CHEM1101 Exam HELL (1 Viewer)

[fush]

im trying to study for chem and it fucking sux, its sooo bloody hard, so much shit to remember

has anyone tried to do the 2004 june exam??? some of the answers are full of shit!! iv been trying all day to do it and i still can figure some questions out.

eg. 2nd question on page 7, how da fuck did they get 56kJ??!

http://firstyear.chem.usyd.edu.au/Pastexams/Chem1101exam(sem1)2004.pdf

http://firstyear.chem.usyd.edu.au/Pastexams/chem1101ans jun2004.pdf


and the fact that i hate chemsitry is not helping

 

[margaret_gn]

Q=mC(dT)

mass = 200g (total volume is 200mL, density is 1g/mL)
C = 4.18J/K/g
dT = 38 - 24.6 = 13.4K

therefore Q = 11202.4J for the reaction

standard enthalpy means you have to express it as KJ/mol

no. of mols = 0.2mol (mol = volume x concentration, and ratio of HCl to NaOH is 1:1)

therefore standard enthalpy dH = 11202.4/0.2 = 56012J/mol = -56.012KJ/mol (-ve cos it's exothermic)

 

[fush]

cool thanx, yea just missed that last part

 

[tennille]

I'm also doing past exam papers on chem. lol. A lot I don't know.

 

[llamalope]

chem1101 i am screwed for......... all hope is lost

 

[~*HSC 4 life*~]

pppfttt atleast you don't need to do organic chemistry in chem1101! it is sooo much harder

 

[SmokedSalmon]

oh stop ur whining... its only first year, first year barely counts towards ur degree. you just need to pass.

 

[fush]

yea passing is the problem

 

[tennille]

pppfttt atleast you don't need to do organic chemistry in chem1101! it is sooo much harder

I know in CHEM1902 we have to do organic chem. Yum.

 

[llamalope]

hmmm...... was anyone here in Bob's lectures?
Anyway, I started studying for chem1101 today, and I got through the first 8 chapters and the first two and a half assignments. I have a loooooooong way to go. :'( Ron's notes were too long and boring. Bob's lecture notes get to the point and make sense.

omg ur in my class!

and u are so far ahead of me already! freakin Ron....... He was so boring

 

[llamalope]

i always nearly fell asleep in ron's lectures... he was so boring....My notes for chem are so long already, and even then i haven't finished.... How am i gonna remember all this shit for the exams????

 

[fush]

ive given up with notes, dont bother

i think its better jsut going through the assignments and the 'typical' questions they usually ask

 

[aquaa123qwe]

so wot?
lucky u not as sick as i am

my life is falling apart

sob sob sob sob

 

[fush]

can anyone do this question? 2004 november exam

Two students determined the specific heat capacity of steel as follows. Student 1 poured 200 mL of recently boiled water into a styrofoam cup calorimeter and measured the temperature of the water to be 90.0 °C. A steel teaspoon (25 g) at room temperature (20.0 °C) was placed into the cup, where it was completely submerged. After equilibrium was presumedly reached, the temperature was measured to be 88.7 °C. Student 2 took the same cup and filled it with 200 mL of water at room temperature (20.0 °C) and placed the same teaspoon from the freezer (– 40.0 °C) into the cup. After equilibrium was presumedly reached, the temperature recorded was 19.2 °C. Determine the specific heat capacity of steel using the experimental data of each student. The specific heat capacity of water is 4.184 J g–1 K–1. Assume the density of water is 1.00 g mL–1 at all temperatures.




Answers are:
Student 1: 0.633 J K–1 g–1 Student 2: 0.452 J K–1 g–1
Student 2's answer is more accurate as the hot water (Student 1's experiment) will lose a lot of its heat to the surroundings. This effect is minimal in Student 2's experiment because the water is at room temperature.




BUT HOW!!!!??????????

 

[margaret_gn]

Student 1:
Q (water) = 200 x 4.184 x (90 - 88.7) = 1087.84
Q (steel) = 25 x c x (88.7 - 20) = 1717.5c
(where c is the specific heat capacity of steel)
Equate both Q's, and c = 1087.84/1717.5 = 0.633

Same method for Student 2.

Good Luck

 

[fush]

wow ur good....

thanx heaps by the way,
try this one

Bromine-containing compounds are even more ozone depleting than the analogous chlorine-containing compounds. In the stratosphere, an equilibrium exists between bromine and the NOx species. One of these equilibrium reactions is:
2NOBr(g) > 2NO(g) + Br2(g)
To study this reaction, an atmospheric chemist places a known amount of NOBr in a sealed container at 25 °C to a pressure of 0.250 atm and observes that 34% of it decomposes into NO and Br2. What is Kp for this reaction?

0.0113 atm is the answers

 

[fush]

im using Kp = Kc (RT)^delta n(gas)

is that what im meant to use, i got that form the lecture notes, i worked out Kc = 0.085 but its prob wrong

 

[tennille]

Initial presure of 2NOBr IS 0.25. At equilibrium, its pressure is 0.66*0.25=0.165.
Say there is a change in pressure of -2x for 2NOBr (because the pressure decreases), +2x for 2NO and +x for Br2. You can work out x from the pressures of 2NOBr.

-2x = 0.25-0.165
x=0.0425

Therefore, the equilibrium concentrations for the products are 0.085 (because it increases by +2x) for 2NO and 0.043 for Br2 (because it increases by +x).

Kp=([NO]^2[Br2])/[NOBr]^2

= (0.085^2*0.0425)/(0.165^2)
=0.0113 atm

 

[Meldrum]

Fucking chemistry, want me to beat up Maschmeyer?

I mean, he just looks so damn happy:

 

[tennille]

lol.

Too happy.