Answer:
5.004kg
Explanation:
Combustion of carbon
C+O2=CO2
from the relationship of molar ratio
mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)
mass of carbon =1000kg
atomic mass of carbon =12
volume of CO2 produced=1000×22.4/12
volume of CO2 produced =1866.6dm3
from the combustion reaction equation provided
CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)
applying the same relationship of molar ratio
no of mole of CO2=no of mole of urea
therefore
vol of CO2\22.4=mass of urea/molar mass of urea
molar mass of urea=60.06g/mol
from the first calculation
vol of CO2=1866.6dm3
mass of urea=1866.6×60.06/22.4
mass of urea=5004.82kg
Given :
Moles of Na : 1.06
Moles of C : 0.528
Moles of O : 1.59
To Find :
The empirical formula of the compound.
Solution :
Dividing moles of each atom with the smallest one i.e 0.528 .
So,
Na : 1.06/0.528 = 2.007 ≈ 2
C : 0.528/0.528 = 1
O : 1.59/0.528 = 3.011 ≈ 3
Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.
So, empirical formula is
.
Hence, this is the required solution.
Answer:c which is 70
Explanation:
i took the test and got it right