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Catwalker

Chemistry II- This week: Gas Volume Problem

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Okay, on to Quantum Mechanics. Quite a fascinating chapter. My book sucks, but that's why there's an internet to learn what the book doesn't teach.

 

Indicate which of the following sets of quantum numbers in an atom are unacceptable and explain why:

 

a. (1, 0, 1/2, 1/2)- unacceptable because the magnetic quantum number is found by (2l+1) and since it's always an integer and 1/2 is not an integer, it is incorrect.

 

b. (3, 0, 0, +1/2)- acceptable as it seems to meet the requirements of each number

 

c. (2, 2, 1, +1/2)- unacceptable as the angular momentum number is found by using the principal quantum number and has possible values from 0 to (n-1). Since 2 is not (n-1), this is not possible.

 

d. (4, 3, -2, +1/2)- I think this is acceptable. The principal and angular momentum numbers seem to check out, the magnetic quantum number checks out and +1/2 is acceptable for the spin.

 

e. (3, 2, 1, 1)- unacceptable because the magnetic spin is always either -1/2 or +1/2.

 

Am I correct in my assessment?

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Anyone know a good site that I can go to help me with Lattice Energy and Enthalpy? Lattice Energy is kicking my ass.

 

Which of the following solids would have the largest lattice energy?

 

a. NaCl

b. NaF

c. CaBr2

d. CsI

e. CaCl2

 

Through some searching on the internets, I found that the answer is c. But, I don't know why or even how to arrive at that answer.

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The section I'm on now isn't too hard, but I've encountered a problem and I can't find how to work it in the book. Sure, I can just look up the answer, but that doesn't help me.

 

What I need, if you can help, is how you arrived at the answer.

 

The OH- concentration in 7.5 x 10-3 M Ca(OH)2 is ___.

 

a. 7.5 x 10-3 M

b. 1.50 x 10-2 M

c. 1.30 x 10-12M

d. 1.00 x 10-7 M

e. 1.00 x 10-14M

 

I already know the answer is b. But I have no idea why. Anyone care to help?

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You get 2 hydroxides (OH-) from one molecule of Ca(OH)2. If your concentration of Ca(OH)2 is 0.0075 moles/L, then your concentration of OH- is twice that: 0.015 moles/L (1.5e-2 M).

 

Also, this answer is incorrect. But none of those other choices are less incorrect.

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I got the same answer... why are you saying that it's incorrect (please don't make me take out my old org. chem. book out.

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I got the same answer... why are you saying that it's incorrect (please don't make me take out my old org. chem. book out.

Ahh, never mind - Ca(OH)2 is a strong base. For some reason, I thought it wasn't.

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You get 2 hydroxides (OH-) from one molecule of Ca(OH)2. If your concentration of Ca(OH)2 is 0.0075 moles/L, then your concentration of OH- is twice that: 0.015 moles/L (1.5e-2 M).

 

 

Thanks, RD. So, it's just a matter of stoichiometry, then.

 

This explanation and equation is not in my book. I hate this book and this class.

 

If that problem is worked that way then this next problem should be worked the same way, right?

 

What is the OH- ion concentration in 5.2 x 10-4M HNO3?

 

a. 1.9 x 10-11M

b. 1.0 x 10-7M

c. 5.2 x 10-4M

d. zero

e. 1.0 x 10-4M

 

If it's worked the same way,the answer is c. Or is this one where we find the ion-product constant? (Kw) If that is the case, we use the equation you showed me and substitute it in.

 

So the equation would be Kw=[H+][OH-] or 1.0 x 10-14/5.2 x 10-4, which would give us 1.9 x 10-11, or a.

 

The only difference between the two problems is in the first we're finding simple concentration. In the second, we're finding ion concentration. I just have to look out for the word "ion," then.

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Thanks, RD. So, it's just a matter of stoichiometry, then.

 

This explanation and equation is not in my book. I hate this book and this class.

 

If that problem is worked that way then this next problem should be worked the same way, right?

 

What is the OH- ion concentration in 5.2 x 10-4M HNO3?

 

a. 1.9 x 10-11M

b. 1.0 x 10-7M

c. 5.2 x 10-4M

d. zero

e. 1.0 x 10-4M

 

If it's worked the same way,the answer is c. Or is this one where we find the ion-product constant? (Kw) If that is the case, we use the equation you showed me and substitute it in.

 

So the equation would be Kw=[H+][OH-] or 1.0 x 10-14/5.2 x 10-4, which would give us 1.9 x 10-11, or a.

 

The only difference between the two problems is in the first we're finding simple concentration. In the second, we're finding ion concentration. I just have to look out for the word "ion," then.

It doesn't have anything to do with the word "ion." It has to do with strong vs. weak bases. In a strong base situation, OH is equal to the molarity. If you have a weak base, you need to calculate the OH concentration using Kb. The same can be said with acids and H.

 

Check here: http://www.saskschools.ca/curr_content/chem30_05/5_acids_bases/acids2_3.htm

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It doesn't have anything to do with the word "ion." It has to do with strong vs. weak bases. In a strong base situation, OH is equal to the molarity. If you have a weak base, you need to calculate the OH concentration using Kb. The same can be said with acids and H.

 

Check here: http://www.saskschools.ca/curr_content/chem30_05/5_acids_bases/acids2_3.htm

 

Yeah, once I got further into the chapter, I saw that. I feel like a dolt now.

 

The link was very helpful. Thank you.

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Yeah, once I got further into the chapter, I saw that. I feel like a dolt now.

 

The link was very helpful. Thank you.

Don't feel like a dolt. Chemistry is tough, especially organic.

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Don't feel like a dolt. Chemistry is tough, especially organic.

 

 

tell me about it. Gen Chem 2 isn't exactly a stroll in the park even with a degree in it, its one of the classes I have to mildly study for when I'm TAing.

 

I'm TAing it this summer, darned classes are at 8am and I'm trying to attend so I'm not a sucky TA. But I'm not sure sleeping through the classes every day is making me any better.

 

*fingers crossed that I only have one more semester of TAing ever ever ever*

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I'm doing just okay in it. Probably a B. But it takes ALL of my time. I also do a lot of searching around on the internet for solutions and help. If it wasn't for the fact that I actually need this class to get into the PA program, I would not have taken it online. I learn so much more and can remember so much more in an actual classroom environment.

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pfft... you'll have another kid and need to TA for another year ;)

 

I'd just like to state that I find this comment...NOT funny!

 

 

Six years for a PhD is moderately tolerable, if I seriously end up going 7 years I might go jump off a cliff in my lab coat and green goggles....

 

I'm doing just okay in it. Probably a B. But it takes ALL of my time. I also do a lot of searching around on the internet for solutions and help. If it wasn't for the fact that I actually need this class to get into the PA program, I would not have taken it online. I learn so much more and can remember so much more in an actual classroom environment.

 

 

Summer classes are usually more fastpaced and difficult, but I'm sorry you have to take it online as well! Although you have to have a slight advantage in the fact that you're not sitting in the 8am lecture.....

 

Do you have to do lab? I have several PA people in class this semester as well, I was surprised by how many there were. Usually we're swarmed with Pre-Pharms...yech.

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No labs, boots. Shame, too, since I think I would learn a whole lot more from that. Plus, I learn extremely quickly in a classroom setting, so I'm also fighting that whole thing.

 

This one seemed easy at first, but I'm finding that to not be the case.

 

What is the pH of a saturated zinc hydroxide solution?

 

A table in my book gives the Ksp of [Zn(OH)2] as 1.8 x 10-14.

 

Since I have to find out how many H+ ions there are to find out the pH (or 14 minus the number of OH), I need to figure out how many OH ions there are.

 

Zn(OH)2 ==><== Zn2+ + 2OH-

 

Working backward it's a ratio of 1:2 for Zn to OH ions. That means that to find the number of OH- ions, I need to take the cube root of 1.8 x 10-14. I think. This gives me 3.7 x 10-3, which is the mol/L of OH- ions in a saturation solution. Again, I think.

 

So, to find the pOH: -log[OH-] which is -log[3.7 x 10-3].

 

Answer: 2.4318. To find pH: 14- 2.4318 = 11.5682.

 

Is this right? Why or why not? And how would I work this to figure it out the correct way?

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No labs, boots. Shame, too, since I think I would learn a whole lot more from that. Plus, I learn extremely quickly in a classroom setting, so I'm also fighting that whole thing.

 

This one seemed easy at first, but I'm finding that to not be the case.

 

What is the pH of a saturated zinc hydroxide solution?

 

A table in my book gives the Ksp of [Zn(OH)2] as 1.8 x 10-14.

 

Since I have to find out how many H+ ions there are to find out the pH (or 14 minus the number of OH), I need to figure out how many OH ions there are.

 

Zn(OH)2 ==><== Zn2+ + 2OH-

 

Working backward it's a ratio of 1:2 for Zn to OH ions. That means that to find the number of OH- ions, I need to take the cube root of 1.8 x 10-14. I think. This gives me 3.7 x 10-3, which is the mol/L of OH- ions in a saturation solution. Again, I think.

 

So, to find the pOH: -log[OH-] which is -log[3.7 x 10-3].

 

Answer: 2.4318. To find pH: 14- 2.4318 = 11.5682.

 

Is this right? Why or why not? And how would I work this to figure it out the correct way?

It's close.

 

KSP=[A]ab, right? If you take [A]=[Zn] & =[OH-], then you get:

 

KSP=[Zn]1[OH-]2.

 

If you assume that [OH-] is some value X, then [Zn]=X/2, so substituting that in gives you:

 

KSP=X/2*X2

1.8e-14=X3/2

3.6e-14=X3

X=3.302e-5 M (This is the concentration of OH-)

 

pOH=-log(3.302e-5)

pOH=4.48

pH=9.52

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Thanks, RD.

 

It makes more sense now. I messed up on a basic part of it, but I'm glad I got the general process correct.

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Do you know how to set up an ICE table? They really help me out on problems like this I set up ICE tables for all problems regarding K values (even very basic problems you don't have to do it for, I do it just to always lay it out clearly for myself and who ever might have been grading it). It helps me get an idea of how the problem is going about and also won't let you mess up mol-to-mol ratios because you have the equation right there at the top.

 

Here's a random similar problem you can look over as another example: http://answers.yahoo.com/question/index?qid=20080207094903AAF3w98

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I've seen them, but I've never set up an ICE table.

 

Why would I need an ICE table for this problem?

 

Some people are visual learners and find this way of solving problems easier and more organized. Basically, write it all out so that you don't miss something (necessary when dealing with more complex equations).

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Some people are visual learners and find this way of solving problems easier and more organized. Basically, write it all out so that you don't miss something (necessary when dealing with more complex equations).

 

 

exactly. But I always tell my students to set them up on these easier problems just to get into the practice so that later when the problems are more complex they're familiar with how to go about it.

 

 

So an ICE table is just setting up a little chart, that shows the concentrations Initially, the Change, and then @ Equilibrium. You use the equation for the top of the table, and then each species is given a column. In this problem, because you're starting with a solid you ignore the reactants because solids don't affect your K value.

 

Ksp of [Zn(OH)2] as 1.8 x 10-14.

 

Zn(OH)2 ==><== Zn2+ + 2OH-

I--------------------------0----------0

C-----------------------+x---------+2x

E-------------------------x----------2x

 

If you set your tables up like this, and you're asked to find "molar solubility" later, when you solve for X that IS your molar solubility, so just keep that in mind in case you see it in the future.

 

So then when you plug it into your Ksp equation, you just pull the values from the Equilibrium line.

 

KSP=X*(2X)2

 

Like I said, unnecessary for this problem, but it helps me visualize and helps me not to make mistakes or forget to use the correct mol-to-mol ratios.

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This is one of the last questions I'll have to do for this class. It seems like it should be easy, but I'm not sure what route I should take to solve it.

 

A concentration of 8.00 x 102 ppm by volume of CO (28.01 g/mol) is considered lethal to humans. Calculate the minimum mass of CO in grams that would become a lethal concentration in a closed room that is 17.6m long, 8.8m wide and 2.64m high. The temperature and pressure are 20.0 C and 756mm Hg, respectively.

 

The first thing is to figure out the volume of the room: 408.88 m3 or 408880L.

 

Also, we're looking at .0008% of the gas will be CO.

 

I've thought about using PV=nRT to figure out how many moles of gas might be in the room. In this case, (.995 x 408880)/n(.082057 x 293), n = 16921.4 moles of gas in the room. (That seems absurdly high.) Then we use Avogadro's number to figure out how many molecules are in the room: 1.019 x 1028. Then I should just have to find out what .0008% of that is to determine the answer: 8.15204 x 1023 molecules of CO, which works out to 1.35 moles of CO. That is .04833g.

 

That seems like it might be right, half of a half of a gram of CO. Is it?

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In case people are thinking "Why do we have to learn these things that we'll never use in real life"

 

I have to do exactly the kind of problems that Cat is working on in my job, and it drives me batshit insane because I never learned to do it in school.

 

Good luck with it Cat, I would have offered my help but we really do have people here much better qualified to assist you.

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First, 1.35 mol of CO is not .04g; you already know that 1 mol = 28.01g.

 

Second, this is a problem where DIMENSIONAL ANALYSIS IS YOUR FRIEND YES IT IS DO IT PUT UNITS IN YOUR EQUATION.

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