<span>A solution is somthing desolved in somthing else. By desolved i mean it needs to have some particals ionized a solid you place in water that dissosiates (ions split apart from each other) makes a solution a good solution you can make in your kitchen is a salt-water solution, Put some regular table salt in a glass and stir it and you will notice the salt "disapears" what happens is the sodium ions and the chloride Ions seperate and 'hide' between water molocules.
In basic terms only some substances can make a solutions others are refered to as insoluble as they can't be seperated in water or another solvent. In actuality however all ionic compounds (compounds that are composed of ions) are at least somewhat soluble, but don't dissociate well at all in some solvents.
Hope that helps</span>
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:
pH = 3.65
Explanation:
given data
pKa of HNO2 = 3.40
nitrous acid (HNO2) = 0.110 M
NaNO2 = 0.200 M
to find out
What is the pH
solution
we get here ph for acidic buffer that is express as
pH = pKa + log(salt÷acid) ........................1
put here value and we get
pH = 3.40 + log(0.200÷0.110)
pH = 3.65
