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| | From:  Albert1145 (Original Message) | Sent: 8/18/2008 2:23 PM |
Hi Steve,
I'm so sorry for the trouble. I doing online pretest, but just can't get these 4 questions out! Could you show me how you convert each values and which times which so I can compare with mine? I have so many versions of answers for each question and I don't know which one is right?
Q2:
The heat of neutralisation of HCl by NaOH is given below:
HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) Δ H= -56.2 kJ/mol
What is the value of qp (in kJ) when 125 mL of 1.750 M HCl is mixed with 184.5 mL of 0.667 M NaOH?
Q3:
0.1326 g of aluminium was completely combusted in a constant-volume bomb calorimeter having a heat capacity of 1410 J/oC.
The bomb calorimeter was surrounded by 800 g of water. The temperature of the bomb and the water increased by 1.726 oC.
Calculate the heat of combustion of aluminium in J/mol. Which of the choices below is the correct equation for the calculation? The specific heat of water is 4.18 J deg-1 g-1.
A. [1410+(800)*(4.18)]*(1.726)*(26.96)/(0.1326)
B. (800)*(4.18)*(1.726)/(26.98)
C. (1410)*(1.726)/(0.1326)
D. [1410+(800)*(4.18)}*(1.726)/(26.96)
Q12
A helium balloon had a volume of 22.0 L at sea level where the atmospheric pressure is 0.951 atm. When the balloon rose to a height at which the atmospheric pressure is 73.3 kPa, the balloon burst. What was the volume of the balloon (in L) before it burst?
Q7
Two gas flasks, A and B, are kept at the same temperature and pressure. Flask A has a volume of 10 L and contains 10 g of methane gas, CH4. Flask B has a volume of 5.6 L and contains carbon dioxide gas. Calculate the mass of carbon dioxide (in g) in flask B.
Thank you very much
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| | | From: ·Steve· | Sent: 8/18/2008 10:16 PM |
Hi Albert, here you go:
Q2. Calculate moles of HCl and moles of NaOH first (moles = volume in liters X molarity).
Then see which is the limiting reactant. That will be the number of moles of HCl neutralized.
The reaction tells us that 56.2 kJ of heat is released per 1 mole of HCl neutralized (the coefficient in front of HCl is one). Therefore your calculation is,
Your moles of HCl neutralized X 56.2 kJ =
1 1 mole HCl
Q3. 1) Calculate the moles of aluminum.
2) Calculate the heat absorbed by the reaction chamber: 1410 J /°C X ΔT
3) Calculate the heat absorbed by the water: (800 g)(ΔT)(4.184 J/g°C)
4) Add the two heats together to get the total heat released by the reaction.
5) Divide the total heat by the moles of aluminum to get the answer in J/mol.
Q12. Convert 73.3 kPa to atm ( 1 atm = 101.325 kPa). Then you can use Boyle's law, P1V1 = P2V2, to calculate V2 (V1 = 22.0 L, P1 = 0.950 atm, P2 = 73.3 kPa converted to atm, V2 = ?).
Q7. Convert the 10 g of CH4 in flask A to moles. The volume is given as 10 L. Now plug into PV = nRT to get the ratio P/T:
P = nR
T V Once you know P/T, plug in the value of V for flask B and solve for the moles of CO2 gas, n, and then multiply by the molecular weight 44.0095 g/mol to convert to grams. 
Steve
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Thank you Steve, it turned out our lecturer hasn't tought us how to solve some of the question.
I also tried these two last night, could you help me?
Question 9
Suppose 42 mL of hydrogen and 69 mL of nitrogen, each at 1 atm pressure and 0C are transferred to the same container. What is the pressure of the mixture in atm at 0C if the volume of the container is 90 mL? (assume ideal gas behaviour)
Question 11
A sample of nitrogen gas occupies a volume of 21 L at 124C and 726 mm Hg. What volume (in L) would the gas occupy at 0C and 1.00 atm? (assume ideal gas behaviour)
Thank you so much |
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Reply
| | | From: ·Steve· | Sent: 8/19/2008 6:06 AM |
Q9. Here are several ways to work this problem: Method 1. Use PV = nRT to calculate the moles of each gas, hydrogen and nitrogen. n = PV / RT where P is the pressure in atmospheres, V is the volume in liters, R is the gas constant 0.08206 L atm mol�? K�?, and T is the Kelvin temperature. Call them nH2 and nN2. Then add them together to get the total moles of gas, ntot. Now you can plug back into PV = ntotRT again (but this time V is 0.090 L) and solve for the pressure. T is still 273.15 K (0°C). Method 2. Alternatively, you can add the two volumes, 0.042 L + 0.069 L, and plug the total volume into PV = nRT, where P is 1 atm and T is 273.15 K, and solve for n. You should get the same value of n this way as ntot above. Plug back into PV = nRT again and solve for the pressure as before. Method 3. Or, since ntot is not changing and T is also constant at 273.15 K, you can simply use P1V1 = P2V2 where P1 = 1 atm, V1 = 0.042 L + 0.069 L, P2 = ?, and V2 = 0.090 L, and solve for P2. This is the simplest way. 
Q11. For this one, you just have to plug into the combined gas law,
P1V1 = P2V2
T1 T2
and solve for V2. Be sure the units of pressure are the same on both sides (either both in atm or both in mmHg) and that the temperatures are both in Kelvins.
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Thank you so much Steve
I just have one more on this module. I think this question is really easy, but I just can't get it!
How many grams of helium are required to fill a 2.3 L balloon to a pressure of 1.10 atm at 25C?
I did PV=nRT, so n= RT / PV
So I went, (0.08206 x 298.15) / (1.1 x 2.3) and I got 9.67
and then 9.67 x 4.003 = 38.68g
But the answer is 0.414g |
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OH MY GOD! What am I doing????
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