Chemistry: Post your doubts here!

[Dania_Yz]

I cant understand the mass spectrometer, my teacher didn't teach us anything about it.. Help!

 

[daredevil]

And check this out :

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_2.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_ms_1 2 3 4 5 6.pdf

isnt the MP graph for question 4A wrong? Since there is a slight increase from P to S in melting point?

it is! :O

 

[daredevil]

I cant understand the mass spectrometer, my teacher didn't teach us anything about it.. Help!

is it needed for AS?

 

[Jinkglex]

Bon enthalpies? You're going to have to elaborate on that

As far as bond breaking and making is concerned, look at the following reaction:

CO + 1/2O2 -> CO2.

Simple enough right? Now all you have to remember is before reacting all atoms must go into their atomic gaseous state. Thus here, the O2 molecule will be broken into 2 O ATOMS. This is bond BREAKING. Then, the O atom will react with the CO molecule to make a CO2 molecule. During this, the C bonds with the extra O atom. And this is bond MAKING. Bond breaking is always ENDOTHERMIC because it always requires energy. Thus, Bond making is automatically exothermic.

Think of bond making as a bank transaction. When you're MAKING a bond you're putting in money. Now theres less money [Energy] in your pocket [surroundings], but more money [Energy] in your bank account [between bonded atoms]. Likewise when you make a withdrawal or BREAK a bond, there is less money in your bank account, but more in your pocket.

 

[knowitall10]

I cant understand the mass spectrometer, my teacher didn't teach us anything about it.. Help!

Which book are you using.. my book explains it pretty well.. i'm assuming you have the same one

 

[Jinkglex]

Mass spectrometer for AS is simple enough. You ionize/vaporize a sample, throw electrons at it until it loses an electron going into the +1 state, and pass it through electric field [to streamline it] and magnetic field [to make it acquire a circular pathway]. Then you change the magnetic field to make it hit your detector, giving you a mass/charge ratio against percentage abundance of said peak.

 

[knowitall10]

Bon enthalpies? You're going to have to elaborate on that

As far as bond breaking and making is concerned, look at the following reaction:

CO + 1/2O2 -> CO2.

Simple enough right? Now all you have to remember is before reacting all atoms must go into their atomic gaseous state. Thus here, the O2 molecule will be broken into 2 O ATOMS. This is bond BREAKING. Then, the O atom will react with the CO molecule to make a CO2 molecule. During this, the C bonds with the extra O atom. And this is bond MAKING. Bond breaking is always ENDOTHERMIC because it always requires energy. Thus, Bond making is automatically exothermic.

Think of bond making as a bank transaction. When you're MAKING a bond you're putting in money. Now theres less money [Energy] in your pocket [surroundings], but more money [Energy] in your bank account [between bonded atoms]. Likewise when you make a withdrawal or BREAK a bond, there is less money in your bank account, but more in your pocket.

Nice... thanks.. it explained everything

 

[knowitall10]

Mass spectrometer for AS is simple enough. You ionize/vaporize a sample, throw electrons at it until it loses an electron going into the +1 state, and pass it through electric field [to streamline it] and magnetic field [to make it acquire a circular pathway]. Then you change the magnetic field to make it hit your detector, giving you a mass/charge ratio against percentage abundance of said peak.

Mass spectrometry is only in A2.. what are you guys talking about?

 

[PANDA-]

Looks like almost everyone here is A2
I'm still AS.

 

[knowitall10]

Looks like almost everyone here is A2
I'm still AS.

But you still manage to embarrass us A2 candidates by answering our (at least mine) stupid basic questions

 

[daredevil]

Looks like almost everyone here is A2
I'm still AS.

Yeah it seemed like that to me to..... nice to find a fellow AS here *hi5

 

[Jiyad Ahsan]

You got it wrong there, all alkali are bases bot not all bases are alkaline. This is because alkali's are bases which are dissolved in water, like NaOH(aq) for example.
Reactivity mainly depends on the effective nuclear charge, atomic radius, and molar mass (not really sure about molar mass lol).

yep thats right^

btw panda... i need a bit of help
in organic chemistry we have oxidising agents KMnO4 and K2Cr2O7
and some reactions state using only one.. is there a difference? similarly LiAlH4 and NaBH4..

 

[knowitall10]

Yeah it seemed like that to me to..... nice to find a fellow AS here *hi5

Check it out.. daredevil likes my depressing reply to Panda

 

[Jiyad Ahsan]

Looks like almost everyone here is A2
I'm still AS.

whoa awesome you knew so much stuff i thought you'd be in A2 .. if not higher up

 

[Gémeaux]

Hey can anyone step by step explain Q1 (c) for MJ12-41 ??
I always get SIMPLE lattice energy or enthalpy change questions in general wrong :/
I'd be forever greatfuk (

I'd suggest to draw the cycle every time for questions like this.

∆H = 148 + 736 + 1450 + 496/2 - 141 + 798 – 3791
= –552 kJ mol–1
Hope this helps.

 

[Jiyad Ahsan]

i guess i'll just google my question?

 

[knowitall10]

Panda probably feels so damn awesome now... don't you
panda
i get that feeling too when i explain the smartypants in my class

 

[knowitall10]

i guess i'll just google my question?

Whats your question?

 

[Jiyad Ahsan]

yep thats right^

btw panda... i need a bit of help
in organic chemistry we have oxidising agents KMnO4 and K2Cr2O7
and some reactions state using only one.. is there a difference? similarly LiAlH4 and NaBH4..

this one ^

 

[PANDA-]

yep thats right^

btw panda... i need a bit of help
in organic chemistry we have oxidising agents KMnO4 and K2Cr2O7
and some reactions state using only one.. is there a difference? similarly LiAlH4 and NaBH4..

KMnO4 is a strong oxidising agent if I'm not mistaken, and K2Cr2O7 is a weak oxidising agent.
You can use K2Cr2O7 for mild oxidation of Alcohol to Aldehydes... Unless a reflux with heat is used, then even K2Cr2O7 can oxidise alcohols directly to carboxylic acids.