(a) To Determine: The respective lines of solution in graph. Calculation: T f = K f I m As, solution B having the greater value of I = 3 , which indicates that its change in boiling temperature is higher than solution A. T 0 f -T f = T f s With larger value of change in freezing temperature, the temperature of solutionwill be lower of red line. From graph, blue line having the higher freezing temperature considered as solution A whereas Solution B havethe lower temperature present with red line. For A, T f s = 14 C 0 For B, T f s = 11 C 0
(a) To Determine: The respective lines of solution in graph. Calculation: T f = K f I m As, solution B having the greater value of I = 3 , which indicates that its change in boiling temperature is higher than solution A. T 0 f -T f = T f s With larger value of change in freezing temperature, the temperature of solutionwill be lower of red line. From graph, blue line having the higher freezing temperature considered as solution A whereas Solution B havethe lower temperature present with red line. For A, T f s = 14 C 0 For B, T f s = 11 C 0
As, solution B having the greater value of I = 3 , which indicates that its change in boiling temperature is higher than solution A. T0f -Tf= Tfs
With larger value of change in freezing temperature, the temperature of solutionwill be lower of red line.
From graph, blue line having the higher freezing temperature considered as solution A whereas Solution B havethe lower temperature present with red line.
For A, Tfs = 14 C0 For B, Tfs = 11 C0
Interpretation Introduction
(b)
To Determine:
Melting temperature of pure solvent A and B
Calculation:
As, both the solutions having the same concentration So, considering the concentrations, m1 = m2 = 1 m
Freezing point of pure solvent is also considered as the melting point.
T0f =Tfs + Tf For A, Tf = Kf I m = 3 * 1 * 1 = 3 C0. T0f = 14 + 3 = 17 C0.
For B, Tf = Kf I m = 3 * 3* 1 = 9 C0. T0f = 11 + 9 = 20 C0.
Thus, for pure solvent A and B, the melting temperature are 17 C and 20 C
14. Calculate the concentrations of Ag+, Ag(S2O3), and Ag(S2O3)23- in a solution prepared by mixing
150.0 mL of 1.00×10-3 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3
Ag+ + S20
Ag(S203)¯
K₁ = 7.4 × 108
Ag(S203)¯ + S20¯ = Ag(S203)
K₂ = 3.9 x 104
ΗΝ,
cyclohexanone
pH 4-5
Draw Enamine
I
I
CH3CH2Br
THF, reflux
H3O+
I
Drawing
Draw Iminium Ion
:0: :0:
Select to Add Arrows
:0:
(CH3)2NH
:0:
■ Select to Add Arrows
:0:
:0:
(CH3)2NH
■ Select to Add Arrows
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