CAA: Christian Anime Alliance

If so, good for you. Temulin despises sciences that are really math with another face.

Would Pb2+ + 2Br- → PbBr2

Still be considered a redox reaction even though both reactants/products end up with the same final charge?

This isn't a redox reaction; however, I must warn you that all charges are always conserved, whether or not any of the species are being oxidized or reduced.

Thank you, thank you. I think I got it now. (hey, you're still alive Ing? I demand a new CAA tournament...)

Yes I am alive, and the Tournaments have been replaced with the Free-For-Alls, which allow more freedom since there are no predetermined brackets.

If you hate math-Chemistry so much, why not look into Organic Chemistry? (Though unfortunately OChem isn't understandable on a meaningful level without knowledge of Schroedinger's wavefunctions, which involves a lot of hairy equations with EEEEEVIL greek letters).

Unfortunately, my degree requires Chem 121 and no substitutes. I highly doubt I'd seek out another chem class after this just for kicks...

*bangs head on desk* Anyone know how to this demonic problem?

A gas sample occupies a volume of 8 L at 20°C. The temperature at which the gas would double its volume while keeping the pressure the same is ________.

I thought the P always changed with V. T__T;; I have 586 K as my answer currently, but I'm wondering if it wouldn't be 273 K.

P does change with V.....and don't forget that you have to convert Celcius to Kelvins.....

273K is basically 0Celcius......

So it looks as if your original answer is correct...


( Man! It's been forever since I did this kinda stuff! @,@)

Yay! Thank you greatly, Phantom.

A gas sample occupies a volume of 8 L at 20°C. The temperature at which the gas would double its volume while keeping the pressure the same is ________.

Okay. You're told the pressure is constant. You are given a volume 8 L, so V1= 8 L. It tells you your second volume will be twice that, thus V2 = 16 L. T1 is 20 C + 273 = 293 K. You are solving for T2. Since PV = nRT, V and T are directly proportional, so if one goes up, so must the other.

V1/T1=V2/T2 which is Charle's Law.

This can be represented by shrinking a balloon in liquid nitrogen. The cold makes the volume contract sharply.

I'm a second year Ph D in chemistry by the way. I teach freshmen chem.

Ah, thanks. That makes sense now. I'll remember you next time a certain physics major abandons me. XD

Help.

The following chemical reaction has reached equilibrium:
2NO + O2 <-> 2NO2
Calculate the equilibrium constant (Keq) if the concentrations of reactants and products are 1.26 × 10-10 M for NO, 2.28 × 10-3 M for O2, and 9.32 × 10-6 M for NO2.

a. 3.28 × 10^-7

b. 2.40 × 10^-12

c. 2.40 × 10^12

d. 2.61 × 10^-17

I vote a. o.O

In the following equation 2 Y (g) + X (g) → C (g) at 100°C the following data were obtained:

Initial concentration Y:
0.010
0.020
0.030
0.030

Initial concentration X:

0.010
0.010
0.010
0.020

Rate:

1.2 x 10^-3

2.4 x 10^-3

3.6 x 10^-3

14.4 x 10^-3

The rate of law for this reaction is __________.

a. k[Y][X]

b. k[Y]2[X]2

c. k[Y][X]2

d. k[Y]2[X]

... I have no clue. C?

Alright. The first problem is already in equilibrium so all you have to do is set up the equilibrium expression and plug in the molarities. The equilibrium expression is products over reactants where the molecules with a 2 as the coefficient are squared (the NO and NO2). So, square NO2, the divide it by O2 and NO squared.

Rate laws are set up by seeing how change in molarity effect the reaction. For Y, every time you increase the concentration, the rate increases proportionally. Double the concentration it goes twice as fast, triple it it goes three times as fast, so the Y is raised to the 1. For X when you doubled the concentration, the rate got 4 times as fast, so X is proportional to the square, meaning the exponent for X is a 2.

Ah. Okay I think I got the second one down. The thing that was confusing me with the equilibrium problem was squaring the powers of ten. I'll give it another try.

So, it's a calculator issue? Anyway, to whenever you raise a power to a power you multiply them, so a (10^-7)^2 is 10^-14.

Well, first I wasn't sure if you still raised them to the coefficient. But I still don't seem to be getting the right answer here.

I did 9.32 x 10^-12/1.26 × 10^-20 x 2.28 x 10^-6 and got 3.24^17 according to my calculator.