Answer: The enthalpy of formation of
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of 
The chemical equation for the combustion of propane follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of
is -396 kJ/mol
Answer: CrO₄⁻ and Ba²⁺
Explanation:
1) Chemical equation given:
2H⁺ + CrO₄⁻ + Ba²⁺ + 2OH⁻ → Ba²⁺ + CrO₄⁻ + 2H₂O
2) Analysis
That is an oxidation-reduction equation (some species are been oxidized and others are being reduced).
The given equation is known as total ionic equation, because it shows all the species as ions that are part of the reaction.
2) Specator ions
Spectator ions are the ions that do not change their oxidation state and are easily identified as they are the same in the reactant and product sides.
Here the ions that are the same in the reactant and product sides are:
CrO₄⁻ and Ba²⁺
3) Addtitional explanation.
Once you identify the spectator ions you can delete them from the equation to obtain the net ionic equation , which in this case turns to be:
2H⁺ + 2OH⁻ → 2H₂O
But this is not part of the question; it is some context to help you understand the use of the spectator ions concept.
Answer: 48.95g
Explanation:
no. of moles of Cl2 = 39/(2*35.5) = 0.55 mol
no. of moles of Al = 34/27 = 1.26 mol
hence, aluminium is in excess so we'll do calculation using no. of moles of Cl2 as it will be the only reactant to be used up completely. So,
no of moles of AlCl3 = 2/3 * (0.55) = 0.367 mol
hence amount of AlCl3 = 0.367 * (27+3*35.5) = 48.95g
Answer:
Q = 306 kJ
Explanation:
Given that,
Mass, m = 60 kg
Specific heat, c = 1020 J/kg°C
The temperature changes from 20°C to 25°C.
Let Q be the change in thermal energy. The formula for the heat released is given by :

Put all the values,

So, 306 kJ is the change in thermal energy.