Question

How many atoms are present in 5.6cc of helium gas at STR​

Answer 1

Answer:

1.505 x $10^{20\\}$

Explanation:

1 $cm^{3\\}$ = 1 ml

1 $cm^{3\\}$ = 1 L

5.6  $cm^{3\\}$ = 5.6 x $10^{-3\\\\}$  

no of moles of helium = $\frac{volume}{22.4}$

                                    = $\frac{5.6 x 10^{-3} }{22.4}$

                                    = 0.25  x $10^{-3\\\\}$  

no of atoms of He =   0.25  x $10^{-3\\\\}$  x 6.023 x  $10^{23}$

                              = 1.505 x $10^{20\\}$