Melting points and boiling points of molecular compound are usually lower than ionic compounds. This is so as only a small amount of energy is required to overcome the weak intermolecular forces of attraction (Van de Waals forces) thus having lower m.p. and b.p.
Ionic compounds require a large amount of energy to overcome the strong electrostatic forces of attraction between the ions, hence having higher mp and bp
It can be made true by changing "cannot" to "can".
The phase's composition is as follows: 27
What is the alloy's composition?
Both have mass fractions of w a = w b = 0.5.
A-B alloy composition; C o = 57 wt% B - 43 wt% A
C b = 87 wt% B - 13 wt% A is the new phase composition.
Using the Lever rule, we can calculate the mole fraction (x i) or mass fraction (w i) of each phase of a binary equilibrium phase as follows:
W a = W b = 0.5
This provides us with;
0.5 = (C b - C o)/(C b - C a)
(87 - 57)/(87 - C a) = 0.5
30/0.5 = 87 - C a
60 = 87 - C a
C a = 87 - 60
C a = 27
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Heat gained by ice cubes would be equal to the - heat lost by warm water
The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol
Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J
Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C
In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:
q(ice/water) = - q(warm water)
moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)
50.5 g / 18.0 g/mol = 2.81 mol
2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)
16916 + 211.3T2 = -669.4 T2 + 53555
36639 = 880.7 T2
T2 = 41.6 C
