How many sulfur atoms would be present in 41.8 moles

Ответ:

hallmansean04

22.01.2022

An atom is the smallest unit of ordinary matter that forms a chemical element.

There are $7 .86*10^{23$ atoms of sulphur are contained in 41.8 g of sulphur

The data is as given:-

Molar mass of sulphur = 32 g/mol Mass of sulphur = 41.8g

The formula we will use is $Number\ of\ moles = \frac{Given \ mass}{ Molar Mass}$ .

So, according to the formula the number of moles of sulphur = $\frac{41.8}{32} = 1.30625 moles$

Therefore, In 1 mole of atoms, there is $6.022 *10^{23$ number of atoms.

So, In 1.30625 moles of Sulphur atom, number of atoms = $1.30625 * 6.022 * 10^{23} = 7.86*10^{23} atoms$

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cece3467

22.01.2022

7 .86×10^23 atoms of sulphur are contained in 41.8 g of sulphur

Explanation:

Molar mass of sulphur = 32 g/mol

Given mass of sulphur = 41.8g

Number of moles = Given mass / Molar Mass

So, number of moles of sulphur = 41.8/32 = 1.30625 moles

We know that,

In 1 mole of atom there are 6.022 × 10^23 number of atoms.

So, In 1.30625 moles of Sulphur arom, number of atoms = 1.30625 × 6.022 × 10^23 = 7.86×10^23 atoms

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glitterpanda2468

11.03.2023

I mean . . .

Explanation:

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