1) Neon gas can diffuse through a particular 8meter thick porous barrier in 5 mins. How many minutes would it take nitrogen to diffuese through this same barrier?

A) square root of (28/20)8^5^2

B) square root of (28/20)5^2 <--------- ANSWER

C) square root of (20/28)8^2

Solution::According to the kinetic molecular theory of gases, the average kinetic evergy of a gas particle is proportional to the absolute temperature. KE=1/2mv2=3/2kT. Upon the assumption of same temperature, both gases will have same KE.

So....let neon gas(mw:20)=a and nitroge(mw:28)=b. From the relationship, 1/2ma(8/5)^2=1/2mb(vb)^2. (8meter/5min=velocity for a). If above equation being simplified, vb=(ma/mb)^(1/2)*(8/5). But we want the time taken. Distance/velocity(speed)=time. Therefore, 8meter/(vb)=answer. If you plug vb in 8/vb and rearrange it, I believe you get the answer (b).

2) In a saturated sol'n of Magnesium hydroxide, the conc of OH ions is 2.4 x 10^-4.

Mg(OH)2 (s)<=======>Mg2+ (aq) + 2OH- (aq)

the answer is (1/2)(2.4 x 10^-4)^3

write down ksp equation first. ksp=[Mg2+]*[OH-]^2. But remember when the reaction ratio of [Mg2+]:[OH-]=1:2 by looking at the constants in front of those reactants.

So [OH]=2.4 x 10^-4 reacts with concentration of Mg2+, which is 1/2(2.4 x 10^-4). Plug the concentration of [OH] and [Mg2+] into ksp equation. then you will get the answer.

3) What is the molar solubility of silver chloride in a 0.1M sol'n of sodium chloride, given that the Ksp for silver chloride is 1.6 x 10^-10?

AgCl (s) <====> Ag+ (aq) + Cl- (aq)

THe answer is 1.6 x 10^-9.

this one is not as hard as you think. VERY VERY SIMPLE~~~

This problem is about common ion effect in ksp. What's common ion. You already have some ions, associated with one of products or reactants in the reaction (here, Cl-). Cl- from AgCl and Cl- from KCl. The solution has the Cl- ion in common from different sources.

Back to the problem I believe you know how to get solubility when only the reaction and ksp are given. Ksp=[Ag-][Cl-]=~~ when s=solubility. But in this question there are Cl- comming from another source(which already in the solution). In the solution, it already contains 0.1M of KCl, which is [K+]=0.1M [Cl-]=0.1M. Therefore, the Ksp becomes ksp=[0.1+s]~~~~=1.6 x 10^-10. But here goes the fun part. You can solve this equation s^2+0.1s- 1.6 x 10^-10=0 to get s. You have time for this ugly equation?? nono...When you carefully observe the ~~~~ from any type of ksp reaction, you find it will be usually less than 10^-4 or 5. So, you just assume 0.1+s=0.1. for example, 0.1+0.00001=0.10001. 0.00001 out of 0.1M is just a tiny number. So just ignore 0.00001. The equation becomes ksp=[0.1+s]~~~~~[0.1]~~~~=1.6 x 10^-10. Answer s=1.6 x 10^-9. YOU SHOULD BE able to get this answer without any caculation....easy..huh? tip: I have never seen common ion effect problem (like this) which requires to solve ugly #s^2+#s+...equation, even in advanced analytical chemistry from my undergrad...so...just ignore and assume [0.1+s]=[0.1]. S is way way smaller than 0.1M even 0.01M, 0.001M ~~