VCE Chemistry Question Thread

[psyxwar]

The attached picture has been named in our notes as 3-ethyl-2methylhexane. Makes sense but couldn't it also be 3-isopropylhexane? There are two ways to get the six carbon backbone...(Image removed from quote.)

Yeah you're correct.

[vintagea]

am i supposed to post my questions here?

A calorimeter containing 100 mL of water is calibrated by
passing a 3.00 A current through the instrument for 36.0 s
at a potential difference of 3.50 V. The temperature rises by
0.82°C.
Potassium hydroxide weighing 0.654 g is added to the
calorimeter and dissolved rapidly by stirring. The temperature
rises from 20.82°C to 22.23°C.
Determine the delta H

i know the calibration factor is 461J/celcius

i got -55.8 kj/mol but the answers say -55.8MJ/mol

did i get it wrong?

[hobbitle]

Hi guys. See attached pic.
Just want to clarify a few things.
Even though we can make an 11-carbon long chain here (incl the substituent), we DONT do this because it's a ring, right? So this is cyclodecyne?
And then the numbering... Do you start at the triple bond, so the methyl group is on carbon 6?
So do we get 6-methylcyclodecyne?

[lzxnl]

am i supposed to post my questions here?

A calorimeter containing 100 mL of water is calibrated by
passing a 3.00 A current through the instrument for 36.0 s
at a potential difference of 3.50 V. The temperature rises by
0.82°C.
Potassium hydroxide weighing 0.654 g is added to the
calorimeter and dissolved rapidly by stirring. The temperature
rises from 20.82°C to 22.23°C.
Determine the delta H

i know the calibration factor is 461J/celcius

i got -55.8 kj/mol but the answers say -55.8MJ/mol

did i get it wrong?

So...dH = heat / number of moles of reaction = 461 J/K * 1.39 K / (0.654/56.1 mol) = 55.0 kJ/mol
Exothermic => dH = -55.0 kJ/mol
Answers seem off

Hi guys. See attached pic.
Just want to clarify a few things.
Even though we can make an 11-carbon long chain here (incl the substituent), we DONT do this because it's a ring, right? So this is cyclodecyne?
And then the numbering... Do you start at the triple bond, so the methyl group is on carbon 6?
So do we get 6-methylcyclodecyne?(Image removed from quote.)

I'm fairly sure you'd start at the triple bond. Triple bonds have higher priority than methyl groups. So this would be 6-methylcyclodecyne

[hobbitle]

Thanks lznxl.

Question:
Which metals in the table will react with water? Write chemical reactions for these equations.



This is my main point of confusion when it comes to electrolysis/galvanic cells.  Water.  I don't even know where to start here...

[lzxnl]

Thanks lznxl.

Question:
Which metals in the table will react with water? Write chemical reactions for these equations.

(Image removed from quote.)

This is my main point of confusion when it comes to electrolysis/galvanic cells.  Water.  I don't even know where to start here...

Look at where water is. It's at 1.23 V and -0.83 V. So anything above 1.23 V will oxidise water, while anything below -0.83 V will reduce water. So it looks like Au 3+ will oxidise water, while Al, Mg, Na, Ca, K, Li will all reduce water.

[vintagea]

thank you lzxnl

can you please help me: Suggest why salt is included in the contents of the bag.

please see attachment

[darklight]

Hi guys,
If a question asks for the combustion reaction of maltose, how do you know that maltose is (s)? Couldn't it also be (aq)? VCAA 2007,  exam 2, q4aii only allows for (s).

[vintagea]

Calculate the energy absorbed when 2.00 g of NH4Cl is
dissolved in:
i 100 mL water ii 1000 mL water
- find mols
-use delta H and the reaction equation...

i get how to do the question but the worked solutions says: The amount of energy absorbed does not depend on the volume of the solution?
why is that?
i know that changing the volume changes the concerntration of the solution because there is a constant 2g of NH4Cl is each solution... wouldn't the less concerntrated absorb less? i'm not sure... please clarify

[Bestie]

hello everyone !!!
i have a question:
why is it necessary to calibrate a calorimeter? don't you have to do it every time you are about to begin the experiment to calculate the heat of reaction, you can't just start it on one day and then finish it off on the next day using the same calibration factor... you have to do it all over again???
something about heat released in absorbed by instrument? how does that affect?
I'm doing the prac next week

thank you

[lzxnl]

thank you lzxnl

can you please help me: Suggest why salt is included in the contents of the bag.

please see attachment

It's a redox reaction => electrons need to flow. Dissolved salts increase the conductivity of the solution and thus speed up redox reactions.

hello everyone !!!
i have a question:
why is it necessary to calibrate a calorimeter? don't you have to do it every time you are about to begin the experiment to calculate the heat of reaction, you can't just start it on one day and then finish it off on the next day using the same calibration factor... you have to do it all over again???
something about heat released in absorbed by instrument? how does that affect?
I'm doing the prac next week

thank you

Well...you want an accurate result from your calorimeter and anything can affect its accuracy, including its temperature (like if it was used before), whether or not the instrument has been affected by heat etc. Basically you recalibrate it to be safe.

Hi guys,
If a question asks for the combustion reaction of maltose, how do you know that maltose is (s)? Couldn't it also be (aq)? VCAA 2007,  exam 2, q4aii only allows for (s).

You're burning it. I dunno why you'd be burning something aqueous as you'd be heating the water as well; not awfully efficient

Calculate the energy absorbed when 2.00 g of NH4Cl is
dissolved in:
i 100 mL water ii 1000 mL water
- find mols
-use delta H and the reaction equation...

i get how to do the question but the worked solutions says: The amount of energy absorbed does not depend on the volume of the solution?
why is that?
i know that changing the volume changes the concerntration of the solution because there is a constant 2g of NH4Cl is each solution... wouldn't the less concerntrated absorb less? i'm not sure... please clarify

The concentration of the solution has no bearing on how much energy is released. The amount of energy released is really only dependent on the number of moles of reactant, not the concentration of reactant.
What WILL change is the temperature change as then there's more water to absorb the energy released. The energy release doesn't change.

[vintagea]

thank you lzxnl, your help is very much appreciated

how do the salts increase the conductivity of the solution? they provide the electrons? how does it work?

[thushan]

Salt solutions contain mobile charged particles. An anion (eg. NO3 - ) travelling in a particular direction is effectively similar to an electron moving in the same direction, or a cation moving in the opposite direction.

It's the presence of mobile charged particles such as dissolved cations and anions, as well as mobile electrons (eg. delocalised electrons in metals), that cause a substance to be electrically conductive.

[hobbitle]

Thanks lznxl for your solution yesterday.

I have another one, please see attached solution which also contains a diagram of the electrolysis setup (two cells in series).  There are four electrodes labelled A B C D and the ions in the aqueous solutions are shown underneath the beaker.  The question is to write what half-reactions are happening at each electrode.

My issue is that my answer for the half-reactions at B and D do not match the answers.  The provided answers say:
At B: 2H₂O(l) + 2e– → H₂(g) + 2OH–(aq)
At D: 2H+(aq) + 2e– → H₂(g)

I think my working outlines my thought process fairly clearly so hopefully someone might be able to see where I have gone wrong.  I have been battling with the solutions that we have been provided, about 10% of them have been incorrect, so it is not outside the realms of possibility that the answer is in error.  I just want to know the truth!!  Thanks

NB.  The second last line should say "At D..." not "At B...".

[lzxnl]

Thanks lznxl for your solution yesterday.

I have another one, please see attached solution which also contains a diagram of the electrolysis setup (two cells in series).  There are four electrodes labelled A B C D and the ions in the aqueous solutions are shown underneath the beaker.  The question is to write what half-reactions are happening at each electrode.

My issue is that my answer for the half-reactions at B and D do not match the answers.  The provided answers say:
At B: 2H₂O(l) + 2e– → H₂(g) + 2OH–(aq)
At D: 2H+(aq) + 2e– → H₂(g)

I think my working outlines my thought process fairly clearly so hopefully someone might be able to see where I have gone wrong.  I have been battling with the solutions that we have been provided, about 10% of them have been incorrect, so it is not outside the realms of possibility that the answer is in error.  I just want to know the truth!!  Thanks

NB.  The second last line should say "At D..." not "At B...".

The reaction you quoted at B requires acid, whereas I only see aluminium sulfate in your solution (which doesn't have H+ as far as you're meant to know from the question; oddly enough aluminium sulfate is acidic which would actually change the entire question but you haven't been told that)
Even so, a bit of fiddling around with Nernst equations suggests that even at neutral pH, reduction of sulfate is more favourable than reduction of water. Sigh.

As for D, the only issue I can see is potentially whether or not you have 1 M sulfate ions there. Otherwise your answer should be ok.