Answer:
C2H2O4
Explanation:
To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :
C = 26.86%
H = 2.239%
O = 100 - ( 26.86 + 2.239) = 70.901%
We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u
The division is as follows:
C = 26.86/12 = 2.2383
H = 2.239/1 = 2.239
O = 70.901/16 = 4.4313
We now divide each by the smallest number I.e 2.2383
C = 2.2383/2.2383 = 1
H = 2.239/2.2383 = 1
O = 4.4313/2.2383 = 1.98 = 2
Thus, the empirical formula is CHO2.
To get the molecular formula, we use the molar mass .
(CHO2)n = 90
We add the atomic masses multiplied by n.
(12 + 1 + 2(16))n = 90
45n = 90
n = 90/45 = 2.
Thus , the molecular formula is C2H2O4
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
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