There would be a direct result as an increase in the solute temperature will result in increase in its solubility. A greater amount of solute molecules will possess more kinetic energy and will be distributed and in container.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
idk
Explanation:
idk cool pee bee mee nee hee gee fee kee
you add the masses of the reactants, because of conservation of mass. if there are two or more products they will ask you to find the mass of only one product or the sum of the mass of all products
Answer:
A3B3
Explanation:
Molecular formula = n x empirical formula
(AB) n = 90
MM of AB = 30 g/mol
30n = 90
Divide both side by the coefficient n i.e 30
n = 90/30 = 3
Molecular formula = n x empirical formula
Molecular formula = n x (AB)
Molecular formula = 3(AB) = A3B3
