Hello, a few questions: Thanks so much!
1. The heat capacities of titanium and water are 0.522 J g-1 K-1 and 4.184 J g-1 K-1, respectively. What is the final temperature of the titanium-water mixture if a 175.0-g sample of titanium, initially at 90.0oC, is added to 50.0 g of water, initially at 30.0oC?
2.
What is w when the volume of a gas expands from 22.3 L to 41.3 L against a constant external pressure of 731 mmHg? (Remember to include a "+" or "-" sign as appropriate.)
3.
What is w when 1.27 kg of H2O(l), initially at 25.0oC, is converted into water vapour at 133oC against a constant external pressure of 1.00 atm? Assume that the vapour behaves ideally and that the density of liquid water is 1.00 g/mL. (Remember to include a "+" or "-" sign as appropriate.)
4. What is the internal energy change for the system if 901 joules of heat flows from the system into the surroundings and 453 joules of work is done on the system by the surroundings?
5.
A sample of gas is compressed from 13.57 L to 3.93 L using a constant external pressure of 2.91 atm. At the same time, 1,759 J of heat flows into the surroundings. What is the internal energy change for the gas?
6.
A 0.016076-mol sample of an organic compound was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 23.7oC to 24.5oC. If the heat capacity of the calorimeter is 4.98 kJ (oC)-1, then what is the constant volume heat of combustion of this compound, in kilojoules per mole? (Remember to include a "+" or "-" sign in your answer, as appropriate.)
7. The enthalpy change for the reaction below at 25oC is -2,051 kJ (per mole of C10H22). What is the internal energy change for the reaction at 25oC?
10 CO(g) + 21 H2(g) --> C10H22(l) + 10 H2O(l)
Enter your answer in kJ (per mole of C10H22), rounded to the nearest kilojoule.
8. The standard enthalpy change for the reaction below is -3908.7 kJ (per mol C7H8). What is the standard enthalpy of formation of C7H8(l)?
C7H8(l) + 9 O2(g) --> 7 CO2(g) + 4 H2O(l)
Data:
Substance
Standard enthalpy
of formation
(in kJ mol-1)
CO2(g) -393.5
H2O(l) -285.9
9. The standard enthalpy of combustion of C2H6O(l) is -1,367 kJ mol-1 at 298 K. What is the standard enthalpy of formation of C2H6O(l) at 298 K? Give your answer in kJ mol-1, rounded to the nearest kilojoule. Do not include units as part of your answer.
Note: The standard enthalpies of formation of CO2(g) and H2O(l) are -394 and -286 kJ mol-1, respectively, at 298 K.
10.
Given the following thermochemical equations,
NO(g) + (1/2) Cl2(g) --> NOCl(g) -37.78 kJ (per mol NOCl)
NO(g) + (1/2) O2(g) --> NO2(g) -56.53 kJ (per mol NO2)
2 NO2(g) --> N2O4(g) -58.03 kJ (per mol N2O4)
what is the standard enthalpy change for the reaction below?
N2O4(g) + Cl2(g) --> 2 NOCl(g) + O2(g)
Give your answer in kJ (per mol N2O4), accurate to two decimal places.
11. When a 6.79-g mixture of methane, CH4, and ethane, C2H6, is burned in oxygen at constant pressure, 369 kJ of heat is liberated. What is the percentage by mass of CH4 in the mixture? The standard enthalpies of combustion for CH4 and C2H6 are -890.3 kJ mol-1 and -1569.7 kJ mol-1, respectively.
Thank you so much again!!
Most of us don't have the time to work 10 or more time consuming questions. I'll do the first one; if you post them separately someone may get to them.
1. heat lost by Ti + heat gained by H2O = 0
[mass Ti x specific heat Ti x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf which is the only unknown.
2. work is p*delta V. The gas is doing th work so it is -.
4. delta E = q + w
q = +901
w = +453
Ah! okay! But thank you so much for answering a few!
1. To find the final temperature of the titanium-water mixture, we can use the principle of energy conservation. The heat gained by the titanium will be equal to the heat lost by the water.
First, calculate the heat gained or lost by titanium:
q_titanium = m_titanium * c_titanium * ∆T_titanium
Where:
m_titanium = mass of titanium = 175.0 g
c_titanium = heat capacity of titanium = 0.522 J g^(-1) K^(-1)
∆T_titanium = change in temperature of titanium = final temperature - initial temperature = final temperature - 90.0°C
Next, calculate the heat gained or lost by water:
q_water = m_water * c_water * ∆T_water
Where:
m_water = mass of water = 50.0 g
c_water = heat capacity of water = 4.184 J g^(-1) K^(-1)
∆T_water = change in temperature of water = final temperature - initial temperature = final temperature - 30.0°C
Since the heat lost by the water is equal to the heat gained by the titanium (assuming no heat loss to the surroundings):
q_water = -q_titanium
Now, equate the two equations and solve for the final temperature:
m_water * c_water * ∆T_water = -m_titanium * c_titanium * ∆T_titanium
(50.0 g) * (4.184 J g^(-1) K^(-1)) * (∆T_water) = -(175.0 g) * (0.522 J g^(-1) K^(-1)) * (∆T_titanium)
Simplify the equation and solve for ∆T_water:
(50.0 g) * (4.184 J g^(-1) K^(-1)) * (∆T_water) = -(175.0 g) * (0.522 J g^(-1) K^(-1)) * (∆T_titanium)
∆T_water = -(175.0 g) * (0.522 J g^(-1) K^(-1)) * (∆T_titanium) / (50.0 g) * (4.184 J g^(-1) K^(-1))
Now substitute ∆T_water into the equation and solve for the final temperature:
final temperature = initial temperature of water + ∆T_water
Plug in the values and calculations into the equation to find the final temperature.