Answer:
1. [tex]n=6.03x10^{-13} mol[/tex]
2. [tex]n=5.90x10^{-2}molKNO_3[/tex]
3. [tex]n=1.48x10^{-2}molCH_4[/tex]
4. [tex]V=2.98x10^{-2}L[/tex]
Explanation:
Hello.
In this case we apply mole-mass-particles relationships by which we can consider that 1 mole of a substance equals 6.022x10²³ particles and also the atomic or molar mass of the substance to be analyzed. In such a way, for each case, we proceed as follows:
1.
[tex]n=3.63x10^{11}atoms*\frac{1mol}{6.022x10^{23}atoms}=6.03x10^{-13} mol[/tex]
2. In this case, the molar mass of KNO3 is 101.1 g/mol, thereby the moles are:
[tex]n=5.96gKNO_3*\frac{1molKNO_3}{101.1gKNO_3}=5.90x10^{-2}molKNO_3[/tex]
3. In this case, the molar mass of CH4 is 16 g/mol, thereby the moles are:
[tex]n=1.50gCH_4*\frac{1molCH_4}{101.1gCH_4}=1.48x10^{-2}molCH_4[/tex]
4. In this case, since the density of KBr is 2.75 g/mL and 1000 mL equals 1 L, the liters are:
[tex]V=82.0gKBr*\frac{1mLKBr}{2.75gKBr}*\frac{1L}{1000mL} =2.98x10^{-2}L[/tex]
Best regards!
