Carbon dioxide is through the plants
Answer:
The answer is 0.01 M
Explanation:
The problem is solved by applying the expression for ionic product of water as follows:
Kw = [H₃O⁺] [OH⁻]
Where Kw is the ionic product of water and it is 1.10⁻¹⁴ M at⁴ 25ºC. As [OH⁻]= 1.10⁻¹² M, [H₃O⁺] will be:
[H₃O⁺]= Kw/ [OH⁻]= 1.10⁻¹⁴M/1.10⁻¹²M= 0.01 M
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
Answer:
Gas
Explanation:
Because Crude oil can usually be found in the ground as a liquid and in the air is gas can be kerosene.
