Answer:
A. Sublimation of camphor
Answer:
sample B contains the larger density
Explanation:
Given;
volume of sample A, V = 300 mL = 0.3 L
Molarity of sample A, C = 1 M
volume of sample B, V = 145 mL = 0.145 L
Molarity of sample B, C = 1.5 M
molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol
Molarity is given as;

The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g
The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g
The density of sample A 
The density of sample B 
Therefore, sample B contains the larger density
Answer:
it must be testable I think that's the answer
The solution for this problem is:
C6H5NH3Cl is a strong salt: C6H5NH3+ + Cl- C6H5NH3+ + H2O <-----> C6H5NH2 + H3O+
K = Kw/ Kb= 1.0 x 10^-14 / 3.8 x 10^-10= 2.6 x 10^-5
=2.6 x 10^-5 = x^2 / 0.240-x
x = [H3O+] = 0.00251 M
pH = 2.60 is the concentration