<span>This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles. </span>
Answer:
The correct options are "b" and "c". A further explanation is given below.
Explanation:
- Saturation temperature can be determined where this enough of some other solution that is incorporated like that can be absorbed by a solvent.
- The formulation is saturated at this same stage, so Ksp could be computed. As well as the solid throughout solution should continue to appear upon freezing below a certain temperature.
The other options offered aren't relevant to the situation described. So the equivalents above are the right ones.
Answer:
lignands, the central atom/metal ion
Explanation:
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
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
