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[Anticitizen One]

There are two questions on this assignment i'm not sure of the answers to:

Which of the following substances is more likely to dissolve in water?
A. O

CH3(CH2)9CH
B. HOCH2CH2OH
C. CCl4
D. CHCl3
E. CH3(CH2)8CH2OH


2. When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is __________.
A. AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq)
B. Ag+ (aq) + I- (aq) → AgI (s)
C. Ag+ (aq) + NO3- (aq) → AgNO3 (aq)
D. Ag+ (aq) + NO3- (aq) → AgNO3 (s)
E. AgNO3 (aq) + KI (aq) → AgI (aq) + KNO3 (s)

 

[shibby]

The first one should be B, because it is more polar than the rest of the options.

The second one should be A.

 

[Al-ibn Kermit]

generally, you should assume that something polar is more likely to dissolve in a polar solution (ie: H2O) and vice versa for something non-polar.

The second one is simpler though because it tells you that you'll have AgI precipitate in solid form so there's only two options, A and B, and only A tells you gives you the full stoichiometry.

 

[Visualante]

I think 1 is a trick question. I'd say E bcause it has two alcohol groups compared to B's one. So it's more likely to dissolve readily in water.

Wait I forgot about the order of dis associativity. I think the CCl4 will go. Chlorine is very kean to give up electrons. It would probably Form HCl gas with the water instantly.

 

[Scrow]

I misread the thread title as, "Anyone here good at Christianity?"

did a double take and lol'd

 

[Husker86]

I think 1 is a trick question. I'd say E bcause it has two alcohol groups compared to B's one. So it's more likely to dissolve readily in water.

Wait I forgot about the order of dis associativity. I think the CCl4 will go. Chlorine is very kean to give up electrons. It would probably Form HCl gas with the water instantly.

E for one contains a very long carbon chain, not good for solubility; and B has two alcohol groups anyway. edit: Wait, E doesn't even have 2 alcohol groups, so...yeah.

Cl wouldn't break a covalent bond for an ionic.

 

[Visualante]

E for one contains a very long carbon chain, not good for solubility; and B has two alcohol groups anyway.

Cl wouldn't break a covalent bond for an ionic.

Wow and I got a B in chemistry. Was going to read it at degree level too. 2 years later it's all fucking gone. :lol

edit. Also it's 4am :|

 

[Costanza]

ask salva

 

[Erudite]

2 is B.

Net Ionic Equation is only the ions that make the solution a solid if I recall correctly.

 

[Al-ibn Kermit]

CCl4 isn't polar.

 

[spermatic cord]

Me

 

[Tornado Condor]

The second one is B for sure

 

[FairyD]

Me

I was thinking the exact same thing.

 

[Lamel]

Number 2 is B. It is asking for NET IONIC equation, not balanced equation.

Edit: Did not mean to quote salva, unless he stealth edited :lol

 

[A Penguin]

2 is B.

Net Ionic Equation is only the ions that make the solution a solid if I recall correctly.

Yup, not sure what the first two posters were thinking.

 

[Visualante]

The second one is B for sure

Yeah principle of black and white photography if I remember correctly. Crystals form the image on the photograph paper.

 

[Tornado Condor]

Yes i messed up by saying A. I just looked at my chemistry notes and yes, it is B.

Stealth ninja editing FTW!

 

[Lamel]

Yes i messed up by saying A. I just looked at my chemistry notes and yes, it is B.

Stealth ninja editing FTW!

Ha, common error. Only reason I remember is because our chemistry midterm was last week.

 

[Tornado Condor]

Ha, common error. Only reason I remember is because our chemistry midterm was last week.

I'm on the second semester of general chemistry, dealing with entropy and some other bullshit. I should know last semester's stuff since i'm going to tutor it!

le sigh...

 

[Dr.Guru of Peru]

I think the CCl4 will go. Chlorine is very kean to give up electrons. It would probably Form HCl gas with the water instantly.

That would be very bad for organic chemists.

 

[shibby]

Yeah sorry, I didn't read the net ionic part of the second question.

 

[watervengeance]

I'm on the second semester of general chemistry, dealing with entropy and some other bullshit. I should know last semester's stuff since i'm going to tutor it!

le sigh...

For a moment there, I thought you were my friend. But he doesn't go on GAF. (or does he?!?!)

 

[liquid_gears]

The answer to the first question is B. Ethylene glycol is a diol and so is much more polar than the other (much larger) molecules. It doesn't look polar when drawn out, but it can form hydrogen bonds between each OH group and so the dipole is concentrated on one side of the molecule making it very polar. This is the stuff that's in antifreeze btw and I've seen tramps drinking this stuff in A&E.

The answer to the second one is B. That one is obvious. It's the only half equation that you are interested in with the right end product.

BTW: I don't know how you Americans do things, but I've got ISIS DRAW on a pendrive. It's hard to come by and it's a very useful/professional piece of software to draw molecules out. It's useful for lab reports/assignments. It's a legal copy so if anybody wants it then just PM me. I also have Scigress, a powerful semi-empirical modelling program.

I'm on the second semester of general chemistry, dealing with entropy and some other bullshit. I should know last semester's stuff since i'm going to tutor it!

le sigh...

My girlfriends brother wants to study Chemistry at College. Do you mind saying what sort of stuff that you are covering on your course. I'm just curious as to the differences between the British and American systems.

 

[Raist]

B

B

 

[Flash]

die general chemistry.

organic chemistry ftw though. especially organic synthesis.

 

[Tornado Condor]

The answer to the first question is B. Ethylene glycol is a diol and so is much more polar than the other (much larger) molecules. It doesn't look polar when drawn out, but it can form hydrogen bonds between each OH group and so the dipole is concentrated on one side of the molecule making it very polar. This is the stuff that's in antifreeze btw and I've seen tramps drinking this stuff in A&E.

The answer to the second one is B. That one is obvious. It's the only half equation that you are interested in with the right end product.

BTW: I don't know how you Americans do things, but I've got ISIS DRAW on a pendrive. It's hard to come by and it's a very useful/professional piece of software to draw molecules out. It's useful for lab reports/assignments. It's a legal copy so if anybody wants it then just PM me. I also have Scigress, a powerful semi-empirical modelling program.



My girlfriends brother wants to study Chemistry at College. Do you mind saying what sort of stuff that you are covering on your course. I'm just curious as to the differences between the British and American systems.

First semester was basic stuff such as stoichemitry, gases (ideal gas law), using light spectrum, Lewis dot structures, etc. Basic stuff.

This semester is more entropy, acids and bases, and well I'm not sure what else, I have yet to see what's on this course.

 

[Flash]

First semester was basic stuff such as stoichemitry, gases (ideal gas law), using light spectrum, Lewis dot structures, etc. Basic stuff.

This semester is more entropy, acids and bases, and well I'm not sure what else, I have yet to see what's on this course.

I'm guessing you'll touch up on rate laws as well.

 

[Aesius]

I was thinking the exact same thing.

That John Denver's full of shit, man.

 

[Dooraven]

There are two questions on this assignment i'm not sure of the answers to:

Which of the following substances is more likely to dissolve in water?
A. O

CH3(CH2)9CH
B. HOCH2CH2OH
C. CCl4
D. CHCl3
E. CH3(CH2)8CH2OH

Start by eliminating options, tetrachloromethane (CCl4) is not polar because all 4 polar sides cancel each other out so C is obviously incorrect (using the like dissolves in like rule).

A is an aldehyde, both aldehydes and ketones are soluble in water but their solubility decreases with increasing Carbon chain length and this aldehyde has an extremely large chain so it is an unlikely candidate for the best soluble. (This is of course assuming that the O is meant to be attached to the right hand side, because attaching it to the left hand side doesn't make any sense as the Carbon would have 3 Hydrogens, 1 oxygen and a R group (for a total of 6 bonds)).

D is tricholromethane, which is a suitable candidate for polarity because one of the C-Cl bonds is not cancelled out by the C-H bond so it has some polarity and it is very short. But the C-CL bond is less electronegative than the O-H bond. So there might be better options

E is a hydroxyl compound (in this case an alcohol) and has a O-H bond which makes it polar but it also has a large chain which significantly decreases its polarity.

Therefore the best answer is B, 1,2 Ethanediol (which is a diol) while longer than tricholomethane has two polar OH groups which offsets the increased length easily.

2. When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is __________.
A. AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq)
B. Ag+ (aq) + I- (aq) → AgI (s)
C. Ag+ (aq) + NO3- (aq) → AgNO3 (aq)
D. Ag+ (aq) + NO3- (aq) → AgNO3 (s)
E. AgNO3 (aq) + KI (aq) → AgI (aq) + KNO3 (s)

Precipitation means we can cancel out all the aq AgIs therefore E and C are cancelled

In Net Ionic equations we only use ions that change state or oxidation number, since none of them actually change oxidation numbers as this is not redox we must consider those that change states

A full ionic equation would be

Ag+ + NO3- + K+ + I- → AgI (s) + K+ + NO3-

we can reduce this down to form the net ionic equation which is:

Ag+ + I- -> AgI(s)

so therefore B is the best answer.

 

[whiteknight]

nevermind

 

[Mystic Theurge]

Carbon tetrachloride has a tetragonal configuration. Although the dipole charge for each chlorine is equilvalent in magnitude, the directions aren't opposite each other. It would have an overall dipole charge downwards on the following picture.

No it wouldn't.

Edit: Caught you.

 

[Dooraven]

Nevermind this as well then

 

[MThanded]

 

[nocode]

Is it bad that I took a lot of chem in college, worked in a chem lab for a couple of years after college, neutralized silver nitrate solutions all the time for disposal at that job, and I had no idea what the answer to those questions were? I have the worst memory of all time.:lol

 

[Flash]

Didn't want to start a new thread but anyone know any good resources (books, online) for learning NMR spectroscopy? Good places for practice problems etc? I'm having some trouble with it due to the ambiguity (well, ambiguous for me, at least), for whatever reason i just can't read the graphs correctly

 

[Evolved1]

Didn't want to start a new thread but anyone know any good resources (books, online) for learning NMR spectroscopy? Good places for practice problems etc? I'm having some trouble with it due to the ambiguity (well, ambiguous for me, at least), for whatever reason i just can't read the graphs correctly

What are you having trouble with? Proton NMR or C13 NMR?

 

[Flash]

What are you having trouble with? Proton NMR or C13 NMR?

Well im working on proton nmr at the moment, will jump into c13 when i figure this out i guess.

 

[Evolved1]

Well im working on proton nmr at the moment, will jump into c13 when i figure this out i guess.

There are probably a lot of places online that explain it... if I knew what in particular you were having trouble with, I could try and explain it.

 

[Ra\/en]

first one has got to be B.

A has quite a long non-polar chain.
B. has two polar ends (alcohols) with a short chain in between (best choice)
C.CCl4 has no polarity.
D. CCl3 has a little bit of polarity, but option B. is still better.
E. again, another long chain with an alcohol at the end. It would dissolve, but not the the same as B.

I think the answer to question 2 is B. You are given the product AgI solid, so the last 3 options are out:lol (no AgI(s)).

Now, I don't remember the stinky rules of gen chem very well, but it seems to me that net equations don't care about the aqueous bits in the end.

I'm going with B. final answer.

Edit: I was right. but I didn't remember why. Listen to Dooravan. .

P.S. we should have more questions about Organic Chem here. I love that stuff.

 

[Ra\/en]

Didn't want to start a new thread but anyone know any good resources (books, online) for learning NMR spectroscopy? Good places for practice problems etc? I'm having some trouble with it due to the ambiguity (well, ambiguous for me, at least), for whatever reason i just can't read the graphs correctly

go to your university library and see if they have this book.

It was one of the textbooks for my upper level Spectroscopic Organic Chem class. Fantastic text. One of the best books of my undergrad.

http://www.amazon.com/dp/0030319617/?tag=neogaf0e-20

 

[Evolved1]

Flash... I did a quick search. There are quit a few tutorials like this one that provide instruction, practice and feedback. On that site, probably best to just do the peak assignments until you get a handle on it.

Lots more like that though... just do a quick search for "proton NMR tutorial".

 

[Flash]

There are probably a lot of places online that explain it... if I knew what in particular you were having trouble with, I could try and explain it.

Well, sometimes the integration is totally unclear and I really don't understand how they got the numbers (i.e. looks like a 4, but turns out to be a 5; integrations are really close to each other and difficult to read). Also complex splitting usually screws me over for whatever reason.

And more importantly, what i think i'm looking for is a general way of tackling problems, what you would look for first, etc.

edit: thanks a lot guys, ill be checking out those resources soon.

 

[bjork]

Au

 

[Ra\/en]

Well, sometimes the integration is totally unclear and I really don't understand how they got the numbers (i.e. looks like a 4, but turns out to be a 5; integrations are really close to each other and difficult to read). Also complex splitting usually screws me over for whatever reason.

And more importantly, what i think i'm looking for is a general way of tackling problems, what you would look for first, etc.

edit: thanks a lot guys, ill be checking out those resources soon.

well for proton NMR, I usually start with the chemical shift ranges... see if I can pick out anything obvious. For example, a Carboxylic acid at around 11-12, or benzene hydrogens.