VCE Chemistry Question Thread

[FarAwaySS2]

Can someone help me with this question. The answer is B: 16.67 mL but i dont understand how.


Thanks!

[Chazef]

the two things are the same concentration (0.0100M)  meaning the volume is a direct indicator of how many moles there are (because n = cV and the c's can cancel out). If you look at the equations, one permanganate ion reacts with 5 iron (II) ions but the dichromate reacts with 6 iron (II) irons meaning you won't need as many moles of chromium to fully react with the iron ions. I always try to think in whole numbers and then generalise to fractions: so if a permanganate ion reacted with one iron (II) ion and the dichromate reacted with two, you would obviously need half the moles of dichromate ions to react with the iron in comparison to permanganate ions i.e. 1/2 (using the coefficients of the iron (II) ions in the equation). In this case the coefficients are 5 and 6 so we need 5/6 as many dichromate ions in comparison to permanganate ions. Because they are both the same concentration, we just need 5/6 * 20.00mL = 16.66mL dichromate ions i.e. 16.66mL of potassium dichromate.

Another way to think of it is, if I want my two reaction equations to be identical in terms of iron (II) reacted, I need to times everything in the bottom equation by 5/6. Meaning 1 mole of potassium permanganate can be switched with one 5/6 of a mole of potassium dichromate and you'd have no change in terms of iron (II) reacted. You can then say, well if I need 5/6 the amount of ions and the permanganate is 20.00mL I only need 5/6 of that volume in dichromate (with an equal concentration) to react with all of the iron (II).

there's no harm converting to moles and back to volume but imo it's good to know where you can avoid unnecessary calculating. Hopefully this has been helpful

[knightrider]

Because electrons don't fill according to "this shell has more space, let's jump in!"
Instead, you need to consider each sub-shell and their relative energy level. It just so happens that the 4s subshell has a lower energy level than the 3d subshell, hence why we only see 8 electrons in the third shell, even though we've started filling up the fourth.

To memory, you won't ever have to worry about subshells or the filling of atoms beyond the 4s subshell, so you can just not worry about this if you want (although I swear you're meant to learn it in unit 1?)

2, 8, 8, 2 represents the electron configuration of Calcium.

The 's' and 'p' subshells are full within the third shell. Before filling the 'd' subshell of the third shell, electrons will fill the 's' subshell of the fourth shell.

But for scandium (atomic number 21), the electron configuration is 2, 8, 9, 2.

Once the 's' subshell within the fourth shell is full, electrons will then begin to fill the 'd' subshell of the third shell.

Refer to Figure 2.21 on page 30 of Heinemann Chemistry 1 for the order that electrons fill subshells.

I hope this helps!

Thanks eulerfan101 and Maths Forever 

[FarAwaySS2]

Thank you so much for that thorough explanation!!

the two things are the same concentration (0.0100M)  meaning the volume is a direct indicator of how many moles there are (because n = cV and the c's can cancel out) and the . If you look at the equations, one permanganate ion reacts with 5 iron (II) ions but the dichromate reacts with 6 iron (II) irons meaning you won't need as many moles of chromium to fully react with the iron ions. I always try to think in whole numbers and then generalise to fractions: so if a permanganate ion reacted with one iron (II) ion and the dichromate reacted with two, you would obviously need half the moles of dichromate ions to react with the iron in comparison to permanganate ions i.e. 1/2 (using the coefficients of the iron (II) ions in the equation). In this case the coefficients are 5 and 6 so we need 5/6 as many dichromate ions in comparison to permanganate ions. Because they are both the same concentration, we just need 5/6 * 20.00mL = 16.66mL dichromate ions i.e. 16.66mL of potassium dichromate.

Another way to think of it is, if I want my two reaction equations to be identical in terms of iron (II) reacted, I need to times everything in the bottom equation by 5/6. Meaning 1 mole of potassium permanganate can be switched with one 5/6 of a mole of potassium dichromate and you'd have no change in terms of iron (II) reacted. You can then say, well if I need 5/6 the amount of ions and the permanganate is 20.00mL I only need 5/6 of that volume in dichromate (with an equal concentration) to react with all of the iron (II).

there's no harm converting to moles and back to volume but imo it's good to know where you can avoid unnecessary calculating. Hopefully I've used the correct terminology, I haven't touched chem in ages haha

[Eiffel]

if the indicator changes too soon or too late can someone explain the effect on the results, specifically if the substance under analysis is in burette  and  one in the conical flask.

Also, how does this effect our results as a whole, e.g. volume/conc etc?

Thanks 

[Eiffel]

A solid is only made up of sodium chloride and potassium chloride, mass 0.8870g. To determine the amount of chloride ions in the solid a gravimetric analysis was performed. The solid was treated with silver nitrate to from a precipitate of silver chloride, mass 1.9113g. Determine the % composition of both sodium and potassium in the sample

[ally12579]

For gravimetric analysis, what are some errors for both an increase and decrease in mass of precipitate collected compared to the manufacture's information?

[zyzz101]

For gravimetric analysis, what are some errors for both an increase and decrease in mass of precipitate collected compared to the manufacture's information?

Increase:
1. Presence of contaminants in the precipitate
2. Precipitate is incompletely dried.
3. Co-precipitations - other ions in the solution forming preciptates.

Decrease:
1. The slight solubility of precipitate.
2. Precipitate is lost during filtration.

[Chazef]

if the indicator changes too soon or too late can someone explain the effect on the results, specifically if the substance under analysis is in burette  and  one in the conical flask.

Also, how does this effect our results as a whole, e.g. volume/conc etc?

Thanks 

lets say you have one mole of a base, NaOH, in the conical flask and you're titrating it with HCl. You're going to need one mole of the acid, HCl, from the burette to neutralise it (all the H+'s from HCl match up with the OH-'s from the NaOH). The equivalence point is where you've put one mole of HCl into the conical flask and it's neutralised the NaOH.

The end point (the point when the indicator changes colour) is ideally the same as the equivalence point because you want to know exactly when the acid neutralises the base.

So, if the end point (the point where the indicator changes colour) is too soon, you will think that you have put a mole of the acid into the conical flask when you really haven't. If the volume of acid you added like a single drop you're gonna think 'WOAH that is one concentrated solution of acid! It already completely neutralised the base'. So an early end point means you overstate the concentration of the acid. If we take the opposite case where the end point is far too late, you'll have put litres of the acid into the conical flask and FINALLY it produces a colour change and you'll think 'WOAH that is one dilute solution of acid!. It took ages to neutralise the base!'. So a late end point means you'll understate the concentration.

If you are analysing the thing in the conical flask (e.g. a base) then if the end point is way too early you'll think 'damn that solution must be pretty dilute if it can be neutralised that quickly' and if the end point is way too late you'll think 'damn that solution must be insanely concentrated to have that much of the base in such a small volume, considering it took so much acid to neutralise it'.

I always consider the extremes of the spectrum when thinking about chemistry stuff because it makes thinking about the less extreme things easier because the trends are the same.

Hopefully this helps

[Maths Forever]

A solid is only made up of sodium chloride and potassium chloride, mass 0.8870g. To determine the amount of chloride ions in the solid a gravimetric analysis was performed. The solid was treated with silver nitrate to from a precipitate of silver chloride, mass 1.9113g. Determine the % composition of both sodium and potassium in the sample

Essentially, there are two precipitation reactions occurring:

NaCl (aq) + AgNO3 (aq) --> AgCl (s) + NaNO3 (aq)

KCl (aq) + AgNO3 (aq) --> AgCl (s) + KNO3 (aq)

For both reactions, Ag⁺ (aq) + Cl^- (aq) --> AgCl (s)


m (KCl) + m (NaCl) = 0.8870 g

m (AgCl) = 1.9113 g

n (AgCl) = m / M = 1.9113 / [107.9 + 35.5] = 0.013328 mol (5 significant figures)

Therefore, since n (AgCl) = n (Cl^-), n (Cl^-) = 0.013328 mol

n (Cl^-) = n (Cl) = 0.013328 mol

Therefore, m (Cl) = n x M = 0.013328 x 35.5 = 0.47316 g (to 5 significant figures)

% composition of Cl = [0.47316 / 0.8870] x 100% = 53.34% (to 4 significant figures)

% composition of Na and K = 100% - % composition of Cl

= 100% - 53.34% = 46.66% (to 4 significant figures)

I think this is correct. I hope it helps!

[IndefatigableLover]

When doing titrations involving dilutions, what would be a use(s) of an electronic balance (given that your products used for titration are already in solution)?

[paper-back]

When doing titrations involving dilutions, what would be a use(s) of an electronic balance (given that your products used for titration are already in solution)?

I believe if you are diluting to say 60ml from 20ml, it's more accurate to use an electronic balance than looking at the volume/estimating the volume from the side of a flask, assuming that the density is 1g/ml

[stockstamp]

Hi,

Was hoping for some clarification here:

What volume of 0.100 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide?

Below is how I worked out the answer - would this be correct?

• NaOH (aq) --> Na⁺ (aq) + OH^- (aq)
• KOH (aq) --> K⁺ (aq) + OH^- (aq)
• n(NaOH) = n(OH^-) = 0.5/40 = .0125 mol
• n(KOH) = n(OH^-) = 0.8/56.1 = 0.0142603 mol
• n(OH^-)total = 0.0125 + 0.0142603 = 0.027603 mol
• Therefore n(H₃O⁺) required to neutralise is also = 0.027603 mol
• H₂SO₄ (aq) --> 2H₃O⁺
• n(H₂SO₄) required to neutralise = 1/2 n(H₃O⁺) = 0.0138015
• V(H₂SO₄) = n/C = 0.0138015/0.1 = 0.138 Litres

Is this correct? Am I working it out the right way?

any help appreciated!

[Chang Feng]

For substitution reactions, why exactly does like OH- swap with CL in an Alkane for example. Like why can't is swap with a hydrogen in the Alkane instead.

[Maths Forever]

For substitution reactions, why exactly does like OH- swap with CL in an Alkane for example. Like why can't is swap with a hydrogen in the Alkane instead.

These are the possible substitution reactions:

Alkane + Cl₂ --> Chloroalkane + HCl (i.e. A hydrogen atom has been replaced with a chlorine atom on one of the side chains). The lost H atom bonds to the other Cl atom to form HCl. Catalyst of UV light is required.

Chloroalkane + H₂O --> Alkanol + HCl (i.e. A chlorine atom has been replaced by an -OH group on one of the side chains). The lost Cl atom bonds to the H atom to form HCl. Catalyst such as NaOH is required.

The reaction between an Alkane and an Alkanol is outside the scope of VCE chemistry. 

I hope this helps!