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.
Say the dereference between the two is that "Exo" means "outside" and "Endo" means "inside" meaning one pulls heat inside (endothermic) and the other expels heat (exothermic)
<span>To
solve this we assume that the gas is an ideal gas. Then, we can use the ideal
gas equation which is expressed as PV = nRT. At number of moles the value of PV/T is equal to some constant. At another
set of condition of temperature, the constant is still the same. Calculations
are as follows:</span>
P1V1/T1 = P2V2/T2
P2 = P1 (V1) (T2) / (T1) (V2)
P2 = 475 kPa (4 m^3) (277 K) / (290 K) (6.5 m^3)
P2 = 279.20 kPa
Therefore, the changes in the temperature and the volume lead to a change in the pressure of the system which is from 475 kPa to 279.20 kPa. So, there is a decrease in the pressure.
Convert to grams to moles, convert g of water to litres, divide moles by litres = M
