Full question need, post more
Answer: 600 mL
Explanation:
Given that;
M₁ = 5.85 m
M₂ = 1.95 m
V₁ = 200 mL
V₂ = ?
Now from the dilution law;
M₁V₁ = M₂V₂
so we substitute
5.85 × 200 = 1.95 × V₂
1170 = 1.95V₂
V₂ = 1170 / 1.95
V₂ = 600 mL
Therefore final volume is 600 mL
Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %