From 48 gram of helium ,6×10^23atom are removed then find the mass of

1.	From 48 gram of helium ,6×10^23atom are removed then find the mass of carbon atom which contain same number helium atom are left
Answer» Answer: We know by Avogadro NUMBER that, 6.023 x 10^(23) no. of ATOMS of Helium make 1 mole that is equivalent to molar MASS of Helium approximately 4 grams. Thus, 6.023 x 10^(24) atoms of Helium = (1/(6.023 x 10^(23) ) x 6.023 x 10^(24) moles = 10 moles. Since 1 mole = 4 grams(approx.) we can easily say 10 moles of helium will have a mass of 40 grams(approx.). Explanation:

From 48 gram of helium ,6×10^23atom are removed then find the mass of carbon atom which contain same number helium atom are left

Answer»

Answer:

We know by Avogadro NUMBER that, 6.023 x 10^(23) no. of ATOMS of Helium make 1 mole that is equivalent to molar MASS of Helium approximately 4 grams.

Thus, 6.023 x 10^(24) atoms of Helium = (1/(6.023 x 10^(23) ) x 6.023 x 10^(24) moles = 10 moles.

Since 1 mole = 4 grams(approx.) we can easily say 10 moles of helium will have a mass of 40 grams(approx.).

Explanation: