The Student Room - Chemistry

Sixth form Official AQA AS Chemistry Unit 1 - 23Rd May 2013

SonamH — Tue, 21 May 2013 17:48:36 GMT

Quote:

Originally Posted by tumblrgirl

i get really confused with bond angles, i'm alright with drawing them out but when it asks you to predict the bond angle, I never know what it is :s help :(

Just remember that the bond angle in a tetrahedral shape is 109.5 and each lone pair reduces the angle by 2.5 so for example a trigonal pyramidal shape has 3 bonded pairs and 1 lone pair so 109.5 - 2.5 = 107 which is the bond angle in the trigonal pyramidal shape. Hopefully that makes sense

Sixth form OCR Chemistry: F321 & F322 revision notes

Rubyturner94 — Tue, 21 May 2013 17:34:30 GMT

Quote:

Originally Posted by herothing11

For anybody doing this exam in the summer, I've comprised a revision file answering the specification.
This should be useful in covering the theory side of Chemistry for anybody who needs it.
Don't forget that these are only revision notes, they will not guarantee you will pass. Some people don't learn from reading the notes of others so I would suggest using the textbook in conjunction with the notes to clarify anything that may seem vague to you.

F321 - Attachment 139834

F322 - Attachment 148502 - and a reactions flow-chart Attachment 193431

F323 is the coursework element, there are no notes for this.

F324 is here http://www.mediafire.com/view/?uwlaxb37l5c0u2c

F325 is also here - http://www.mediafire.com/view/?vyw614p0pr5cxyg

I cant seem to open the file :( what format is it in?

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Sixth form Anyone know any methods to remember coloured compounds?

Scienceisgood — Tue, 21 May 2013 17:22:26 GMT

Quote:

Originally Posted by igobyursula

I've memorised all of the colours and the equations and the best piece of advice an A2 student gave me last year was not to leave it to the last minute. I learnt them before Easter, but there's a month left and plenty of time. I learnt it by making a table and writing down all of the equations and colours formed. I did it systematically. For example, spend time learning the Aluminium reactions and then moving on to Chromium and Iron (III) etc.

Here's a handy table my teacher made for us! It doesn't show the equations, but you can make one which does as I did. Happy revising :)

Thanks. :)

Sixth form OCR Salters F331 23/05/13 discussion

LAZYABDI — Tue, 21 May 2013 17:15:58 GMT

I suck at chemistry revision even though the concepts are piss easy but this book makes me fall a sleep every time. Started today covered all of Elements of Life with story line plus the book and hopefully 2 papers this afternoon. Should be ready for thursday i hope.

Sixth form Sugar as an organic product

shengoc — Tue, 21 May 2013 17:10:35 GMT

Quote:

Originally Posted by KKP109

hwat is the importance of studying sugar as a chemical organic product

Sugars are family of complex macromolecules - i.e. carbohdyrates

Biochemistry deals mainly with the study of these macromolecule, carbs (sugars included) - metabolism/energy/mimicking nature - all of which drives interest and research in this area

Sixth form Chemistry- Chains, Energy and Resources thread

shengoc — Tue, 21 May 2013 17:08:30 GMT

Quote:

Originally Posted by sarathellama

Just a thread to discuss Chemistry unit 2 for AS
Does anyone care to explain mass spectrometry haha

Google: chemguide mass spectroscopy/spectrometry

Secondary Alcohols

shengoc — Tue, 21 May 2013 17:03:58 GMT

Quote:

Originally Posted by jackhadrill

Is the process of reacting ethene with water in the presence of phosphoric acid to form ethanol known as the hydration of ethene?

I.e.
Attachment 218015

Thanks,
Jack

it is nothing more than an addition reaction, the type commonly associated with alkenes and alkynes due to the presence of unsaturation

hydration is just the specific name for addition of water across these unsaturation

Sixth form Chemistry A-level difficulty

Tha Realest — Tue, 21 May 2013 16:53:33 GMT

Quote:

Originally Posted by 344302

Not sure. I think OCR A isn't that bad. The practical is easier than the AQA isas because there are 3 strands and you can have 3 goes at them each so that's 9 times to perfect your coursework mark. One is like a titration, the second type is like just a few experiments and writing down what you see and the third type is like a mini topic test. They aren't strictly timed either.

The OCR textbook is all you need. It has everything. It's the best textbook out there honestly.

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Ok cool

Sixth form How many structural isomers?

StUdEnTIGCSE — Tue, 21 May 2013 15:32:03 GMT

Quote:

Originally Posted by Magzter

http://www.ocr.org.uk/Images/59185-m...es-january.pdf

You think again.

EDIT: Sorry for bluntness, we're making isomers of 3-methylheptane here not just octane. The methyl group makes every cl position an independent isomer.

Oh yes, I thought again and got that awkward moment.

You caught me in my own trap! :o

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Secondary endo and exothermic chemicals?

Magzter — Tue, 21 May 2013 15:25:58 GMT

We've both made statements which aren't absolute truths or semantically unambiguous, let's move on.

Sixth form Describe the bonding in sodium

LeonVII — Tue, 21 May 2013 14:21:57 GMT

Quote:

Originally Posted by KKP109

Describe the bonding in sodium

Sodium's a metal, therefore metallic bonding...

Positive sodium ions are immersed in a sea of negative electrons, therefore it's the ionic attractions between the two.

Sixth form AQA AS Chemistry Unit 2: Specification Q

lindseyyy — Tue, 21 May 2013 14:15:47 GMT

Quote:

Originally Posted by Magzter

This came up in a past paper, ill try to find it now.

I think what it's basically saying is that the more a molecule absorbs IR radiation the more of a greenhouse gas it is and so the bigger contributor to global warming it will be.

So on the graph, the 'deeper' fingerprints will mean a higher absorption and a stronger greenhouse gas.

Thank you!

Sixth form Self-teaching Chemistry A-level (as a private candidate)?

Rascacielos — Tue, 21 May 2013 14:01:52 GMT

Quote:

Originally Posted by spwills

A year :)

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Does that include the AS as well? Are you just taking the A2 exams?

Secondary Titration Calculations Made Easy! :D

ismailjan3 — Tue, 21 May 2013 13:09:02 GMT

Quote:

Originally Posted by Jackboy25

Showing this in lesson, neat Attachment 217940

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Oh my god, you're really showing my video in your lesson? :O
Thank you....I feel awesome :')

Sixth form how many full orbitals in an atom of sulfur

tazmaniac97 — Tue, 21 May 2013 11:34:40 GMT

Quote:

Originally Posted by illusionz

The 4 3p electrons are spread over the 3 p orbitals, so one is full and two are singly occupied.

Quote:

Originally Posted by eggfriedrice

Aha of course. But then surely they would all be unpaired since the orbitals fill up one by one first?

EDIT; nvm, my brain isn't working xD been doing too many transition metals. The guy below is right.

Quote:

Originally Posted by Gnome :)

The electron configuration is 1s2 2s2 2p6 3s2 3p4
So from 1s2 2s2 2p6 and 3s2 you get 6 full orbitals
In 3p4 it's slightly different. Electrons "prefer" being on their own in this context due to electrostatic repulsion. The 3p subshell can have 3 pairs of electrons, or 3 orbits, so 3 electrons will go into separate orbitals. The remaining electron then has to join one of these electrons. So the 3p subshell is made of 3 orbitals; 2 with a single electron, one with a full orbital.
Therefore you have 7 full orbitals, and 2 half filled orbitals :)

Quote:

Originally Posted by JMaydom

The electrons will try to align their spins as this minimises electron repulsions as in the all spin parallel configuration there is a zero probability of finding a pair of electrons at a separation of zero.
Now 4 electrons cannot align all spin parallel in 3 orbitals, so the final electron must pair up with one of the other spins. This leaves two unpaired spins forming a triplet state.

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Oh I see what I did wrong, I forgot they fill the orbitals one by one before pairing up. I was rushing that question because I thought it was easy and didn't really think about it :p

Sixth form Chemistry - Methyl Group Activating the Benzene Ring

mojtaba85 — Tue, 21 May 2013 11:11:44 GMT

Quote:

Originally Posted by illusionz

Whatever effect there is due to the electronegativity difference between an sp2 and an sp3 carbon (which is minimal) is completely dwarfed by the effect of sigma conjugation (or hyperconjugation as you like to refer to it).

Carbon is more electronegative than hydrogen (approximately 10% more electronegative, which is conveniently pretty much equal to the difference between sp2 and sp3 given in Corey's book) . Therefore if cation stability was significantly effected by polarisation of sigma bonds then we would expect primary carbocations to be stable due to polarisation of the C-H bonds stabilising the positive charge. You and I both know this is not the case. The reason that the tertiary cation is stable is due to sigma conjugation. Not the miniscule effect of having the sigma bond slightly more polarised towards the cation. Also note that the cation itself is vastly more electronegative than a normal sp2 carbon, but the increased polarisation of the sigma bonds is not the result of the stability/or lack of.

I'm starting to think that you are not understanding what I'm trying to say properly, as forgive me for saying it... but english is clearly not your first languge. All I was trying to argue was that thinking about electronegativity difference is pointless... not that tertiary cations are unstable.

I think both of us have the same idea: both hyperconjugation and electronegativity affect on ring activation (by stabilizing the carbocation intermediate) but the hyperconjugation effect is dominant. I think you are agree with this. and sorry for my poor english.

Sixth form Describe the bonding in carbon dioxide

LeonVII — Tue, 21 May 2013 10:48:13 GMT

CO2 has 2 carbon-carbon double bonds, linear molecule, no overall polarity/dipole.

Sixth form Enthalpy changes (unit 2 energetics)

Aiden11 — Tue, 21 May 2013 10:42:32 GMT

I think with these questions you need a method that you can apply every time. Mine is as follows:
1) Write down the balanced equation
2) Find out the enthalpy change of all the reactants (take positive value)
3) Find out enthalpy change of all the products (take positive value)
4) Reactants - products = enthalpy change of the reaction
I think it's easier to keep all the values positive, because when you do reactants - products it will sort the sign out for you since you're measuring the energy in the bonds of the reactants and the energy in the bonds of the products. If the energy in the bonds of the reactants is higher, then you'll get a positive value and thus and endothermic reactions. This makes sense because an endothermic reactions occurs when you break strong bonds and form weak bonds. Whereas when the energy in the bonds of the reactants is lower then you'll end up with a negative value, which works because an exothermic reaction is when you break weak bonds and form strong bonds.

Sixth form Van der waals forces

Aiden11 — Tue, 21 May 2013 10:21:36 GMT

I think the best thing to do would be to find a question on it in a past paper and look at the mark scheme. That way you can get an answer you are 100% confident will give you the marks! I'm not 100% sure, but If I had to guess I'd say that you wouldn't need to mention them when talking about polar molecules, especially nitrogen/fluorine/oxygen with hydrogen since their intermolecular forces are strong enough to make vdw seem irrelevant. But like I said, trust the mark scheme, not me! :)

Sixth form How do you identify a macromolecular crystal?

Magzter — Tue, 21 May 2013 10:15:48 GMT

I wondered this myself. The best way in the context of A-Levels, I found, is to realise that there are actually very few giant covalent molecules so just learn them. Diamond, graphite and silicon dioxide are the most frequently come across examples maybe there are a couple more you will need to know but I can't think of them offhand. Anything ionically bonded will form a macromolecular crystal when not in solution and those are easily identified too (metal/non metal).

Sixth form What is the highest grade I can get?

CheeseCrackers — Tue, 21 May 2013 10:02:01 GMT

I'm doing AS Chemistry OCR Salters

And for Jan 2013 these are the following results I got:

F331 - C
F332 - (D BUT RETAKING)
F333 (practical work) - D

I wanted to know what the highest possible overall grade I can achieve by the end of my a-levels if I aim for A's in my retakes and A/A* for my A2... I know it might not be realistic to some but I had a lot going on this year which didn't allow me to achieve my best and I do believe I can get A's in the papers to come...

Thanks for taking your time to read this I really do appreciate it!

*I have posted a similar question regarding my maths and biology results in the Maths and Biology sub forums

Undergraduate Flash photolysis - Beer Lambert law

JMaydom — Tue, 21 May 2013 09:51:30 GMT

Quote:

Originally Posted by DonnieBrasco

OK thanks, but provided I can convert my units correctly, does the calculation look plausible?

You got the ratio of I/I0 right but it looks to me that you got it the wrong way :p
So should be correct from then onwards, apart from units of course!

Sixth form Solubility of alcohols.

LeonVII — Tue, 21 May 2013 08:56:37 GMT

Quote:

Originally Posted by Hi, How are you ?

Can't the hydrocarbon chain produce London forces with the water thou??
Or it does London force bond with water, but to such a small extent, that it is insignificant??

Yes, there are London forces between water molecules, but as water is polar, these forces don't contribute much to overall attraction between water molecules. The most important attraction will be hydrogen bonding.

Secondary chemistry thursday !!!!

jannattulfirdaus — Tue, 21 May 2013 08:39:45 GMT

hi I have my chemistry on Thursday but I really needs some help, can anyone give me some notes on: purifying copper electrolysis and the Harber Process

Other/Not applicable Molar volumes help?

mrdoovde1 — Tue, 21 May 2013 06:16:12 GMT

Quote:

Originally Posted by chrisawhitmore

Usually these problems will either give a molar volume (a number of decimeters cubed per mole of substance in given conditions) or assume an ideal gas.

At SATP (25 C 1 atm) an ideal gas occupies 24dm cubed. This is the one I usually see used.

So for 1, you take your volume in mL and convert it into decimeters cubed. (divide it by 1000, as 1dm cubed is 1 litre.) That means you've got 1.34 dm3 of gas. Assuming ideal gas behavior (note when you do this in the exam) that's 1.34/24= 0.056 mol of SO2 and with a molar mass of 64 g/mol that's 0.056*64= 3.58g

For 2, it's simpler. 24 moles per decimeter cubed means 34.53/24=1.44 decimeters cubed. Assuming SATP and ideal gas behavior (a popular exam question is to ask what assumptions are made, and depending on your level, how valid they are for gas x)

Wait for Q1, why are you dividing the mass by 24?

I was under the impression that 1L = 1KG so 1.34 x 10^3ml of SO2 = 1.34 x 10^3g of SO2?

Sixth form Chem1 question help

cllio — Tue, 21 May 2013 00:48:05 GMT

m/z is mass over charge ratio. So you need to go look at the periodic table and see which element has a relative mass of 64. And it will the answer will be that element with a positive charge because for mass spec to work you need to knock an electron off. Also as you know the element that has been detected you'll have to put the isotope with the right amount of neutrons and protons and write that down in your answer

Sixth form Describe the bonding in Magnesium Chloride

natninja — Mon, 20 May 2013 22:23:09 GMT

Quote:

Originally Posted by JMaydom

Ah, lincoln!!! Going to Queens ball this year :)!!!!

A lot (well virtually all) of that goes over my head.... My knowledge regarding this is MO theory which (to my knowledge) doesn't work for gases, I've only ever seen it for solids and for the bonds in discrete molecules.

I guess 'm lucky that I don't have to do plasma's in depth unless I take that short option (it was just a waves question that was in a past paper... I'm a first year btw) get to do all the stat mech with funny shaped gas molecules next year...

Queen's ball should be good, I'm going too, bit on the pricey side though.... My gf is a chemist so I just got a 45 min chem mini-lecture off her...

Sixth form Identifying Radicals

Dr_of_chemistry — Mon, 20 May 2013 20:54:30 GMT

Think of the electronic configuration of nitrogen (1s2 2s2 2p3) and for oxygen (1s2 2s2 2p4)........so nitrogen has three unpaired electrons....oxygen two upnpaired electrons......waiting to form bonds... so NO (nitric oxide) must have an unpaired electron on nitrogen making it the radical

Sixth form Platin and Haemogloblin

chembob — Mon, 20 May 2013 20:46:07 GMT

Quote:

Originally Posted by eggfriedrice

Could someone give me model answers for;
How haemoglobin undergoes ligand substitution and how this helps the transfer of oxygen to cells.
In terms of stability constant how carbon monoxide reduces the amount of oxygen transferred.
How platin is used for cancer treatment.

Thanks. (:

You mean cisplatin, PtCl2(NH3)2 right?

Upon entering the body, one of the chloride ligands is slowly displaced by water forming the complex [PtCl(H2O)(NH3)2]+. This water ligand is itself easily displaced, and as such the complex can easily bind to DNA bases, predominantly guanine. After binding to a guanine moiety, the remaining chloride ligand can be displaced and the complex can cross-link with another guanine base. This crosslinking can occur in several ways, and it interferes with cell division (mitosis). The damaged DNA invokes bodily mechanisms for DNA repair; these can then cause apoptosis (basically killing off cells; programmed cell death or PCD) if repair is not possible.

Sixth form OCR Chemistry Notes F322

AjishJoshySorry! — Mon, 20 May 2013 19:14:31 GMT

Here are some notes I had made I may complete the others if I have time :P

Sixth form A-level without a practical?

Dynamo123 — Mon, 20 May 2013 18:30:43 GMT

If you are not from uk, go for CIE.

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Undergraduate Organic question problem

cllio — Mon, 20 May 2013 15:41:34 GMT

Yeah me and a friend went through it last night at stupid o'clock and got A and C it's good to get the answer to check its right too. Cheers guys.

edexel chemistry unit 1 sample assessment material HELP!!

kiara139 — Mon, 20 May 2013 14:29:42 GMT

Use the data about four fuels given below to answer this question.
Fuel Formula Name
Enthalpy change of combustion /kJ mol–1 , Molar mass/g mol–1
A CH4 methane –890 16
B CH3OH methanol –726 32
C C3H8 propane –2219 44
D C4H10 butane –2877 58
(a) Which fuel, A, B, C or D, produces most energy per gram on complete combustion?
A
B
C
D

Sorry the question is not clear coz i just did copy-paste but its from sample assesment material. Can someone please explain it and what is the effect of having more negative ΔH value?

Sixth form Chemistry questions

Plato's Trousers — Mon, 20 May 2013 14:06:20 GMT

Quote:

Originally Posted by otrivine

you mean cm3 ?

ml is the same as cm³ !

Undergraduate NMR, Cancelling out H2O

illusionz — Mon, 20 May 2013 12:21:01 GMT

Quote:

Originally Posted by cllio

but it going to be the same peak all the time isn't it? like pretty much exactly the same so why can't you just cancel it out? just make an algorithm for it

I guess you've never run an NMR sample of a decidedly wet product. The water peak can be very broad and it is easy for several peaks to be lost underneath it.

Also, as everyone else has said, the ratio of solvent peak to product peak would be insanely large and you could not calibrate the machine without knowing exactly how much solvent was used.

Sixth form Question on past paper about concentration ?

RoseBrilliante — Sun, 19 May 2013 23:01:45 GMT

Quote:

Originally Posted by User32432432

Thanks for the reply ?
What formula is this ? I have never seen it nor used it

No. of moles = Mass in g/Mr

One of the first equations I ever used in GCSE/AS Level.

Sixth form Good at Chemistry? Try this pH calculation..

Doppel — Sun, 19 May 2013 21:32:43 GMT

Quote:

Originally Posted by physicso

Sorry about the title (awful attempt at trying to get a genius to explain the theory behind this calculation!):

20cm^3 of 0.015mol dm^-3 HNO3 is mixed with: 10cm^3 of 0.015mol dm-3 NaOH.

Calculate the pH of the following solution.
I am normally pretty good with calculations but i have no idea where to start here. Could anyone help?
Thanks!

Spoiler:

Show

Hello.

Since HNO3 is a strong acid, the reaction is as follows:

HNO3 --> H+ + NO3-

In other words, the acid dissociates completely into its' ions. Thus, [H+]=[HNO3].

Now, I think you have made some error in typing out the question, because the concentration after adding HNO3 would not change in this case. That is because you are adding more HNO3, but with the same concentration as before. You are thus neither diluting, or concentrating the HNO3.

You can also confirm this mathematically:

C=n/v --> n=CV.

Total (n)= n1 + n2 --> 0.015*0.02 + 0.015*0.01 = 0.00045

C = Total (n) / Total (v) = 0.00045/0.03=0.015M, which is the same as you had in the beginning.

Now, pH=-log[H+] = -log[0.015]. I do not have a calculator on me or nearby, so you will have to do this last part yourself. Keep in mind that the base of the logarithm is 10.

EDIT: It seems like my slow typing enabled someone to give you hints. I will thus post the solution as a form of spoilers to allow you to try to solve it using the above hints. Good luck.

Sixth form Zwitterion and Ph

charco — Sun, 19 May 2013 21:26:35 GMT

Quote:

Originally Posted by KKP109

how is zwitterion effected by a low ph and how is it effected by high ph

low pH = excess H+ ions. This causes the amine groups to protonate.

high pH = excess OH- ions. This causes the carboxylic acid groups to lose a proton and become carboxylate ions.

Sixth form Random Chem Equation Q

charco — Sun, 19 May 2013 21:19:12 GMT

Quote:

Originally Posted by Chloe_White

Thank you, I thought because it was a solution you divide by 1000 but that makes more sense

So, just to clarify I divide by 18 = 0.0933..

And my ratio would be 0.2:1 (tiimes them both by 5 so that the CuSO4 is equal to 1)

CuSO4 5H2O
X=5

Secondary C2 Exam

omfgm0lly — Sun, 19 May 2013 20:27:47 GMT

Hiya! Does anyone have any ideas of what the 6 mark question will be tomorrow for the GCSE C2 exam? I know January was on gas chromatography but I'm really nervous of what will come up D:

Sixth form Giant and Simple covalent structure help

dippers — Sun, 19 May 2013 20:10:31 GMT

Quote:

Originally Posted by munrot07

ii) the most electrons that can be in the cloud of delocalised electrons, the stronger the metallic bonding so look for the element in period 3 that is a metal that can give the most electron when it is an ion :)

iii) look for the element that can receive 3 electrons to get a full outer shell so that it has the same number of electrons as a neon atom does

iv) the element which requires the least amount of energy to remove the third electron. This means an atom with 3 electrons in its outershell that has the most shielding

v) look for the big jump in ionisation energies and work out how many electrons there are in the outershell :)

i hope that helps :)

thank you so much!

Sixth form DNA as an Organic Product

Type 052D — Sun, 19 May 2013 19:06:27 GMT

Quote:

Originally Posted by illusionz

The biochemists here know **** all chemistry.

Hey! We (I'm future Biochemist) know alot more than just bio-molecules. We do go further in Organic Chemistry.

Sixth form WJEC Ch4 revision help

SamJHill — Sun, 19 May 2013 18:43:07 GMT

so I'm resitting ch4 after achieving an abysmal mark in January, and no matter how hard I try I just cannot get all the organic information in order to learn/memorise. I've made countless mind maps, straight notes, highlighted textbooks, discussions with classmates, nothing seems to make it stick

anyone with advice on how they tackled Ch4 would be much appreciated, not sure why it's just this module that's tripping me up as so far Ch5 seems like a doddle! Got an A and a B in Ch1 and Ch2 respectively so a D in Ch4 doesn't really follow my results.

thanks for your time :H

Sixth form Conjugate Acids and Bases

grassgrazers01 — Sun, 19 May 2013 17:39:47 GMT

Thanks, I think I get it now :)

Secondary Chemistry 2 AQA - GCSE 20th May

zhang-liao — Sun, 19 May 2013 16:05:14 GMT

Any questions about Chemistry 2 for tomorrow
Let me start - do you have to know the procedure of gas chromtography and the retention line and molecular ion peak.....

Secondary help with molecular structure please?

Sherlockedd — Sun, 19 May 2013 15:51:34 GMT

Quote:

Originally Posted by ThePremierLeague

identify one way in which the molecular structure of cholesterol is similar to the molecular structure of a carbohydrate

They will have the same number of atoms in the same ratio. :)

Sixth form Why//mutation//...........

ILoveUSA — Sun, 19 May 2013 15:29:15 GMT

Why does a narrow gene pool decrease the chance of mutation?

Sixth form how is this reaction possible

charco — Sun, 19 May 2013 14:37:09 GMT

Quote:

Originally Posted by celina10

No in the mark scheme it has 2 equations that we were supposed to write and apparently Fe2+ does react with I-, I'm really confused and don't understand how that's possible especially by looking at the electrode potentials :confused::
2I^- + 2Fe²⁺ => 2Fe³⁺ + I₂
S₂O₈^2- + 2Fe²⁺ => 2SO₄^2- + 2Fe³⁺

The first equation is incorrect. The charges do not add up on both sides ...

... and both species are oxidised!!!!

Sixth form Hydrogen bonding with no d+ hyrdogen?

JMaydom — Sun, 19 May 2013 13:33:10 GMT

The polar O will interact with the water favourable but the apolar alkane chains interact very poorly with the water, so it is poorly soluble overall.

Sixth form Why do bonding and lone pair of electron repel each other?

LeonVII — Sun, 19 May 2013 12:09:17 GMT

They both contain negatively charged electrons. Like charges repel.

Secondary Acid-Metal Reactions

LeonVII — Sun, 19 May 2013 11:43:11 GMT

Quote:

Originally Posted by fisika

Hey,

I don't understand why reactions of metals and acids are not neutralisation

For example:
Mg + 2HCl --> MgCl2 + H2

The H+ ions no longer exists, does it? Doesn't that mean it's a neutralisation reaction?

Thanks in advance :)

It's not neutralisation because for that to be the case, you need H+ ions to react with OH- ions to form water. there are no OH- ions in this reaction.

Sixth form Amino acids as an organic product

thegodofgod — Sun, 19 May 2013 11:43:06 GMT

Quote:

Originally Posted by KKP109

Why is it imprortant to study amino acid as an organic product ? (Biological molecules)

Well, amino acids are the monomers involved in making proteins (the polymers), and there is a whole host of uses of proteins from enzymes to keratin, to transport proteins (e.g. haemoglobin), to hormones, e.g. insulin.

Sixth form Deducing the formula of a tertiary amine

charco — Sun, 19 May 2013 11:35:43 GMT

Quote:

Originally Posted by Joannee

The Mark Scheme show 2 methyl groups on the Nitrogen? It's off JUN12

If the reagent is bromoethane they're wrong...

... however, if you meant to write bromomethane then they're correct!

Secondary empirical formula?

charco — Sun, 19 May 2013 11:34:23 GMT

Quote:

Originally Posted by shaylux

I've got the formula pretty covered, but I'm confused in which order the atoms/molecules are written in once you've figured out how many of each is in the substance..
For example, Cu3 and N. Does that become NCu3 or Cu3N? Is there a way of defining which comes first in the final formula?!

Thanks!

It's not very important, but there is a CHONS convention for carbon, hydrogen, oxygen, nitrogen, sulphur ...

Moles question?

HesitantCicada — Sun, 19 May 2013 11:24:56 GMT

Quote:

Originally Posted by Soloman

Hi,
Bit confused but this is the question,
1, A solution of NaOH contains 10 gdm^-3

a) What is the conc. in moldm^-3

How can you convert these two units?

Since Relative Formula Mass of NaOH is 40, and Moles = (Real Mass orMass in grams) / (Fake mass or RFM) , you can just divide 10/40 (which gives 0.25mol/dm^3)

Secondary Chemistry and Physics triple science AQA

DomTownrow — Sun, 19 May 2013 11:21:29 GMT

Hey guys. Just wondering how you guys are getting on with P3 and C3. I have Chem 3 and then Physics on the 23rd. How are you guys revising cos I am struggling with P3 a little. Thanks :)

Secondary Rate of Reaction

Tristian Fox — Sun, 19 May 2013 10:26:15 GMT

The amount of Sodium Thiosulfate that has to react to make enough of the coloured compound, sulfur, to block out the cross is the same each time but we don't know what the amount is. As rate is equal to change in number of moles (which could also be change in mass or volume) over time we are unable to work out the rate exactly but because it is the same each time the exact value doesn't matter because the final result is just a comparison between them so the rate can be calculated by dividing a set number by the time. We just use one for simplicity.

Sixth form Ions and Partial charge Help, and IMF q too.

Hi, How are you ? — Sun, 19 May 2013 10:02:07 GMT

Quote:

Originally Posted by JMaydom

Yes... But the entropy of dissolution is very favourable. So really its a spectrum. All compounds are theoretically soluble to a certain extent. Remember the equilibrium constant is dependant upon delta G

Posted from TSR Mobile

I'm only up to as chemistry, not a2 or uni chem yet.

Sixth form Synthesis haloalkane to nitrile

JMaydom — Sun, 19 May 2013 09:53:35 GMT

Quote:

Originally Posted by CasualSoul

Seems a bit complicated for AS but from here it seems that the answer is due to the solvent stabilising it. Anyway I've given you all rep for trying :D JMaydom I tried to give you rep but it says please rep another user : s

Yeah its a pretty dumb system

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Sixth form Polymerisation nomenclature

CasualSoul — Sun, 19 May 2013 09:38:47 GMT

Quote:

Originally Posted by Dynamo123

You can

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thanks, run out of rep today. Promise to give tomorrow :D

Sixth form Ez isomerism

CasualSoul — Sun, 19 May 2013 09:37:35 GMT

Thanks to both of you- gave rep like I said I would :)

Quote:

Originally Posted by Munrot07

have to be opposite sides of the double bond :) two main rules are there must be C=C double bond, preventing rotation of the molecule and there must be two different groups on each carbon :)

Quote:

Originally Posted by _HabibaH_

Hi,

It would be neither since the same group (the H) is on the same Carbon atom.

The rule of E/Z isomerism is that there:

Must be a double bond restricting rotation
Must be two different groups on each Carbon atom

Sixth form Chem 1 electronic configuration

JMaydom — Sun, 19 May 2013 08:09:06 GMT

4s electrons are lost first when you ionise the atom. Now confusion occurs because people think this is equivalent to moving back one left, i.e. to Mn in this case. IT ISN'T!!!!
For reasons way above A-level, the 3d orbitals are stabilised more upon ionisation than the 4S, so by the time you reach Fe2+ the 3d orbitals are significantly lower in energy than the 4s. The electron ionised may not have come from the 4s, but this is then an excited state, so the electrons will rearrange to the lowest energy state, which is the 3d^n configuration.

Sixth form Retention time and solubility

Zzzyax — Sun, 19 May 2013 08:00:38 GMT

In F324

Gas chromatography, with a liquid stationary phase,
Can someone explain why greater solubility of the sample reduces retention time?
http://www.ocr.org.uk/Images/61007-m...is-january.pdf

And can someone explain why greater solubility in the liquid stationary phase means a high retention timehttp://www.chemguide.co.uk/analysis/...raphy/gas.html

Spoiler:

Show

Jan 12 ,Q4 bii) http://www.ocr.org.uk/Images/79471-q...d-analysis.pdf

Markscheme , http://www.ocr.org.uk/Images/61007-m...is-january.pdf

markscheme says:

Alcohol would have short retention time AND
alkane would have long retention time

Sixth form Percentage error

Lunch_Box — Sat, 18 May 2013 23:45:00 GMT

Quote:

Originally Posted by charco

Yes, but there were two weighings taken, one for each experiment ...

Oh okay thanks!

Sixth form F321 ABG Successive I.E

andyb3947 — Sat, 18 May 2013 21:53:51 GMT

hey guys,
my notes are bad for this certain area.Why do successive ionisation energies increase? I want the answer best to match the OCR mark scheme.

cheers

TSR Chemistry Society

3mma_gal96 — Sat, 18 May 2013 18:47:13 GMT

Hi, please could someone help me out regarding the limestone cycle?
I'm writing out the equations (word and symbol) for the reactions which take place. I know slaked lime,Ca(OH)2 reacts with H20 to give lime water, but what would the symbol equation be? Any help greatly appreciated

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Sixth form Half equation

AquisM — Sat, 18 May 2013 18:37:26 GMT

In a half equation both the charge and the number of atoms on both sides have to be balanced. If we put 10 e- on the left hand side the resulting charge would be 10+ (from the H+) -2 (from the NO3-) -10 (from 10e-) = -2, but on the other side we only have neutral compounds, so the charge must be 0. Therefore, 10e- cannot be the answer. Since you found out you have 2 negative charges too many, you reduce 10e- to 8e-. For any half equation, once you've balanced the atoms, you simply have to count the charges on both sides and try to balance them. :)

GCSE Chemistry Question Help? :3

charco — Sat, 18 May 2013 17:13:32 GMT

Quote:

Originally Posted by sparrows

Hi - I was also having difficulty with this question.
In the book it says:
Energy change in kJ/mol can be calculated by multiplying the energy change by the relative formula mass of the substance.

200x4.2x17 = 14,280J/g
Moles in 5.6g of iron = 5.6/56 = 0.1
If the answer is 142.8, why has 14,280 been divided by 0.1?

Could someone please explain?
Thanks

The energy is worked out for only 0.1 moles of reacting iron. The answer is required "per mole",. so it must be scaled up by multiplying by 10 (or dividing by 0.1, which is the same operation)

Sixth form What do I need for an overall A?

Chloe_White — Sat, 18 May 2013 15:50:58 GMT

Revise...a lot

Sixth form Rate Determining Step

charco — Sat, 18 May 2013 15:08:33 GMT

Quote:

Originally Posted by wrnicholls

I don't quite understand what's in the textbook. I'm doing AQA.

Please explain how you work out the rate determining step. In simplest form please don't confuse me!! :)

Here's a little doc I prepared on reaction rates and mechanisms. It may help.

And here's the video that goes with it ...

Sixth form Suggest the formula of sodium chlorate(VII)

palashnick — Sat, 18 May 2013 15:00:18 GMT

as we know that nitrate group overall has a oxidaton no. (-1) . so this gives us the oxidation no of SR in reactants side to be (+1) , while on the product side the oxidation no. of Sr is +2 which shows that Sr has been oxidised. Now Oxidation no. of Nitrogen in nitrate is +5 but in product it is +4 which shows that nitrogen has been reduced. Also since i=oxidation no. of oxygen increases from -2 in reactant to 0 in products , ww conclude that oxygen has also been oxidised. its simple when there is an increase in oxidation no. its called oxidation otherwise it is reduction . and it doesnt hurt to learn some general oxidation no.s showed by certain elements for an example Oxygen tends to show (-2) in most of its compounds. All alkali metals show (+1) always.

Secondary ocr 21st century c3

joostan — Sat, 18 May 2013 12:36:52 GMT

Quote:

Originally Posted by chloe lummis

if you could help what do you know about
-limestone, coal and salt
- reacting and making alkalis
- uses of chlorine
-electrolysis
- industrial chemicals
thank you !

PMed :)

Sixth form why do equal volumes have an equal number of molecules

abdullah hayyan — Sat, 18 May 2013 12:36:26 GMT

Quote:

Originally Posted by celina10

and you commented because...?

Because I like to

Secondary Chemistry 3 gcse help?

GeekisChic — Sat, 18 May 2013 12:31:11 GMT

Quote:

Originally Posted by the1akshay

Breaking bonds does require energy. As larger alcohols have more bonds, they require more initial/activation energy with which to break bonds. That's why you can't just put a match to diesel or heavier alcohols.

However, more bonds are formed after combustion. You know enthalpy, right, or bond energies? Basically, you input the energy to break each bond and subtract it from the energy released by the formation of bonds on the other side of the equation. Although it requires more energy to break the initial bonds as alcohols get bigger, more energy is released from the formation of bonds on the other side, simply because there's more stuff.

For example:

CH2H5OH (ethanol) + 3O2 -> 2CO2 + 3H20

In the ethanol, there's 5 C-H bonds to break, 1 C-C, 1 C-O and 1 O-H. In the oxygen, there's 3 sets of O-O bonds to break. On the formation side, there's 4 C-O bonds to form and 6 O-H bonds.

Moving to propanol:

2CH3CH2CH2OH + 9O2 -> 6CO2 + 8H20

In the propanol, there's 14 C-H bonds to break, 6 C-C, 2 C-O and 2 O-H. In the oxygen, there's 9 O-Os. On the formation side, there's 12 C-O bonds to form and 16 O-H bonds.

C-H is about 413kJ/mol. C-C is 154kJ/mol, C-O is 360kJ/mol and O-H is 366kJ/mol. For ethanol, that's 4(360)+6(366) - (5(413) + 154 + 360 + 366 ). This produces -3701kJ/mol. For the propanol, it's 12(360) + 16 ( 366) - (14(413)+6(154)+2(360)+2(366). That's about -6770kJ/mol.

To summarise, you're completely right that bonds require energy to break and release energy on formation. However, as the number of bonds to break increases, the number of bonds to be formed increases by a higher rate. Thus, larger alcohols produce more energy on combustion. It's actually a linear graph.

Attachment 216992

Thank-you so much! I had a real light bulb moment there :) You've explained it so well, I honestly can't thank you enough!

Sixth form OCR chemistry coursework

Mr boombang — Sat, 18 May 2013 12:09:45 GMT

hi everyone,

Does anyone know how much is the coursework worth for ocr AS chemistry?!?!

thanks

Secondary Chemistry; Titration Calculation

zhang-liao — Sat, 18 May 2013 12:09:40 GMT

Please help. C3 Titration Calculations. Is there any easier way in calculating concentration of the other substance....

Undergraduate pseudo first order rate constant determination

DonnieBrasco — Sat, 18 May 2013 11:50:25 GMT

Quote:

Originally Posted by Bradshaw

What happens when you take the log of a ratio?

Or alternatively, how would you derive the first order rate equation:

Rate=-d[A]/dt=k[A]

Separation of variables:

d[A]/[A]=-kdt

Integration:

ln[A]=-kt + Constant

To determine the constant, take the concentration at t=0 ([A0]):

ln[A0]=constant

ln[A]=-kt + ln[A0]

ln ([A]/[A0]) = -kt

Thus a plot of ln([A]/[A0]) vs t is exactly what you want!

Brilliant, thanks! :)

Sixth form Reaction between aldehydes and monohalogenoalkanes

Bonzo10 — Sat, 18 May 2013 11:29:09 GMT

₃

Ok, so I know that alcoholic alkoxides react with monohalogenoalkanes to form ethers. I was unsure of naming Y, but it turned out to be an ether. Is this a little bit of knowledge that has passed me by? Can somebody explain the mechanism to me? I know the carbonyl group is polar and the carbon in it is susceptible to nucleophilic attack, but can somebody explain it in this context?

Thank you

EDIT: Just realised that the O on the C₃H₇O is actually O^-. Makes more sense now.

AQA Chem 1 June 2011 paper + MS

loknath — Sat, 18 May 2013 11:25:30 GMT

Anyone have the examiners report on this...urgent...thanks

Sixth form OCR A AS Chemistry, old spec past papers?

3.1415927 — Sat, 18 May 2013 09:55:42 GMT

Could someone please send the link to find the old spec past papers for F322 chemistry (It will have a different name) .. but from the earliest till jan 09 ? That will be a big big help, thanks :$

Sixth form More than likely easy, help, chem question! OCR A/S.

teamvezon10 — Fri, 17 May 2013 22:02:59 GMT

Don't worry XD I figured it out, after a long time, realised it was v = n/c * 1000 (cm^3) not v = n/(c*1000) -.-

I feel so stupid now -.- On another note, does anyone have a good revision tool for oxidisation because I fail at it, and a place where can I find legacy papers?

Other/Not applicable (Urgent) Help With A-Level Questions:

sandrabondaruk — Fri, 17 May 2013 18:58:05 GMT

aren't these the chemistry ISA exam questions?

Sixth form can someone help me with this quick question?

Gordon1985 — Fri, 17 May 2013 18:49:57 GMT

Quote:

Originally Posted by mabz123

unfortunately i didn't quite understand.

If you had 1 mol of Mn04- and 1 mol of V2+, then each MnO4- molecule would take 5 electrons from each V2+ molecule, meaning the oxidation states of both molecules change by 5.

But, you have more V2+ than you have Mn04-, so the Mn04- can't take 5 electrons from every V2+ because each MnO4- can only accept 5 electrons.

Don't think in mols, think in individual molecules. If you had 1 MnO4- molecule and 1 V2+ molecule. The Mn04- would take 5 electrons and that would be that. If you had 1 Mn04- molecule and 5 V2+ molecules, than the one Mn04- could only take 1 electron from each V2+. So the oxidation state of Mn04- would change by 5 but the oxidation state of V2+ would only change by (the amount of Mn04- / the amount of V2+)* the number of electrons Mn04- can accept, which is (1/5)*5 = 1. This makes sense as you know each V2+ molecule only gave 1 of it's electrons to the Mn04-.

Basically the oxidation state of V2+ chnages by....

(The amount of Mn04- / the amount of V2+) * the number of electrons Mn04- can accept. This is (1/5)*5 = 1.

Going back to mols, you have 1 Mn04- and 1.67 V2+, and Mn04- can accept 5 electrons. So just doing the same thing as above...

(1/1.67)*5 = 3.

Chemistry Revision

Yoruichi Shihoin — Fri, 17 May 2013 17:44:20 GMT

A way to remember the general formulas of alkanes and alkenes in organic chemistry is to remember that alkanes have two 'a's, and that the second 'a' stands for 'addition', i.e alkanes=CnH2n+2. I know it kind of sounds complicated at first, but once you learn the general formulas of alkanes and alkenes, then everything else can just be worked out from there.

Sixth form solvent extarction

Dynamo123 — Fri, 17 May 2013 16:50:25 GMT

Simply put, its when two immiscible sovents are shaken together. One of them has the solute which passes on to the other solvent that hasBETTER capability of dissolving it. Since they are immiscible so the solute cant go back and is extracted.

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Sixth form Chemistry A Level in a year? (OCR)

Nathan2995 — Fri, 17 May 2013 16:14:51 GMT

My friend did A level Chemistry (OCR) in a year, he is coping absolutely fine (he is actually doing better than me tbh). Having said that he is quite intelligent and it will be tough, but by no means impossible. :)

Secondary nanotube question

JMaydom — Fri, 17 May 2013 14:34:21 GMT

Quote:

Originally Posted by leosco1995

Oops, sorry. I added it now.

Absorbed h2 is in equilibrium with gaseous h2... Maybe try looking up absorption isotherms to supplement your knowledge.
As you lower the pressure of h2, hydrogen will (overall) desorb from the carbon as the pressure of h2 gas has dropped. This acts to maintain a near constant pressure of h2 in the container.
I'm sure you can state le chataliers principle!

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Sixth form Enthalpy change on neut

MBT Bot — Fri, 17 May 2013 13:18:01 GMT

It's been a while since you posted and nobody's replied yet...maybe you should check out MarkedbyTeachers.com, TSR's sister site. It has the largest library of essays in the UK with over 173,000+ coursework, essays, homeworks etc.. all written by GCSE, A Level, University and IB students across all topics.

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Secondary Good Revision Sites Covering all of AQA GCSE C2 and C3?

BayHarborButcher — Fri, 17 May 2013 13:15:44 GMT

Quote:

Originally Posted by decb24

Yes I do, as I stated I have a textbook I just much prefer the format of sites like Bitesize.

Then no you don't have a revision guide, you have a text book. Two different things... Maybe have a look at get revising, you'll probably find good revision notes there.

Sixth form need help

Alkranite — Fri, 17 May 2013 10:40:27 GMT

that is useful thank you :)

Sixth form Chemistry AS unit one help?

JMaydom — Fri, 17 May 2013 10:18:04 GMT

Quote:

Originally Posted by Cosmic_Chique

Hi! I was wondering if anyone could give me some help on bond angles and shapes? I just can't get my head around it!

Its all about the most efficient packing of ligands around the central atom. Learn these important geometries and how to predict the number of lone pairs. Once you have that its really easy!

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Undergraduate Chemical Thermodynamics (First year undergrad.)

JMaydom — Fri, 17 May 2013 10:15:17 GMT

Quote:

Originally Posted by jrphol

That's brilliant thank you. Thought I was but have seen a couple of points on there that have corrected me. So will give a proper read now.

One of the few good things about thermo is that you can write most of it down in a couple of pages!

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Secondary Downward and Upward delivery

chessraceryousuf — Fri, 17 May 2013 10:06:19 GMT

thanks for that

Undergraduate Which mechanism? & Prediction of products?

illusionz — Thu, 16 May 2013 22:52:05 GMT

Quote:

Originally Posted by LeonVII

Indeed, but the percentage of the 1,2 product would be dwarfed by the other, no?

Minor product indeed but any exam answer should, in my opinion, state that both are formed as it gives a more detailed description of the reaction and shows you understand what is going on.

Sixth form Kinetics question

JMaydom — Thu, 16 May 2013 21:42:46 GMT

Quote:

Originally Posted by Big-Daddy

The activation energy barrier increases as the average energy of the particles increases?! But I was taught that if you increased the temperature then a greater number of particles have energy > activation energy?

What? Is that something you've been taught, or did you think I said that??? Activation energy is determined by the energy of the transition state relative to the ground state. To a reasonable approximation this should be constant with temperature.
To say that in general the energy barrier increases as the average energy of the particles increases seems very dodgy, if anything i would say the other way round. You do get what I mean by absolute energy right?

Yes at higher T more particles have the required energy, but you're thinking solely in terms of simple collision theory. Have a look at the eyring equation in contrast to SCT and you'll see just how basic the model you know is.

Sixth form Chemistry question (A level equiv)

Dan1909 — Thu, 16 May 2013 18:56:51 GMT

Deionised water is just water, but with nothing else in it. Pure water.

Ammonium is the NH4+ ion for ammonia. If you take normal Ammonia, and attach another H+ to it you get ammonium.

Sixth form Chemistry A-Level Home School Help

emilylaurajenkin — Thu, 16 May 2013 18:52:18 GMT

Thanks for the reply! How would I be able to do this? I'm so confused and stressed out :(

Undergraduate How did they get this answer?

Dr. Django — Thu, 16 May 2013 18:17:23 GMT

Quote:

Originally Posted by R1C3W1N3

Sodium?

You did a funny

Sixth form Where is this coming from

LeonVII — Thu, 16 May 2013 17:08:44 GMT

Quote:

Originally Posted by ILoveUSA

That still doesn't give the 2:5 ratio, like the mark scheme has

It does. You have the ratio 1:2.49... which is essentially, 1:2.5

Multiply by 2 and you get 2:5

Secondary Haber process help needed please!

zef1995 — Thu, 16 May 2013 16:41:27 GMT

Quote:

Originally Posted by 3mma_gal96

Yeah, thank-you, makes much more sense now!:D

That's great to hear, glad we could be of help :D

Sixth form Is my answer right

charco — Thu, 16 May 2013 16:22:59 GMT

Quote:

Originally Posted by ILoveUSA

A solution of phenol has the concentration of 38gmdm-3
Ka of phenol is 1.31*10-10 moldm3
Calculate pH

MY ANSWER

moldm-3 = 38/94
= 0.404 moldm^-3
Square root of (1.31*10-10 * 0.404)
= 0.00000725
-log ^
= 5.1

Sixth form Thermal decomposition of copper (II) carbonate

bbrain — Thu, 16 May 2013 15:53:10 GMT

You are given basic copper (II) carbonates of the form Cu(OH)₂.xCuCO₃. The value of x can be determined by gravimetric analysis, i.e. by measuring mass changes.

Thermal decomposition equations:

CuCO₃ --> CuO + CO₂
Cu(OH)₂ --> CuO + H₂O

Question: Explain how you would process your data to determine the value of x. NB this is a planning exercise; no data are actually obtained or given.

Mark scheme:
Work out M_r for CuCO₃ =123.5, Cu(OH)₂=97.5 and CuO = 79.5
Relative mass loss for CuCO₃ = 35.6%; that for Cu(OH)₂ = 18.5%.
(% mass loss - % mass loss for Cu(OH)₂) : (% mass loss - % mass loss for CuCO₃) gives ratio CuCO₃:Cu(OH)₂
Divide by value of Cu(OH)₂ to obtain x.

I don't understand the bold bit. If both copper compounds decompose together, how can one work out the composition?

Ref: 9791_y08_sp1.pdf
Cambridge Pre-U Specimen for use 2008 onwards

Sixth form molar ratios etc. need some answers checking.

Sezra — Thu, 16 May 2013 15:05:53 GMT

I would just like someone to check my answers for me. Im not sure my molar ratios are right. I'm going off of the info found here

http://www.ehow.com/how_8149226_calc...lar-ratio.html

The only thing im missing is the molar ratio for Potassium + Iron (II) Sulphate. Any help on that is welcome.

Write properly balanced equations for the following reactions. State the molar ratios of the reactions.

Sodium Hydroxide + Phosphoric Acid
3NaOH + H3PO4 → Na3PO4+3H2O
Molar ratio =3:1

Methane + Oxygen
CH4 + 2O2 → CO2 + 2H2O
Molar ratio = 2:1

Zinc + Sulphuric Acid
Zn + H2SO4 → ZnSO4 + H2
Molar ratio= 1:1

Ammonia + Hydrochloric Acid
NH3 + HCl → NH4Cl + H2O
Molar ratio = 3:1

Sodium bicarbonate + Phosphoric Acid
H3PO4 + 3NaHCO3 → Na3PO4 + 3CO2 + 3H2O
Molar ratio = 3:1

Potassium + Iron (II) Sulphate
MnO4- + 8H+ + 5Fe+2 → Mn+2 + 5Fe+3 + 4H2O
Molar ratio = ???

Iron + Hydrochloric Acid
Fe + 2HCl → FeCl2 + H2
Molar ratio = 2:1

Nitric Acid + Magnesium Hydroxide
2HNO3 + Mg(OH)2 → Mg(NO3)2 + 2H2O
Molar ratio = 1.5:1

Calculate the molecular weights of the following compounds
CuNO₃= 125.551 g/mol
KOH = 56.105 g/mol
Fe = 55.845 g/mol
H₃PO₄= 97.995 g/mol
CO₂ = 44.009 g/mol
CH₄ = 16.042 g/mol
Na₂SO₄ = 142.042 g/mol
LiOH = 23.948 g/mol
Ca₃(PO4)₂ = 310.176 g/mol

Sixth form Chem2: Appearance of Sodium Halides and Potassium halides

charco — Thu, 16 May 2013 14:42:52 GMT

Quote:

Originally Posted by lindseyyy

Anyone have any idea what the appearance of:

- Sodium/ Potassium Bromide
-...iodide
-...chloride

are in aqueous solution? The colours?

Thanks

They are all white crystalline solids that form colourless solutions ...

Sixth form state symbols

CurtainrailMan — Thu, 16 May 2013 14:34:51 GMT

You won't be marked down for putting them there so long as they are correct.

AS Chemistry intermolecular forces

joostan — Thu, 16 May 2013 13:44:54 GMT

Quote:

Originally Posted by carrellii

Thank you! :)

Any time :)

Sixth form F322 ocr jan 2013 paper and ms :)

Angryification — Thu, 16 May 2013 12:24:13 GMT

Where did you find them and do you happen to have the F212 biology paper for Jan 13? :)

Secondary C3 questions - alcohols and equilibrium

Extricated — Thu, 16 May 2013 10:44:03 GMT

Quote:

Originally Posted by indigorain

It's C3 on Monday!! I have a few revision queries I'd like to share...

It says on the spec that alcohols should be represented in the following form:

(ethanol) CH3CH2OH

I understand this is the structual formula, and represents how it's drawn. However, I have always been taught to write it like this:

C2H5OH

Do people think this would still get marks, or do I have to start writing structural formulae?

Also... I've realised that the concept of equilibrium is causing a few problems. I understand the idea of the reactions working at exactly the same rate in each direction, but I don't see how that produces a yield with 'relative amounts of each reacting substance at equilibrium'.

Thanks everyone and good luck!

1) It depends on the mark scheme, they can be quite nit picky but both are correct.

2) The reaction is going forwards as well as backwards hence when the reaction is going to the left, all of the reactants are also being produced in the ratio in which they are reacting.

E.g 2h2+o2 -- > 2h2o

if the reaction goes to the left hand side, the hydrogen and oxygen will be produced in relative amounts of 2:1

Sixth form are these questions from the ocr endorsed book? help please!

mabz123 — Thu, 16 May 2013 10:02:04 GMT

Quote:

Originally Posted by Tha Realest

http://www.freeexampapers.com/index....2006_jun_w.pdf

You want question 4(a) from the above link bro

i owe you bro, lol thanks again man.

Sixth form Any Edexcel AS A2 Chemistry questions for revision apart from past papers?

richardchan001 — Thu, 16 May 2013 05:40:05 GMT

Hi all,
Would there be any Edexcel AS A2 Chemistry questions for revision apart from past papers?
Thanks.
Richard

Sixth form Maddening Buffer Question!

WaqasK — Wed, 15 May 2013 22:15:52 GMT

Can anyone help me with the whole of the chapter 3 , maths isn't my strong point at all. When I look at the past papers I feel clueless

Sixth form A2 Chemistry Practical CCEA

bee1232000 — Wed, 15 May 2013 21:08:28 GMT

I did my exam today, it went relatively well i think i lost a maximum of 10 marks out of 70. Does anyone have any experience or knowledge on the grade boundaries? Everyone said it went well so will it be standardized down? Thank you all in advance!

Secondary Need help with this question!

student_slasher — Wed, 15 May 2013 19:24:50 GMT

a) short chain molecule, smaller molecule so experiences less intermolecular (VDW) forces which are easier to break.

Sixth form Kinectics- Temperature

No-Idea. — Wed, 15 May 2013 16:58:25 GMT

Cheers mate.

Undergraduate Gases Question

ThisIsOurDecision — Wed, 15 May 2013 14:46:14 GMT

Quote:

Originally Posted by Piguy

Surely though most of the CO2 will solidify like the original gas - The melting point of CO2 is 195K, 170K is well below that - so only the same proportion that remained a gas originally will remain gas from the 5 mol you added..

i.e. 1.42x10^-3mol remained from 20mol of CO2... so adding 5 mol, only 1.42x10^-3/4 = 3.55x10^-4mol of it will actually remain a gas so overall there's only 1.775x10^-3mol of gas...

That makes much more sense and gives a more reasonable answer too, thanks for the help!

Undergraduate Rather tricky pericyclic reaction

Midlander — Wed, 15 May 2013 14:42:09 GMT

Quote:

Originally Posted by illusionz

Ah, so simple once you spot it! Cheers

I find it's so easy to get lost down a rabbit hole with these sorts of questions if you get the first step wrong...

Pericyclic stuff tends to be like that-my last ever exam at uni is on radicals and carbenes, so I share your pain!

Sixth form F332 Jan 2013 ?

krisshP — Wed, 15 May 2013 14:11:52 GMT

Quote:

Originally Posted by User32432432

Does anybody have a link to the Jan 2013 paper ? or a unofficial mark scheme ?

Thanks

http://www.thestudentroom.co.uk/show...5#post42297175

Other/Not applicable Temperature and state

charco — Wed, 15 May 2013 13:46:47 GMT

Quote:

Originally Posted by Big-Daddy

This is a bit of a random question but it occurred to me so let's ask:

Let's say we have a sample of a large molecule or an amino acid or something (not relevant for our discussion). And then we raise the temperature in which this sample is held. And then keep raising it. Eventually we will end up with nuclei and electrons floating around.

In what order does this happen? i.e. do the bonds between the atoms break, then the ionizations begin? Or are the electrons removed first?

Logically the least energy demanding process should occur first. Bond enthalpy terms are usually of the order of 200 - 600 kJmol^-1 (ish). A C-H bond is (on average) 412 kJ mol^-1.

Ionisation energies for atoms usually higher (obviously it depends on the element), but as you don't have individual atoms you would need to think about molecular orbitals and the promotion /ionisation of electrons there.

A real life example would be the mass spectrometer. In this the molecules are ionised first and then the ions fragment. However, this isn't with thermal energy, rather with a beam of high energy electrons.

However, it seems likely to me that you would first get ionisation of the molecule and/or fragmentation, depending on the species under consideration.

Sixth form AQA Chem 4 (New Syllabus) Dual Equivalence points

Jake Shake — Wed, 15 May 2013 13:45:14 GMT

In Chem 4 what do we need to know about Titrations with dual equivalence points? As I did a legacy paper and it wanted me to write the reaction that occured at the first, and the concentration of the Acid at the second... I looked in the textbook for the new curriculum and it doesn't mention it anywhere... Can anyone help?

Sixth form Oxidisation question

redtortoise — Wed, 15 May 2013 13:22:43 GMT

Thanks I was having a bit of a stress out this morning as my teacher didn't give me the mark scheme

Sixth form Le Chatelier's Principle

jones107 — Wed, 15 May 2013 13:14:45 GMT

i think it is getting at the fact that the higher initial concentraton of reactant means there is a higher ratio of reactants to products, so more products are made.

that leads to more producs that reactants, so the equiliburium shifts back to the left,

so eventually you get the same ratios as before, just the concentratiuons of each chemical/compound is greater

that or that an equilirium will more to the side with less mols of chemicals used
eg.
H2O < > H+ + OH-

this is how water acts as a weak acid,

there are more mols of product (H+ and OH-)=2 than reactant (H2O)=1
so the equilibrium shifts to the left sore more H20 is produced :)

hope this helps

Undergraduate Kinetics - effective 1st order rate constant

DonnieBrasco — Wed, 15 May 2013 11:34:28 GMT

Quote:

Originally Posted by Big-Daddy

No worries, happy to help. It helps me make sure I understand too!

Yup, t[sub[1/2[/sub]=ln(2)/k_1mod for a pseudo-first order reaction. And for a pseudo-cth order reaction, you just use the modified rate constant in the place of where you would use the normal rate constant for a normal c-order reaction.

The thing to bear in mind is that this isn't the best approximation unless [A] and [ B] are vastly greater than [C] (because obviously, if their concentrations change, your effective rate "constant" is not constant!) Other than that, just remember to calculate the effective constant properly as I showed in my above post.

Hope this helps! :)

Yep got it :)

Thanks again

Sixth form AQA AS Level Redox question

MEPS1996 — Wed, 15 May 2013 10:46:34 GMT

Quote:

Originally Posted by charco

The issue is that the half equations fare different for acidic and basic conditions.

Quote:

Originally Posted by charco

You solved it as if it were in acidic conditions, but as you can see from the fact that it's reacting with sodium hydroxide it must be in base. In this case you use water to balance the half equations:

Cl₂ + 6H₂O --> 2ClO₃^- + 12H⁺ + 10e
Cl₂ + 2e --> 2Cl^-

equalise the electrons and add:

Cl₂ + 6H₂O --> 2ClO₃^- + 12H⁺ + 10e
5Cl₂ + 10e --> 10Cl^-
-------------------------------- add
Cl₂ + 6H₂O + 5Cl₂ --> 2ClO₃- + 12H⁺ + 10Cl^-

collect terms

6Cl₂ + 6H₂O ---> 2ClO₃^- + 12H⁺ + 10Cl^-

divide through by 2

3Cl₂ + 3H₂O ---> ClO₃^- + 6H⁺ + 5Cl^-

Now add base ions (hydroxide) to remove the hydrogen ions to both sides and cancel down water.

3Cl₂ + 3H₂O + 6OH^- ---> ClO₃^- + 6H⁺ + 5Cl^- + 6OH^-

3Cl₂ + 3H₂O + 6OH^- ---> ClO₃^- + 6H₂O + 5Cl^-

3Cl₂ + 6OH^- ---> ClO₃^- + 3H₂O + 5Cl^-

Thanks a lot Thats great cheers :)

Due to popular demand, AQA Chemistry revision notes (AS and A2).

ooyehoh7 — Wed, 15 May 2013 08:14:20 GMT

Thank you!! These are brilliant!

Sixth form Jan 13 Chem2 Paper

KSmith15 — Wed, 15 May 2013 06:40:39 GMT

Thank you Sooo Much!!!!!! :D

Sixth form Why is it important to study organic chemistry

Dan1909 — Wed, 15 May 2013 00:45:43 GMT

Quote:

Originally Posted by KKP109

Thanks guys is there any reasons that I can link to amino acids & sugars.

Posted from TSR Mobile

This may (or may not) be a useful starting place :)

http://www.piercenet.com/browse.cfm?...2-238BB9C7270F

It's about peptide synthesis.

Sixth form Quick complex ion question

cheesypuff — Tue, 14 May 2013 21:05:31 GMT

brah, yes

Sixth form Chemistry F321 & F322

chigyy — Tue, 14 May 2013 20:43:00 GMT

Please does anyone have the January papers for these? With the answers if you can. Thanks you :D

Sixth form OCR F321 january 2013 paper

Happy03 — Tue, 14 May 2013 19:29:21 GMT

Quote:

Originally Posted by GeneralOJB

http://imgur.com/a/nM6Kj

All I could find.

thankyouu :)

Sixth form How to revise OCR Chemistry Salters B Your thoughts?

SyedaK — Tue, 14 May 2013 19:04:22 GMT

Do a lot of exam papers, that's what i do! and the CGP book helps a lot. why do you find Chemistry boring. I love it! Is it the teacher?

chem...

jannattulfirdaus — Tue, 14 May 2013 17:28:58 GMT

Sixth form Why isn't HCl(g) acidic?

JMaydom — Tue, 14 May 2013 17:03:22 GMT

Quote:

Originally Posted by ThrashMetal

Nothing is acidic until it becomes aqueous.

wow, sweeping statement!!! I would be very cautious about saying that. Probably true for most bronstead acids, but almost certainly not the case for other types. Think about the field of inter-stellar chemistry. It's all in the gas phase and there must be 'acid and base' like properties observed.

Undergraduate Chem Eng Help.

illusionz — Tue, 14 May 2013 16:58:47 GMT

I don't think there are any chemical engineers who post on the chemistry forum, sorry. Try the engineering one.

Sixth form Is this equation correct

drahc — Tue, 14 May 2013 16:48:02 GMT

What equation?!

Secondary Is it too late?

drahc — Tue, 14 May 2013 16:44:26 GMT

Speak to your exam officer (or your teacher if you don't have one).

I don't think it's too late - but you'll have to pay for the resit AND a late entry fee.

Sixth form Gold as a metal

LeonVII — Tue, 14 May 2013 16:32:33 GMT

Quote:

Originally Posted by KKP109

Why use gold instead of magnesium or sodium ? in terms of jewelry, goods etc

Magnesium and sodium would react with moisture pretty quickly to form other compounds...

Gold is very unreactive and so is perfect. Also, gold holds its value :D

Undergraduate Anomers and Epimers

JMaydom — Tue, 14 May 2013 16:03:23 GMT

Quote:

Originally Posted by Waqar.

Hey guys,

I know to find a chiral carbon atom in an enantiomer, you can simple find the carbon with 4 different functional groups. Is there an easy way/shortcut to tell if a monosaccharide is an anomer or epimer?

Just because the carbon has 4 different constituents doesn't mean it is chiral... be careful. The only way to be certain is to actually compare it to the molecules mirror image.

Sixth form Lattice enthalphy definition

Locked_box — Tue, 14 May 2013 14:16:21 GMT

I've found chemguide really useful for things like this. The guy who made the website also writes the calculations book for chemistry as/a level. Anyway here's the link http://www.chemguide.co.uk/physical/...s/lattice.html
On this page he states that they're basically the same but it depends on the question, the born haber cycle that you're looking at and what you're trying to find.

i suggest learning both so that when you're faced with the question all you need to do is check the sign of the enthalpy change, or the direction of the arrows.

Sixth form Reducing and oxdising agents help, + rep.

LeonVII — Tue, 14 May 2013 11:53:09 GMT

Quote:

Originally Posted by Hi, How are you ?

So HOBr is the oxidising agent, than opposed to the Bromine the compound?:)

Yes, but not the HOBr as a whole, only the bromine component.

Undergraduate Organic Questions

ThisIsOurDecision — Mon, 13 May 2013 18:57:55 GMT

Quote:

Originally Posted by illusionz

You require an Sn1 reaction because the centre you're trying to substitute is tertiary, and as O- is a poor leaving group I'd use HBr to activate it and supply the bromide ions at the same time, making it a faster reaction. It might not be necessary though...

OK cool, thanks for your help.

Undergraduate Optical isomerism in complexes

JMaydom — Mon, 13 May 2013 17:02:52 GMT

Quote:

Originally Posted by Big-Daddy

Thanks. That was quick.

What types of isomerism can the trigonal-bypramidal or square-pyramidal (each has 5 ligands) exhibit? (Between optical and geometric isomerism only, for the time being)

Not sure what you would call it, but there definitely are isomers. The axial and equatorial positions aren't equivalent but are fluxional cia berry pseudorotation at a temperature which I suppose is dependant on the metal and ligands.
The different environments can be frozen out in NMR.

Posted from TSR Mobile

Sixth form Textbook

ThisIsOurDecision — Mon, 13 May 2013 16:37:56 GMT

Quote:

Originally Posted by Sheikh Awie

Hey guys,any of you used the Engineering Maths textbook by KA Stroud? I wonder if there is a chemistry textbook like that? it would be a great help.

I have Maths for Chemistry by Monk, and to be perfectly honest it isn't that great. Khan Academy is useful, and I have been told about http://discovermaths.rsc.org/ but have never used it.

Sixth form Unit 2 Chemistry Energetics Question

joostan — Mon, 13 May 2013 16:01:19 GMT

Quote:

Originally Posted by lindseyyy

Fantastic, thanks (:

Any time :)

Sixth form AS Chemistry past paper help needed

krisshP — Mon, 13 May 2013 15:59:32 GMT

Quote:

Originally Posted by freetown

Maybe the attached images could help you visualize the reason why you have to get those two answers.

This is what I always draw in my exams :P
The purple lines represent "H-" bond.

Quote:

Originally Posted by xfallingstar

Make sure that for each C atom, there are 4 bonds attached to it. You may be adding too many Hydrogens to some C atoms. Especially the ones with the double bonds in the 1st example - they wouldn't have any H atoms attached to them.

For the c)ii) just remember - when the Bromine is added, the C=C opens up to form single bonds with a Br atom. So since there's 2 C=C bonds in rotundone, then that's 4 single bonds that can be formed. Therefore, it's 2 !

Thanks so much for the nice help :D

Sixth form Half-Equations

xfallingstar — Mon, 13 May 2013 12:56:10 GMT

That is correct.

Have you heard of OILRIG?

It's a silly acronym but it works xD

Oxidation Is Loss Reduction Is Gain

Secondary Long questions/answers

AquisM — Mon, 13 May 2013 12:42:29 GMT

I did IGCSE Science (Edexcel, so if you're with CIE, I'm not sure), and my teacher told me that while paragraphs are preferred, if you are running out of time, putting something down is better than putting nothing down, and the examiners are probably lenient enough to give you the marks (since mostly for the long answers they're looking for specific key words used, so if you put them down you should be fine :) ). Still, I recommend writing in full sentences, or if absolutely necessary to write in bullet points, don't make it too abbreviated, but rather close to standard English.

Sixth form How many past papers?

krisshP — Mon, 13 May 2013 07:24:03 GMT

Quote:

Originally Posted by Melissa.S.

I've started doing past papers for my subjects (biology , chemistry and maths) about 2 weeks ago. Was just wondering how long it is before you guys start getting consistent good grades in them? I'm getting C's in them at the moment, got 4/5 weeks till my main exams, hopefully I can improve in that time!

If you have spare time after doing ALL papers, do them again.

Secondary transition metals (exam on wednesday)

bubakazouba — Sun, 12 May 2013 22:34:03 GMT

Quote:

Originally Posted by charco

Ligand just means 'linked to'.

In hydrated salts, the water molecules surround the metal ions in the same way as in complex ions. There may be some adjustment to incorporate the anions into the structure, but the fundamental structure is still there.

This can be shown by the colour of the crystals when hydrated and dehydrated.

The very fact that copper(II) sulphate crystals are the same colour as the solution demonstrates that the ions are surrounded by the water ligands and the 'd' orbitals are split by the same amount.

hmm ok I think I understood this, may you answer 2 and 3 please

Sixth form Why are azo groups coloured?

AlishaMcC — Sun, 12 May 2013 21:45:48 GMT

Thanks very much :)

Sixth form Alanine Complex

illusionz — Sun, 12 May 2013 20:06:31 GMT

Quote:

Originally Posted by String.

Hmm no I was looking at a past paper question and it said to add Copper (II) Sulphate, and that a dark blue solution was formed; in the mark scheme it said that the deduction was that Glycine was present. I don't understand how they were able to rule out alanine?

A colour change like that isn't enough to rule out alanine (or another amino acid, or even a completely different sort of ligand) as far as I'm concerned.