I'm having trouble answering this problems...please help!?

Question

i have 86 chemistry problems to solve out of those 86 problems I don't understand 6 if you could please help me...I appreciate thanks
1).How much heat must be removed to condense 15.75g H2O(g) at 100 degree C to produce 15.75g h2o(l) also at 100 degree C? Give answer in cal units.
2).A sample of N2 gas at STP occupies a volume of 1250mL. If the temperature and pressure of the gas is raised to 28degree C and 1.35atm respectively, what volume in liters will the gas occupy? Assume that the number of moles of gas is constant.
3)A 2.50L sample of nitrogen gas is at standard pressure and 21 degree celcius. What celcius temperature is required to increase the pressure to 115kPa and reduce the volume to 2.35L? Assume that the amount of gas is constant.
4)Calculate [H+] of a solution of HCl in water havong a pH=3.8. of the solution.
5)If 4.8375kJ of heat is required to warm a solid gold bar from 21degree C to 36 degree C, what is the mass of the bar in grams?

6) A medical institution requests 1.00 gram of Bi-214, which has a half life of 20 min, from a medical radiology pharmacy. How much grams of Bi-214 must be prepared by the pharmacy if the shippping time is 2 hrs?
7)Calculate the amount of heat in "cal" required to warm water from 25.0degree C to 95.0C to prepare a cup of hot tea? Assume that 1.0cup=240mL and that the density of H2O =1.0g/mL.

I still don't get some of the problems for example problem 2 and 3 and 6 plese help me!

Answer 1

1.
Using Q = mcdeltaK
Delta k = 0
therefore Q = 0
However, the enthalpy change from gas to liquid is -44 kJmol^-1
There is 15.75/18 = 0.875 moles(H2O).
This will involve a heat transfer of -44 x 0.875 = -38.50 kJ
-38.50 kJ = -38500 J
-38500/4.184 = -9202 Cals.

4.
pH = 3.8
[H+] = 10^-3.8
[H+] = 1.585 x 10^-4 mol l^-1 = 0.0001585 mol l^-1

5.
Specific Heat(Gold - Au) = 130 J kg^-1 K^-1
4.8375 kJ = 4837.5 J
Temperature difference = 36 -21 = 15 K

Using Q = mcDeltaT
then m = Q/cDeltaT
m = 4837.5 J / (130 J kg^-1 K^-1 x 15K)
m = 2.481 kg
m = 2481 g.

Answer 2

These problems seem designed to scare you. don't let them.

1) I assume you are given the molar heat of vaporisation of water. Multiply that by the number of moles you have in 15.75 g to get your answer. If your input data is in J, just use

1 cal = 4.184 J

2) and (3): Use PV = nRT, so since n is constant, you can use

P1V1/T1 = P2V2/T2

In each case, rearrange using symbols so that the one thing you want to find out is the subject of the equation, and then plug in the numbers. Or, if you prefer, use the first equation to find n and take it from there.

4) Use the definition pH = - log(base 10)[H+]

and invert the math to give you

[H+] = 10^(-pH)

5) I have to assume you have been given the specific heat of gold.

mass (g) x specific heat (J/g) x temperature change = heat required = 4,837.5 J

6) 2 hr is 6 half-lives. So they need to ship 2^6 (= 64) x as much material as is needed at the far end. I don't believe a word of it, but that's the right answer.

7) Just like (5). The specific heat of water is 1 cal degree-1 g-1 (original definition of the calorie).

Chemistry is fascinating. Don't let anybody put you off it.

Over to you, and good luck!

Answer 3

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