Can you please show the work, step by step, on how to complete these Exercises R
ID: 914452 • Letter: C
Question
Can you please show the work, step by step, on how to complete these Exercises Refer to the following phase diagram, and answer the following questions: Which phase(s) occur at point B? Which phase(s) occur at point D? Which phase(s) occur when the pressure is 1.00 atm and temperature of 32 degreeC? What is the normal melting point of the substance? From this phase diagram, which phase is most dense (A, B, or C)? Water has a normal boiling point of 100.0 degreeC and a normal freezing point of 0.00 degreeC. The specific heats of the solid, liquid, and vapor are 2.09, J/g- degreeC, 4.184 J/g- degree C, and 2.03 J/g-degreeC, respectively. The heats of fusion and vaporization for water are 334 J/g and 2260 J/g, respectively. How much heat (kJ) is need to vaporize 2.50 mol H_2O at its normal boiling point? How much heat is needed to raise the temperature of 5.00 kg of water from 10.0 degreeC to 85.0 degreeC? Benzene (C_6H_6) has a normal boiling point of 80.1 degreeC. How much energy (in kJ) would be required to change 450.0 g of liquid benzene at 21.5 degreeC to a vapor at its boiling point? The specific heat of liquid benzene is 1.74 J/g-degreeC and its heat of vaporization is 395 J/g. The specific heat of silver is 0.237 J/g-degreeC, and its melting point is 961 degreeC. Its heat of fusion is 11 J/g. How much heat is needed to change 9.72 g of silver from solid at 24 degreeC to liquid at 961 degreeC?Explanation / Answer
Solution :-
Q1).
a)Phase occur at point B is liquid phase
b) phases occur at point d are solid , liquid and gas. because it is critical point.
c) at 1.00 atm and 32 C the phases will be liquid and gas.
d) 15 C is the normal melting point
e) solid phase is more dense.
Q2) a) heat needed to vaporize 2.50 mol H2O at its normal boiling point
q= Delta H vap * mass of H2O
= 2260 J per g *(2.50 mol * 18.0148 g per mol )
= 101784 J
101784 J * 1 kJ / 1000 J = 101.8 kJ
b) heat needed to raise 5.00 kg water from 10.0 C to 85.0 C
solution :-
5.00 kg * 1000 g / 1 kg = 5000 g
q= m*c*delta T
= 5000 g * 4.184 J per g C * *(85.0 C – 10.0 C)
= 1.57*10^6 J
1.57*10^6 J * 1kJ / 1000 J = 1570 kJ
Q3) first liquid benzene will rise intemeprature and then it will evaporate at constant temperature.
Q total = (m*c*delta T) +(delta H vap* m )
= (450.0 g * 1.74 J per g C *(80.1C-21.5C)) + (395 J per g * 450.0 g)
= 223634 J
223634 J * 1 kJ / 1000 J = 223.6 kJ
Q4)
First silver will rise in temperature from 24 C to 961 C and then it will melt at constant temperature
Q total = (m*c*delta T) +(delta H vap* m )
= (9.72 g * 0.237 J per g C * (961 C -24 C)) +(11.0 j per g * 9.72 g)
= 2265 J
2265 J * 1 kJ / 1000 J = 2.265 kJ