Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
An electron because that is the only part able to be lost or gained without nuclear action needed
I Think that the answer is 15.2096 Kilograms, but I might be wrong.
The balanced chemcial equation is B. So to do single, we need to look at where would Li stand on the metals chart (For my class we have a chart to see which element is the highest on the metals chart) . So since Li is higher, F needs to go with Li, leaving Ba alone.
I hope this helps you!
Answer:1) Volume of 
required is 55.98 mL.
2) 0.62577 grams of 
is produced.
Explanation:
1) Molarity of 
Volume of 
Molarity of 
Volume of 
According to reaction, 1 mole of 
reacts with 3 mole of , then, 0.0041985 moles of will react with:
moles of that is 0.0125955 moles.
Volume of required is 55.98 mL.
2)
Number of moles of 
According to reaction, 3 moles of gives 1 mole of , then 0.004485 moles of will give:
moles of that is 0.001495 moles.
Mass of =
Moles of × Molar Mass of
= 0.001495 moles × 418.58 g/mol = 0.62577 g
0.62577 grams of is produced.
