Answer:
a) 3.969
b) 3.489
Explanation:
a) Calculate the pk, value of the acid HA
PH of salt hydrolysis
P
= 1/2 ( pkw + pka + logC )
8.7 * 2 = 14 + log ( 0.27 ) + Pka
∴ Pka = 3.9686 ≈ 3.969
b) Calculate the PH of a solution containing 0.3 M HA and 0.1 M NaA
PH = Pka + log ( salt / acid )
= 3.9686 + log ( 0.1 / 0.3 )
= 3.9686 - 0.48 = 3.489
Answer:

Explanation:
Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations. Only the species which are present in aqueous state dissociate. So, the net ionic equation of aqueous solution of ammonia is shown below as:-

Answer:

Explanation:
Hello!
In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

Best regards!
Mol= mass (grams) /Mr
Mr of Sulfuric Acid (H2SO4): 98
mol= 329/98
=3.36 moles
Answer:
I think it would be the last answer
Explanation: