**Chemistry...Before/After

Mista

ei.sderob wrote: »

Not really a fan of the Winkler method or any water experiment to be honest. Question: what are the pieces of evidence that the addition of bromine across ethene has occurred?

When the reaction occurs in water containing sodium chloride.. 1-bromo-ethanol, 1-bromo-2-chloro-ethane are formed, and something else. These are evidence for the carbonium ion :pac:

ei.sderob

Hayezer wrote: »

When bromine and ethene react in water with sodium chloride, 2-bromoethanol is formed. (Apparently this is evidence :L)

Yeah, I saw that in the book......I've got no clue how it can be classified as evidence though:P

barryprp

ok im just going to stay up late studying the marking schemes of past exams...someone please tell me it will help and its a good idea!...otherwise im totally screwed!!!!!!!!

Laura.12

Dreading this exam All nighter is on the cards i think...

dfx-

reznov wrote: »

Best way to see if you're capable is to sit a university chemistry exam (3rd year recommended). If you achieve a high grade, you're ready for the leaving.

What university has a 'chemistry' exam bundled together in third year? It's divided into separate physical chemistry, organic chemistry, analytical chemistry, inorganic chemistry, environmental chemistry papers by that stage - or at least my degree was.

Leaving Cert level is 1st year.

ChemHickey wrote: »

Electrochemistry is lovely. I really like it.

Is there much electrochemistry on the course? It is a very difficult subject..

Pepperr

Only thing I can see in my book to do with Electrochemistry is the Electrochemical series..which is a lot to learn for small marks :P

*Edit: Durrr I don't think you were talking about the LC -.- It's too early!

Hayezer

I dont get bonding at all ? How are we meant to know if something has Van Der Waals, Dipole-Dipole etc? And how do you know how many sigma and pi bonds

paddyhes

Hayezer wrote: »

I dont get bonding at all ? How are we meant to know if something has Van Der Waals, Dipole-Dipole etc? And how do you know how many sigma and pi bonds

Van Der Waals can occur between any molecules (i think). They are induced weak dipole-dipoles.

Dipole-Dipoles exist in bonding where one part is partially negative and the other part is partially postive (polar covalent or ionic).

Hydrogen bonding can only occur where Hydrogen is bonded with either; Fluorine, Oxygen or Nitrogen.

In all triple bonds there are 2 pi bonds and 1 sigma bond.
In all double bonds there is 1 sigma and 1 pi.

hotfuzz2

No van der waals can only occur between the same type of non polar molecule, eg Cl2

Spattersonox

hotfuzz2 wrote: »

No van der waals can only occur between the same type of non polar molecule, eg Cl2

Nooo. Van Der Waals occurs between ALL molecules. dipole dipole btween polar molecules and Hydrogen bondings when Hydrogen is bonded to N,O or F

paddyhes

hotfuzz2 wrote: »

No van der waals can only occur between the same type of non polar molecule, eg Cl2

Van Der Waals exist in non polar molecules.
Most commonly in molecules like Cl2 and H2 etc.
But they can also exist in any other non polar molecule such as Butane (C4H10) or Propane (C3H8).

Hayezer

Thanks lads

DepoProvera

Van der Waal is the only type of intermolecular forces between molecules of gases.

hotfuzz2

Spattersonox wrote: »

Nooo. Van Der Waals occurs between ALL molecules. dipole dipole btween polar molecules and Hydrogen bondings when Hydrogen is bonded to N,O or F

Yes I know, what I'm trying to say is that Cl isn't bonded to nitrogen by a van der waal bond, cl can only be bonded to another cl by a van der waal

reznov

Intermolecular forces are:
Van Der Walls, Dipole - Dipole and Hydrogen Bonding.

Van Der Walls are the weakest form of attraction and generally occurs between similar molecules. e.g. Cl2

Dipole - Dipole forces are forces of attraction which occur between polar molecules. Do not get mixed up with Hydrogen bonds as hydrogen bonds are forces of attraction which occur between Hydrogen and either Flourine, Oxygen and Nitrogen. The hydrogen acts as a bridge between two polar molecules. The example of course is water.

Boiling points are as follows:
Vane Der Waals (or London forces) < Dipole Dipole < Hydrogen Bonds

Hydrogen bonds can form In Carboxylic Acids and Alcohols. This gives them their increased boiling points over the dipole dipole forces that form between weak polar molecules with a carbonyl group such as aldehyde and ketone. It also increases their solubility, but remember that solubility decreases as the non-polar chain increases!!!!!

Intramoleculer forces:
Covalent and Ionic bonds.

I'm assuming you're alright with them.
Covalent is sharing of electrons, ionic is attraction between oppositely charged ions.

DepoProvera

Regardless, I'd advise all of yous away from Q5(unless it's your strongest topic).. It's less hard to be definite that you're correct with your answers.

Spattersonox

Do you know the expt where you're testing for the conc of dissolved O2, what is the exact purpose of putting in the potassium iodide and Magnanese Sulfate? and why do you run conc sulphuric acid. I know you do these things, but I don't know WHY

paddyhes

Spattersonox wrote: »

Do you know the expt where you're testing for the conc of dissolved O2, what is the exact purpose of putting in the potassium iodide and Magnanese Sulfate? and why do you run conc sulphuric acid. I know you do these things, but I don't know WHY

The Iodide ions from the Potassium Iodide and the Manganese(III) Hydroxide, from the Manganese (II) hydroxide (which is from the Mn2+ ions from the Manganese sulfate), react to liberate Iodine (which is your red/brown colour).
The sulfuric acid is what causes the Iodide ions and Manganese(III) Hydroxide to react.

DepoProvera

Spattersonox wrote: »

Do you know the expt where you're testing for the conc of dissolved O2, what is the exact purpose of putting in the potassium iodide and Magnanese Sulfate? and why do you run conc sulphuric acid. I know you do these things, but I don't know WHY

The Mn2+ ions react with the OH- ions from the alkali to form a white manganese(II) hydroxide( Mn(OH)2 ), which then reacts with the dissolved oxygen to form Mn(OH)3. This is the oxidising agent that liberates iodine from the potassium iodide. You then titrate to find out how much iodine was liberated and hence how much O2 was there.

That is my understanding of it anyways :L

Pepperr

Edit* Depo did a better job explaining.

DepoProvera

paddyhes wrote: »

The potassium Iodide and Manganese sulfate react to liberate Iodine (which is your red/brown colour).
The sulfuric acid is what causes the Potassium Iodide and Manganese Sulfate to react.

Well more specifically the acid gives H+ ions required for oxidation.

Spattersonox

DepoProvera wrote: »

The Mn2+ ions react with the OH- ions from the alkali to form a white manganese(II) hydroxide( Mn(OH)2 ), which then reacts with the dissolved oxygen to form Mn(OH)3. This is the oxidising agent that liberates iodine from the potassium iodide. You then titrate to find out how much iodine was liberating and hence how much O2 was there.

That is my understanding of it anyways :L

THANK YOU! My teacher just kind of said you need this for it to happen.. but I have no know WHY these things happen. If this expt come up today, I'll think of you :L

DepoProvera

And remember lads, clean water has ~9ppm dissolved oxygen content, so if you get in and around that, chances are you're right!

Hayezer

Think youd get away with just saying : The potassium Iodide and Manganese sulfate react to liberate Iodine (which is your red/brown colour).:L
And that sulfuric acid provides H+ ions for the oxidation?

DepoProvera

Hayezer wrote: »

Think youd get away with just saying : The potassium Iodide and Manganese sulfate react to liberate Iodine (which is your red/brown colour).:L
And that sulfuric acid provides H+ ions for the oxidation?

I'd mention something to do with Manganese(III) hydroxide reacting with the iodide ions though!

paddyhes

Hayezer wrote: »

Think youd get away with just saying : The potassium Iodide and Manganese sulfate react to liberate Iodine (which is your red/brown colour).:L
And that sulfuric acid provides H+ ions for the oxidation?

Doubt it because they don't react directly to liberate Iodine.

Coeurdepirate

Guys, could someone PLEASE summarise the prep. of ethanoic acid experiment for me? I've lost my notes and I can't find them anywhere

Pepperr

Coeurdepirate wrote: »

Guys, could someone PLEASE summarise the prep. of ethanoic acid experiment for me? I've lost my notes and I can't find them anywhere

It's similar to the ethanal one, except you have excess acidified dichromate and you don't distil off the product.

You set it up in reflux, making sure oxidising agent (Dichromate) is in excess.
The Orange dichromate turns green as the ethanol oxidises.

The set up starts with ethanol and water in a dripping funnel, dripping through a liebig condenser into a pear shaped flask. The flask is in a water bath, and in the flask is the dichromate, dilute sulfuric acid and anti bumping granules. After 30 mins, you are done! (Might have missed something, that is most of it though.)

paddyhes

Coeurdepirate wrote: »

Guys, could someone PLEASE summarise the prep. of ethanoic acid experiment for me? I've lost my notes and I can't find them anywhere

Add sodium dichromate, sulfuric acid and antibumping chips to a pear shaped flask.
Add ethanol and water to dropping funnel.
Set up for reflux.

Add contents in dropping funnel to pear shaped flask drop by drop whilst the pear shaped flask is in cold water (as reaction is very exothermic).

After its all added reflux the contents of the pear shaped flask for 30 minutes.

After this set up for distillation.

Distill of ethanoic acid using a bunsen burner (as it's boiling pt. is 118 degrees Celsius).

Coeurdepirate

Thank you so much!