Question

calculate the number of moles present in A) 12.044x10^23 number of Helium atom and B) 52 grams of helium​

Answer 1

Answer:

I Mole =

$6.23 \times {10}^{23}$

then 12.044*10^23 / 6.022 *10^ 23 =2 Moles

B. I Mole of Helium = 2 amu

52 grams of Helium is

$52 \div 2 = 26moles$