Answer:
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Explanation:
...[1]
..[2]
..[3]
The standard enthalpy of formation of ethanoic acid :
..[4]
Using Hess's law to calculate :
2 × [1] + 2 × [2] - [3] = [4]
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Answer:
False. The balanced equation should be
2KClO3-->2KCl + 3O2
it is a decomposition reaction.
This is an ideal gas problem. The gas inside the balloon is considered ideal. Ideal gas equation is a function pressure, temperature, amount and volume. Note: amount is constant since the balloon ins closed. Pressure is maintained constant since the walls are flexible. Ideal gas equation is: PV=nRT. Put all constant in one side and variables in one.
P/nR=T/V. To find the answer to the question equate the constants of both situation
T1/V1=T2/V2
(25+273.15)/3=(x+273.15)/2
x=-74.38 degC
