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## How many atoms are in 5 moles of sulfur?

1 mole is defined as Avogadro’s number (6.02*10^23) of something. So 1 mole of sulfur (or any other element) is 6.02*10^23 atoms of sulfur. 5 moles is just 5 times that amount.

## How many grams are in 5.0 moles?

Therefore, 5 moles × 32.0 g/mol = **160 grams** is the mass (m) when there are 5 moles of O2 .

## Are there in 5.5 mole of sulphur?

Answer: So 1 mole of sulfur (or any other element) is 6.02*10^23 atoms of sulfur. 5 moles is just 5 times that amount.

## How many grams are in 2 moles of sulfur?

The mass of a mole of sulfur atoms is **32.06⋅g** .

## How many atoms are in 5 moles?

Avogadro’s number: the amount of atoms in 1 mole of carbon 12 is 6.022 x 10^23. So if you have 5 mole, you have: 5 x (6.022 x 10^23) = **3.011 x 10^24 atoms**.

## How many atoms are in 1 gram of sulfur?

Explanation: Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1 . Because the units all cancel out, the answer is clearly a number, **≅2×1023** as required.

## How many grams are in 4.3 moles of sulfur?

That means that one mole of sulfur weighs **32.06 grams** (32.06 g/mol). Based on that information, to convert 4.3 moles of sulfur to grams, we multiply 4.3 moles of sulfur by 32.06.

## How many mL is 5 grams?

Grams to mL conversions (water)

Grams to mL | Grams to mL |
---|---|

3 grams = 3 mL | 150 grams = 150 mL |

4 grams = 4 mL | 200 grams = 200 mL |

5 grams = 5 mL |
250 grams = 250 mL |

6 grams = 6 mL | 300 grams = 300 mL |

## Is moles and grams the same?

The mass of a mole of substance is called the **molar mass** of that substance. The molar mass is used to convert grams of a substance to moles and is used often in chemistry.

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Applications of the Mole.

Known Information | Multiply By | Result |
---|---|---|

Mass of substance (g) | 1/ Molar mass (mol/g) | Moles of substance |

## How many atoms or molecules are in 5.0 moles of O?

Hence, number of moles of Oxygen in 5 Moles of CO = 5. The number of oxygen atoms present can be obtained by multiplying the moles of oxygen by Avogadro’s Number. So, number of oxygen atoms present = 5 x ( 6.023 x 10^23 ) = **30.115 x 10^23 nos**.