4.48
pH=pKa+log([A-/HA])
25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.
4 = pKa + log
4 - log
= pKa
4.48=pKa
Answer : The molal freezing point depression constant of liquid X is, 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of liquid X (solvent) = 450 g = 0.450 kg
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X = 
i = Van't Hoff factor = 1 (for non-electrolyte)
= Molal-freezing-point-depression constant = ?
m = molality
Now put all the given values in this formula, we get
Therefore, the molal freezing point depression constant of liquid X is,
The correct answer is B. The shape of a molecule where three pairs of electrons are shared is a trigonal planar. This is characterized by one central atom and three atoms forming an equilateral triangle which is bonded to the central atom.
Answer:
B) exothermic.
Explanation:
Hello!
In this case, we need to keep in mind that exothermic reactions release heat, so they increase the temperature as the final energy is less than the initial energy; in contrast, endothermic reactions absorb heat, so they decrease the temperature as the final energy is greater than the initial energy.
In such a way, when a dissolution process shows off a negative enthalpy of dissolution, we infer it is an exothermic process due to the aforementioned; therefore, the answer is:
B) exothermic
.
Best regards!
