VCE Chemistry Question Thread

[sweetcheeks]

Can someone please help me with these questions?

1. Calculate the final concentration of 500mL of 0.500M hydrochloric acid after the addition of 200mL of water.

2. Calculate the final concentration of 1000mL of 0.500M potassium iodide after the evaporation of 10.0mL of water.

Before we answer these questions, how would you try and solve them? What formula's would you use?

[leiva-acuna.s]

I

Before we answer these questions, how would you try and solve them? What formula's would you use?

I tried doing this for the first question, but I'm not sure if it's right:

 c1v1 = c2v2
c1 = c2v2/v1
= 0.500 x 0.2/0.5
[HCl] = 1.25M

I then thought maybe I should have used c=n/v or done something along these lines:

500mL + 200mL = 700mL

and then used the 700mL to try and solve it

I would have then tried to use c1v1 = c2v2 for the second question.

[sweetcheeks]

I

I tried doing this for the first question, but I'm not sure if it's right:

 c1v1 = c2v2
c1 = c2v2/v1
= 0.500 x 0.2/0.5
[HCl] = 1.25M

I then thought maybe I should have used c=n/v or done something along these lines:

500mL + 200mL = 700mL

and then used the 700mL to try and solve it

I would have then tried to use c1v1 = c2v2 for the second question.

You are on the right track for the first one. The final volume should be 700 mL or 0.7 L rather than 0.2 L, because the final volume of the solution is going to be the initial 500 mL of HCl solution plus the 200 mL of water.

Another method to solve this is c=n/V. During dilutions, the number of moles is going to stay constant (this is actually how C1V1=C2V2 is derived).

The same principle applies for the second question, except that the volume will be decreasing, but the moles of iodide will be constant.

[leiva-acuna.s]

You are on the right track for the first one. The final volume should be 700 mL or 0.7 L rather than 0.2 L, because the final volume of the solution is going to be the initial 500 mL of HCl solution plus the 200 mL of water.

Another method to solve this is c=n/V. During dilutions, the number of moles is going to stay constant (this is actually how C1V1=C2V2 is derived).

The same principle applies for the second question, except that the volume will be decreasing, but the moles of iodide will be constant.

Thanks

So if I use c=n/v, do I use n=c x v to calculate the moles of HCl?

[sweetcheeks]

Thanks

So if I use c=n/v, do I use n=c x v to calculate the moles of HCl?

Yep.

[EllingtonFeint]

A brown black compound of the element thallium was found to contain 89.5% Tl and 10.5% O by mass. What is the oxidation state of thallium in this compound.



The combustion of octane in excess air can be represented as
\(\ce{C8H18}\) (g) + \(\ce{12 1/2 O2}\) \(\ce{->}\) \(\ce{8CO2}\) (g) + \(\ce{9H2O}\) (g)

To produce 176 g of \(\ce{CO2}\) in his reaction
A. 1.0 mol of \(\ce{C8H18}\) needs to react with 25 mol of \(\ce{O2}\)
B. 0.500 mol of \(\ce{C8H18}\) needs to react with 6.25 mol of \(\ce{O2}\)
C. 0.500 mol of \(\ce{C8H18}\) needs to react with 3.125 mol of \(\ce{O2}\)
D. 1.0 mol of \(\ce{C8H18}\) needs to react with 3.125 mol of \(\ce{O2}\)

I’ve never come across these kinds of quests before, could somebody please explain them to me?

[fun_jirachi]

Hey there!
For the first question, in general you'd look at the ratio of thallium:oxygen. Set one of the sides to equal 1, usually the smaller one, by dividing the small number into the bigger number (in this case for thallium:oxygen it's roughly 8.5:1). Then, divide this into the molar mass of thallium to find the corresponding amount of oxygen required in the compound. Thallium has molar mass 204.38, so you divide 8.5 into that to get roughly 24. Note that 24 isn't a multiple of 16, but it divides reasonably nicely. 24/16=1.5, so you have the ratio 1:1.5 or 2:3 of thallium:oxygen given the mass ratios. From there, you know that the formula is going to be Tl₂O₃. Since the oxidation state for oxygen is usually -2, then the oxidation state of thallium is +3.

For the second question, use the formula n=m/MM. The number of moles of CO₂ that you have is going to be 176/44.009 which you can just round off to 4. By the molar ratios, you can find that you need half a mole of oxygen, and at least 6.25 moles of oxygen (since the reaction is in excess air). So the answer is B.

Hope this helps, and if you have any more questions (ie I didn't explain properly) feel free to ask!

[EllingtonFeint]

Hey there!
For the first question, in general you'd look at the ratio of thallium:oxygen. Set one of the sides to equal 1, usually the smaller one, by dividing the small number into the bigger number (in this case for thallium:oxygen it's roughly 8.5:1). Then, divide this into the molar mass of thallium to find the corresponding amount of oxygen required in the compound. Thallium has molar mass 204.38, so you divide 8.5 into that to get roughly 24. Note that 24 isn't a multiple of 16, but it divides reasonably nicely. 24/16=1.5, so you have the ratio 1:1.5 or 2:3 of thallium:oxygen given the mass ratios. From there, you know that the formula is going to be Tl₂O₃. Since the oxidation state for oxygen is usually -2, then the oxidation state of thallium is -3.

For the second question, use the formula n=m/MM. The number of moles of CO₂ that you have is going to be 176/44.009 which you can just round off to 4. By the molar ratios, you can find that you need half a mole of oxygen, and at least 6.25 moles of oxygen (since the reaction is in excess air). So the answer is B.

Hope this helps, and if you have any more questions (ie I didn't explain properly) feel free to ask!

Yessss, I finally figured out question 2 and I got the same answer!  Thank youuuu

I’m still a little bit confused about the first question. Are these types of questions called oxidation states? I might look a YouTube video. Also, unfortunately, the possible answers are +1, +2,+3 and +4
 

I’m not sure what to do.

[fun_jirachi]

oof, that's a typo on my part. That should read +3, not -3 , my bad. Will edit that now

EDIT: I probably should've addressed your confusion in this post, so uh...

I'm not really sure what they're called actually (I also do HSC, not VCE so I don't know how it works over your side ). The first part seems like a regular figuring out formula from percentage mass then they just shoved in oxidation states to make you do one more step. Think about it as more a two step problem where you figure out the formula first then do the oxidation state part after. Oxidation state questions are usually pretty simple because they involve knowing a bunch of common oxidation states which are usually constant, then writing a one line addition with one unknown, then figuring out the unknown.

[Tatlidil]

Hey!
If I (pretty much) failed chemistry 1/2 but still get accepted to do 3/4 should I do it? I basically didn't set good foundations.

[AngelWings]

Hey!
If I (pretty much) failed chemistry 1/2 but still get accepted to do 3/4 should I do it? I basically didn't set good foundations.

If you're intent on courses like med (and some related health science/ allied health courses), chemical engineering or biomed courses after VCE, then you'll need to do Chem as a prereq for your tertiary studies. (You can check here by typing Chemistry into the search bar.) If you're not into these areas at all, then you can probably consider dropping Chem if you dislike it or think you can't handle it. You can still do well, as long as you revise the key concepts from Chem U1/2, so stuff like stoichiometry, key formulas and so forth. You might also want to check out the start of the Unit 3 curriculum and study design, which is more about thermo and kinetics (including fuels), if you have time.

Some articles I dug up that will be relevant to you right now:

Articles for starting U3/4 Chem

- Unit 4 Chem: Food Molecules by Janna
- Some tips for VCE Chem exam by zsteve (Even though it's way too early, some tips are great to watch out for from Day 1.)
- Exam Strategy and Note Taking for Chem by Rachid (See above statement.)
- How I got a raw 50 in VCE Chem by zsteve

EDIT: Please note it's not all courses in those areas and not all areas that may require Chem are listed above. Some will still require other prereq.s and the predominant areas are the areas listed above.

[Tatlidil]

If you're intent on courses like med (and some related health science/ allied health courses), chemical engineering or biomed courses after VCE, then you'll need to do Chem as a prereq for your tertiary studies. (You can check here by typing Chemistry into the search bar.) If you're not into these areas at all, then you can probably consider dropping Chem if you dislike it or think you can't handle it. You can still do well, as long as you revise the key concepts from Chem U1/2, so stuff like stoichiometry, key formulas and so forth. You might also want to check out the start of the Unit 3 curriculum and study design, which is more about thermo and kinetics (including fuels), if you have time.

Some articles I dug up that will be relevant to you right now:

Articles for starting U3/4 Chem

- Unit 4 Chem: Food Molecules by Janna
- Some tips for VCE Chem exam by zsteve (Even though it's way too early, some tips are great to watch out for from Day 1.)
- Exam Strategy and Note Taking for Chem by Rachid (See above statement.)
- How I got a raw 50 in VCE Chem by zsteve

Thanks AW, ill check them out as well!

[huity]

Hey!
If I (pretty much) failed chemistry 1/2 but still get accepted to do 3/4 should I do it? I basically didn't set good foundations.

Hi Tatlidil

Following on from what AngelWings said, Chem is a prereq for most science-based courses and health sources e.g med and dent. It's important that you check the prerequisites and scores required for ALL courses you are interested in. Don't forget interstate courses too!

I nearly failed Chem 1/2. Less than 50% in assessments and on the exams, but I still managed a raw 50. It's definitely tough but I think if you approach 3/4 with a positive mindset, put in the hard work and understand what topics from 1/2 are relevant, you can definitely do it. 

[Monkeymafia]

With electrolysis, does the strong reductant always need to be above the strongest oxidant in the electrochemical series?

Or, can the strongest reductant still be below the strongest oxidant in the electrochemical series in an electrolysis situation?

[huity]

With electrolysis, does the strong reductant always need to be above the strongest oxidant in the electrochemical series?

Or, can the strongest reductant still be below the strongest oxidant in the electrochemical series in an electrolysis situation?

Hey Monkeymafia 

The strongest reductant always needs to be above/ at the same level as the strongest oxidant.

The strongest reductant can NOT be below the strongest oxidant (negative gradient), or else this wouldn't be electrolysis.

Hope that helps