missing the following information";</em>

In the reaction of the given compound, $4Hf (g) \ + \ SiO_2 (s) \ --> \ SiF_4(g) \ + \ 2H_2O(l)$ , what mass of water (in grams) is produced by the reaction of 23.0 g of SiO2?

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