Answer:
0.075 moles n=m/M so divide the mass (m) by the molar mass (M) to get the n which is the number of moles
Explanation:
Density is mass by volume
1 mL of nitric acid solution has a mass of 1.36 g
therefore 1000 mL has a mass of 1360 g
the molarity of solution is 2.48 mol/L
1 L of solution has 2.48 mol
the mass of 2.48 mol of HNO₃ is - 2.48 mol x 63 g/mol = 156.24 g
this means that 156.24 g of the solution consists of nitric acid
therefore percentage by mass - mass of nitric acid / total mass x 100%
concentration as percentage by mass = 156.24 g / 1360 g x 100%
percentage = 11.5%
Answer:
atomic mass=Protons+Neutrons=29+35=64
Explanation:
please complete question.
Answer:
pOH = -log[OH – ]
pH + pOH = 14
Explanation:
I did not understand the question but hope this help!
When the titration of HCN with NaOH is:
HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)
So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1
we need to get number of mmol of HCN = molarity * volume
= 0.2 mmol / mL* 10 mL = 2 mmol
so the number of mmol of NaOH = 2 mmol according to the molar ratio
so, the volume of NaOH = moles/molarity
= 2 mmol / 0.0998mL
= 20 mL
and according to the molar ratio so, moles of CN- = 2 mmol
∴the molarity of CN- = moles / total volume
= 2 mmol / (10mL + 20mL ) = 0.0662 M
when we have the value of PKa = 9.31 and we need to get Pkb
so, Pkb= 14 - Pka
= 14 - 9.31 = 4.69
when Pkb = -㏒Kb
4.69 = -㏒ Kb
∴ Kb = 2 x 10^-5
and when the dissociation reaction of CN- is:
CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq)
by using the ICE table:
∴ the initials concentration are:
[CN-] = 0.0662 M
and [HCN] = [OH]- = 0 M
and the equilibrium concentrations are:
[CN-] = (0.0662- X)
[HCN] = [OH-]= X
when Kb expression = [HCN][OH-] /[CN-]
by substitution:
2 x 10^-5 = X^2 / (0.0662 - X)
X = 0.00114
∴[OH-] = X = 0.00114
when POH = -㏒[OH]
= -㏒ 0.00114
POH = 2.94
∴PH = 14 - 2.94 = 11.06
