VCE Chemistry Question Thread

[RuiAce]

C17H36(g /l) → C15H32(g/l) + C2H4(g)

from the 2007 VCAA Unit 3 exam - do you need to be consistent with states?

500th Post

Fairly sure catalytic cracking is gaseous.

i reckon keep it all as gaseous!
and ahh i was told the catalyst for H2SO4 for esterification is concentrate, and had to be liquid :/ idk- will check it up once im back to school xD

Pretty sure it is concentrated but it's not THAT concentrated. Liquid hydrogen sulfate is 18M sulfuric acid

[bananabreadbelle]

Just wanted clarification on chromatographic principles:

What is the 'priority' of factors that affect a substance's rate of adsorption and desorption/retention time - does size of the sample molecule ever take precedence over the polarity of the sample molecule?
Or does this differ between HPLC and GC methods?

Thanks so much in advance! 

[taylorjj57]

Just wondering about naming organic molecule chains. I know where you start if it's an alkanol/carboxylic acid (that functional group) or if there's only one side chain (end closest to functional group) and to name functional groups at the start in alphabetical order (alcohols excepted) .

But what if there's two side chains both the same distance from the ends away?
Eg. CH3CH2CH(CH3)CH(CH2CH3)CH2CH3
&
CH3CNHCH2CH(CH3)CH3

I would call them 3-methyl-4-ethylhexane and 2-methyl-4-pentanamine but as I said earlier I'm not sure.

Could someone please explain some rules to naming these or suggest ways that may help in working out names

Thanks heaps

[sweetcheeks]

Just wondering about naming organic molecule chains. I know where you start if it's an alkanol/carboxylic acid (that functional group) or if there's only one side chain (end closest to functional group) and to name functional groups at the start in alphabetical order (alcohols excepted) .

But what if there's two side chains both the same distance from the ends away?
Eg. CH3CH2CH(CH3)CH(CH2CH3)CH2CH3
&
CH3CNHCH2CH(CH3)CH3

I would call them 3-methyl-4-ethylhexane and 2-methyl-4-pentanamine but as I said earlier I'm not sure.

Could someone please explain some rules to naming these or suggest ways that may help in working out names

Thanks heaps

Your first molecule should be called 3-ethyl-4-methylhexane. The things out the front should go in alphabetical order.
As for naming amines, I hate their nomenclature, it is quite complicated. Your molecule should be named 2 amino 4 methylpentane

[taylorjj57]

Your first molecule should be called 3-ethyl-4-methylhexane. The things out the front should go in alphabetical order.
As for naming amines, I hate their nomenclature, it is quite complicated. Your molecule should be named 2 amino 4 methylpentane

Thanks sweetcheeks that sorta clears things up

[sweetcheeks]

How is everyone going with practice exams?

I did the 2016 TSSM exam at school yesterday and I feel that it really missed most of the course content, it was very organic heavy (no enthalpy/combustion question, no real gravimetric/volumetric analysis).

[Individu]

Could someone help me out with this question

Galvanic cells eventually cease producing electrical energy. Which of the following is true at this point:

• no ions will be present in the salt bridge
• the cell reaction will be at equilibrium

[Swagadaktal]

Could someone help me out with this question

Galvanic cells eventually cease producing electrical energy. Which of the following is true at this point:

• no ions will be present in the salt bridge
• the cell reaction will be at equilibrium

Ions will be present in salt bridge
The cell reaction will be at equlibrium --- was gonna type a more in-depth response but i gotta take a dump rn so might come back to this later

[larissaaa_]

40.0ml of 0.200M HCl reacts with 20.0ml of 0.100 M NaOH. Calculate the pH of the resulting solution.

After caculations, we find that there are 0.008 mol of HCl and 0.002 mol of NaOH. From here I get lost, that means that HCl was in excess and in reality only 0.002 mol of HCl reacted with the NaOH because it's a 1:1 ratio. Consequently I thought that would produce 0.002 mol H₃O⁺ because again, 1:1 ratio so I subbed that into c = n/v to find the concentration. But in the answers why is the 0.006 mol excess HCl used in the c = n/v equation? Did I make a really silly mistake? Thaaankss

[nadiaaa]

Hiii
Idk why but i cant seem to work out whether there will be more or less particles on the right hand side... I could only work out that the reaction was exothermic..soooo need some help thank you

[Swagadaktal]

Hiii
Idk why but i cant seem to work out whether there will be more or less particles on the right hand side... I could only work out that the reaction was exothermic..soooo need some help thank you

For the same temperature at a higher pressure, we get higher yield. This means that there's a net forward reaction. With a pressure increase, the equilibrium will favour the side with fewer particles to partially oppose the increase in pressure (as this would lead to a partial decrease in pressure).
Hence, we can conclude that there are fewer particles on the right hand side (products), and by including your observation that the reaction is exothermic we should get D

[sweetcheeks]

40.0ml of 0.200M HCl reacts with 20.0ml of 0.100 M NaOH. Calculate the pH of the resulting solution.

After caculations, we find that there are 0.008 mol of HCl and 0.002 mol of NaOH. From here I get lost, that means that HCl was in excess and in reality only 0.002 mol of HCl reacted with the NaOH because it's a 1:1 ratio. Consequently I thought that would produce 0.002 mol H₃O⁺ because again, 1:1 ratio so I subbed that into c = n/v to find the concentration. But in the answers why is the 0.006 mol excess HCl used in the c = n/v equation? Did I make a really silly mistake? Thaaankss

I think you are getting confused with H3O+ forming. When an acid reacts with a base, water is produced, not H3O+. The reaction is HCl + NaOH --> H2O + NaCl.
HCl is a strong acid, it ionises almost 100%. HCl + H2O --> H3O+ + Cl-. ). 0.008 mole of HCl will form 0.008 mole of H3O+. In the reaction with NaOH, the limiting reagent being the NaOH, only 0.002 mole of the 0.008 H3O+ will react, leaving 0.006 mole H3O+/H+ left in the solution.

The pH will be the -log[H3O+] concentration. This concentration is derived by going [0.006mole H3O+]/[0.060L]

[larissaaa_]

I think you are getting confused with H3O+ forming. When an acid reacts with a base, water is produced, not H3O+. The reaction is HCl + NaOH --> H2O + NaCl.
HCl is a strong acid, it ionises almost 100%. HCl + H2O --> H3O+ + Cl-. ). 0.008 mole of HCl will form 0.008 mole of H3O+. In the reaction with NaOH, the limiting reagent being the NaOH, only 0.002 mole of the 0.008 H3O+ will react, leaving 0.006 mole H3O+/H+ left in the solution.

The pH will be the -log[H3O+] concentration. This concentration is derived by going [0.006mole H3O+]/[0.060L]

THAT MAKES SENSE THANK YOU

[taylorjj57]

Does anyone know what the 6 general equations to do with acids and bases are? I've seemed to misplaced them but I think it's sorta like
Acid + base -> H20 + CO2 etc

If anyone could help be out that'd be sweet

[sweetcheeks]

Does anyone know what the 6 general equations to do with acids and bases are? I've seemed to misplaced them but I think it's sorta like
Acid + base -> H20 + CO2 etc

If anyone could help be out that'd be sweet

Acid + Base --> Salt and Water
Acid + Metal oxide --> Salt and Water
Acid + Metal Hydroxide --> Salt and Water
Acid + Metal Carbonate --> Salt and Water and Carbon Dioxide
Acid + Metal Sulfide --> Salt and H2S
Acid + Metal --> Salt and Hydrogen gas