Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
The mass of lime that can be produced from 4.510 Kg of limestone is calculated as below
calculate the moles of CaCO3 used
that is moles =mass/molar mass
convert Kg to g = 4.510 x1000 =4510g
= 4510 / 100 =45.10 moles
CaCO3 = CaO +O2
by use of mole ratio between CaCO3 to CaO (1:1) the moles of CaO is also= 45.10 moles
mass of CaO = moles x molar mass
45.10 x56 = 2525.6 g of CaO
It is clean and is blue when you take away air hope i helped.
N=N₀*2^(-t/T)
N₀=200 g
T=10 d
t=30 d
N=200*2^(-30/10)=25 g
25 g will remain
