VCE Chemistry Question Thread

[bobisnotmyname]

Hey everyone. Just wondering can we round during our working out (not just at the answer), if so by how much (do we get a little lee weigh on our answers due to rounding) or is it best to just keep thing in fractions until the end. Thanks

[grannysmith]

Hey everyone. Just wondering can we round during our working out (not just at the answer), if so by how much (do we get a little lee weigh on our answers due to rounding) or is it best to just keep thing in fractions until the end. Thanks

Don't round during your working, only at the very end. Always use the unrounded value given by your calculator. However, if you're writing down your working and showing intermediary steps, then it's helpful to round according to significant figures. This way it's easier to see the number of significant figures your final answer should be and you're less likely to make an error in this way. BUT always use the unrounded value stored on the calculator throughout working through the question.

[RazzMeTazz]

Would it be correct to say that when a metal is oxidised, it always forms its respective cation?
So for e.g. If Na metal was oxidised it would form Na1+ not Na2+ or anything else?

Also, is it correct to say that when you are finding the  formula of a hydrated compound  you do not need to be given the molar mass of the compound and instead the formula of the hydrate is always equal to its empirical formula?

Thanks

[lzxnl]

Would it be correct to say that when a metal is oxidised, it always forms its respective cation?
So for e.g. If Na metal was oxidised it would form Na1+ not Na2+ or anything else?

Also, is it correct to say that when you are finding the  formula of a hydrated compound  you do not need to be given the molar mass of the compound and instead the formula of the hydrate is always equal to its empirical formula?

Thanks

If only chemistry were that simple.
For group 1 metals, you'd think it's always the case. Which it is until you get to like Rb and Cs in which you can oxidise them to ridiculously unstoichiometric compounds like Cs₉O₂. Look up 'metal suboxide' for more details.

Then, transition metals have multiple oxidation states. Fe(II) and Fe(III) both exist and are stable under different conditions. It can also be oxidised to a +6 oxidation state in the form of FeO₄^2-. Manganese can exist in many oxidation states from +2 to +7, while others like osmium can be oxidised all the way to +8. It's not always a given that you'll get a particular oxidation state.

And then, you may not even get a simple cation. If you oxidise iron metal in the presence of excess fluoride ions, you'll form FeF₆^3-. Similarly, chromium can be oxidised all the way to dichromate. There are also other possibilities, like having NiCl₄^2-. Chemistry is incredibly complex, much more than what VCE makes you think.

Would it be correct to say that when a metal is oxidised, it always forms its respective cation?
So for e.g. If Na metal was oxidised it would form Na1+ not Na2+ or anything else?

Also, is it correct to say that when you are finding the  formula of a hydrated compound  you do not need to be given the molar mass of the compound and instead the formula of the hydrate is always equal to its empirical formula?

Thanks

Not true. If you were to hypothetically hydrate something like Co₂CO₈ with 2 water molecules, its empirical formula would be CoCO₄.H₂O, which would be completely wrong.
Again, it depends on the actual example.

[RazzMeTazz]

Thanks so much lzxnl! That was really helpful

[chocolate.cake.1]

Hello
I'm having trouble with these two questions which are from the lisachem book

For question 93, why isn't one of the reactants I- (since the iodine solution is aqueous??)

Thanks

P.s. If you want the answers question 93 is D and 136 is C

[IndefatigableLover]

Hello
I'm having trouble with these two questions which are from the lisachem book

For question 93, why isn't one of the reactants I- (since the iodine solution is aqueous??)

Thanks

P.s. If you want the answers question 93 is D and 136 is C

93: Though it is aqueous, it won't just contain negative ions since they'd just repel each other and all. Most of the time there'd be some positive ions there to balance the solution out (like with NaCl, when they're dissolved in water, it'll form positive and negative ions to balance each other out). One way to rectify the whole positive/negative thing is if the molecule is neutral which is why the reactant is I₂ rather than I-.

136:

Pretty tricky (I got the wrong answer at the start since I made an assumption and got it wrong LOL)

What should be noted is that we don't actually know what 'M' is so we can't assume that it's a pure substance of Magnesium and go from there (which was my mistake). It won't do much justice but the question has been asked a few years back so here's the solution for it (thanks to Mao):

This question is quite mathematically involved, but not impossible.

n(H2) = 0.250 / 2 = 0.125 mol

Since Mg and Zn both react on a 1:1 ratio with H2SO4 to release 1:1 H2 gas, this implies

n(Mg) + n(Zn) = 0.125 mol  ----------[1]

Also, the mass of Mg and Zn add up to 5.00g

n(Mg) * 24.3 + n(Zn) * 65.4 = 5.00  ------------[2]

We see that's a set of simultaneous equation, multiplying [1] by 24.3 and eliminate

n(Zn) * 41.1 = 1.9625
n(Zn) = 0.04775 mol
m(Zn) = 0.04775 * 65.4 = 3.12g
%(Zn) = 3.12/5.00 * 100% = 62.4%

%(Mg) = 100% - 62.4% = 37.6%
:. Option C

[was this from TSFX?]

[Chang Feng]

quick question regarding equilibrium:
when for instance you add a reactant to a reaction, why does both the forward and back reaction increase, but the forward reaction increase a greater amount???

[keltingmeith]

quick question regarding equilibrium:
when for instance you add a reactant to a reaction, why does both the forward and back reaction increase, but the forward reaction increase a greater amount???

Assuming I've understood your question properly, it's due to Le Chetalier's principle.

[cosine]

If 1.104g of Sodium Carbonate is dissolved in 250mL of water, and 20.00mL aliquot a of this solution were tittered with nitric acid. An average titre of 23.47mL was found.

Work out the concentration of Nitric Acid?

Can someone tell me how to go about doing this? I can work out no. Mol for nitric acid, but I can't seem to find another piece of info like concentration or volume of nitric acid. Thanks

[alchemy]

If 1.104g of Sodium Carbonate is dissolved in 250mL of water, and 20.00mL aliquot a of this solution were tittered with nitric acid. An average titre of 23.47mL was found.

Work out the concentration of Nitric Acid?

Can someone tell me how to go about doing this? I can work out no. Mol for nitric acid, but I can't seem to find another piece of info like concentration or volume of nitric acid. Thanks

Na2CO3 (aq) + 2HNO3 (aq) --> 2NaNO3 (aq) + H2CO3 (s)
n(Na2CO3)=1.104/106=0.0104 mol
In the 20mL aliquot: n(Na2CO3) = 20/250 * 0.0104 = 0.000832 mol
At equivalence point n(HNO3) = 2*n(Na2CO3) = 0.001664 mol
C(HNO3) = n(HNO3)/V(HNO3) = 0.001664/0.02347=0.07M

EDIT: Btw, you said you couldn't find the volume of nitric acid in the question, but it says 23.47mL which equals 0.02347L which I used in the last step, just so you know. Also I haven't paid attention to sig figs

[cosine]

Na2CO3 (aq) + 2HNO3 (aq) --> 2NaNO3 (aq) + H2CO3 (s)
n(Na2CO3)=1.104/106=0.0104 mol
In the 20mL aliquot: n(Na2CO3) = 20/250 * 0.0104 = 0.000832 mol
At equivalence point n(HNO3) = 2*n(Na2CO3) = 0.001664 mol
C(HNO3) = n(HNO3)/V(HNO3) = 0.001664/0.02347=0.07M

EDIT: Btw, you said you couldn't find the volume of nitric acid in the question, but it says 23.47mL which equals 0.02347L which I used in the last step, just so you know. Also I haven't paid attention to sig figs

Wait so when it says an average titre of 23.46mL was found, it said previously that 20.00mL aliquots of this solution (Sodium Carbonate), doesn't that mean they used 23.46mL of socdium carbonate to titre the nitric acid?

Sorry for the simple questions.

[RazzMeTazz]

Wait so when it says an average titre of 23.46mL was found, it said previously that 20.00mL aliquots of this solution (Sodium Carbonate), doesn't that mean they used 23.46mL of socdium carbonate to titre the nitric acid?

Sorry for the simple questions.

I think they mean that  the 20.00mL aliquot of Na2CO3 solution was titrated with 23.46mL of HNO3. (So 23.46mL of HNO3 was added to the 20.00mL aliquot of Na2CO3 )

[RazzMeTazz]

When writing a combustion reaction of a hydrocarbon, in which CO2 and H2O are produced, would the H2O  be written in the liquid or gaseous state?

Or would the question specify which of the two states it is?

Thanks

[alchemy]

Wait so when it says an average titre of 23.46mL was found, it said previously that 20.00mL aliquots of this solution (Sodium Carbonate), doesn't that mean they used 23.46mL of socdium carbonate to titre the nitric acid?

Sorry for the simple questions.

Titre refers to the nitric acid which came out of the burette, in this instance 23.47mL.
Aliquot refers to the sodium carbonate which was pip-petted into the beaker, so 20mL.
Draw a diagram, it'll help you see it more clearly.