Answer:
The answer is supposed to be "Electron cloud" or "Electon".
Answer:
872.28 kJ/mol
Explanation:
The heat released is:
ΔH = C*ΔT
where ΔH is the heat of combustion, C is the heat capacity of the bomb plus water, and ΔT is the rise of temperature. Replacing with data:
ΔH = 9.47*5.72 = 54.1684kJ
A quantity of 1.922 g of methanol in moles are:
moles = mass / molar mass
moles = 1.992/32.04 = 0.0621 mol
Then the molar heat of combustion of methanol is:
ΔH/moles = 54.1684/0.0621 = 872.28 kJ/mol
Answer:
2nd one down
Explanation: distance divided by time interval
M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol
m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol
2C + O₂ → 2CO
1 mol 0.625 mol 1 mol
0.625-0.5=0.125 mol
2CO + O₂ → 2CO₂
0.250 mol 0.125 mol 0.250 mol
n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g
n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
