The missing part of the question is shown in the image attached
Answer:
C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)
V= 70.4L of CO2
Explanation:
Equation of the reaction is:
C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)
Number of moles of decane = mass/ molar mass
Molar mass of decane= 122gmol-1
n= 0.370×10^3g/122gmol-1= 3.0 moles
T= 13°C +273=286K
P= 1atm
R= 0.082 atmLK-1mol-1
From :
PV= nRT
V= nRT/P
V= 3.0×0.082×286/1
V= 70.4L of CO2
Answer:
C11H25SO4
Explanation:
The total mass of the compound is 253.4 g, so, the mass of each element will be:
C: 52.14% of 253.4 = 0.5214x253.4 = 132.12 g
H: 9.946% of 253.4 = 0.09946x253.4 = 25.20 g
S: 12.66% of 253.4 = 0.1266x253.4 = 32.08 g
O: 25.26% of 253.4 = 0.2526x253.4 = 64.00 g
The molar mass are: C = 12 g/mol, H 1 g/mol, S = 32 g/mol, and O = 16 g/mol
So, to know how much moles will be, just divide the mass calculated above for the molar mass:
C: 132.12/12 = 11 moles
H: 25.20/ 1 = 25 moles
S: 32.08/32 = 1 mol
O: 64.00/16 = 4 moles
So the molecular formula is C11H25SO4
Explanation:
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