Question

Please help! Will medal!

Calculate each of the following quantities:

A.) mass in grams of 0.68 mol of KMnO4
B.) moles of O atoms in 8.18 g of Ba(NO3)2
C.) number of O atoms in 7.3 x10^-3 g of CaSO4 x 2H2O

Answer 1

A) Mass of compound = amount of compound (in moles) x molar mass\[m=n _{KMnO _{4}}\times M _{KMnO _{4}}= 0.68\times 158.09\approx107.5(g)\]
B) As you can see, each Ba(NO3)2 molecule contains 6 O atoms. Therefore, \[n _{O}=6n _{Ba(NO _{3})_{2}}=6\frac{ m }{ M }=6\times \frac{ 8.18 }{ 216.34 }\approx 0.227(mol)\]
C) As you know, 1 mole of a compound will contain 6.022x10^23 molecules. At first, we will calculate the amount of CaSO4x2H2O. You can also see that each molecule contains 6 O atoms. Now you can easily calculate the number of O atoms. \[n _{CaSO _{4}.2H _{2}O}=\frac{ m }{ M }=\frac{ 7.3\times 10^{-3} }{ 172.70 }\approx 4.2\times 10^{-5}(mol)\] \[\rightarrow n _{O}=6n _{CaSO _{4}.2H _{2}O}= 2.52\times 10^{-4}(mol)\]
Now you can calculate the number of O atoms \[N _{O}=n _{O}\times 6.022 \times 10^{23}= 2.52\times 10^{-4}\times 6.022\times 10^{23}\approx 1.52\times 10^{20}(atoms)\]

Answer 2

Thanks so much :) @tinybookworm

Answer 3

It's my please.

Answer 4

:)