VCE Chemistry Question Thread

[Elizawei]

Thought I'd add on this topic:
- There is a common method of doing your calculations to an arbitrary degree of precision and then rounding down to the lowest number of sig figs in the input data. Strictly speaking, this is WRONG and only works for questions involving multiplication only.
- In general, the best approach is to write intermediate steps down to the correct number of sig figs, storing values in your calculator for later use.

Sadly I'm a noob at my calculator and I can't store XD
But I do go back and redo the steps sometimes

[zsteve]

Sadly I'm a noob at my calculator and I can't store XD
But I do go back and redo the steps sometimes

Elizawei: highly recommended to use the TI-30XB. And it's not *that* late to begin using it
The TI-30XB allows you to copy/paste previous calculations. Very worth.

[Elizawei]

Elizawei: highly recommended to use the TI-30XB. And it's not *that* late to begin using it
The TI-30XB allows you to copy/paste previous calculations. Very worth.

arhhhhhhhhhhh hahaha probably is too late? XD okay I'll try look for it in Officeworks ;_;
Nah oh well i've been using my EL-531XH for two years I'm pretty loyal XD
Thanks tho

[Muchos Help]

For this http://m.imgur.com/XUWI5DZ, how do we actually determine what is more reactive? I'm trying to use the electrochemical series to determine it but I'm probably doing it wrong. I'm assuming the higher the voltage, the more reactive it is...?

Thanks

[jyce]

For this http://m.imgur.com/XUWI5DZ, how do we actually determine what is more reactive? I'm trying to use the electrochemical series to determine it but I'm probably doing it wrong. I'm assuming the higher the voltage, the more reactive it is...?

Thanks

Metals act as reductants, i.e., they give away electrons not take them. So, when we're talking about the "reactivity" of a metal in this context, we are talking about how strongly it acts as a reductant or gives away electrons. More reactive metals are stronger reductants and when looking at an electrochemical series reducing strength increases down the series. Therefore, for the question you've attached, what you need to do is select the metal in each paring that is further down the electrochemical series as the more reactive metal.

Hope this helps 

[Muchos Help]

Thanks jyce

[Muchos Help]

I have another question but about redox http://imgur.com/9TvtKk9

I'm trying to find out which thing is undergoing oxidation. For the first equation, I know that the F arrow is the oxidation equation but I tried finding the oxidation number of 2CuI to make sure I knew that this was undergoing reduction, but I'm stuck. I'm not sure what the oxidation number of I is, so can't solve for the oxidation number of Cu.

The 2nd equation, I'm just unsure how to do this so I kinda guessed. What do I do if theres 2 things that I don't know the oxidation numbers for? Those things being K and Cl that is in KClO?

[ayesha2011t]

I have another question but about redox http://imgur.com/9TvtKk9

I'm trying to find out which thing is undergoing oxidation. For the first equation, I know that the F arrow is the oxidation equation but I tried finding the oxidation number of 2CuI to make sure I knew that this was undergoing reduction, but I'm stuck. I'm not sure what the oxidation number of I is, so can't solve for the oxidation number of Cu.

I'm not completely sure about this but ik the oxidation number of any compound is 0, so you don't need to deal with 2CuI separately.

The oxidation number of I (aq) is -1 and the oxidation number of I (s) is 0. That is an increase in oxidation number and an increase in oxidation number means oxidation is happening.

Now moving back to 'confirm' if E is undergoing reduction. The oxidation number of Cu (aq) is +2 and the oxidation number of CuI (s) is 0. This is a decrease in oxidation number which means reduction is happening.

I'm just a year 11 student just grasping the concept, if I made any mistake in my post, pls let me know!

[Sine]

How many marks could you drop on across both Unit 3 and Unit 4 VCAA 2008-2012 exams to get a 50 (or 45+)  These exams were roughly /75 each

[HopefulLawStudent]

So I got back my SACs for chem on FRI and I'm only just getting around to checking it out now. In one of our SACs, we did a back titration and used NaOH. My teacher said that to get full marks, we needed to talk about how NaOH presented a source of error because it absorbs CO2.

How exactly is that a source of error? Could someone please clarify?

[Gogo14]

How exactly is that a source of error? Could someone please clarify?

You cannot weigh it accurately. Because it absorbs water from the atmosphere, you get inaccuracies in the mass of NaOH because it could contain H2O

[sweetcheeks]

How exactly is that a source of error? Could someone please clarify?

You make a standard solution using the NaOH, something with a known, precise concentration for attaining accurate values. You do this by weighing out a precise amount and calculating the moles from that. In the case of NaOH (and other hydroxides and oxides), they will react with atmospheric CO2 to form carbonates. When you have NaOH it is unlikely to be a pure sample, instead a mixture of NaOH and Na2CO3. This presents a problem when using it as a standard solution, as even in a solution the NaOH can become Na2CO3. So when you are trying to weigh out a mass of NaOH, you won't be able to accurately determine the moles.

A solution to this problem is to standardise the solution. This involves a titration with an acidic standard solution (e.g. using a 1.000 mole HCl solution) and working out the concentration from the titration results.

[Adequace]

What am I doing wrong here? http://imgur.com/a/7iIV9 The book has 9.032M

Thanks

[sweetcheeks]

What am I doing wrong here? http://imgur.com/a/7iIV9 The book has 9.032M

Thanks

I get 10.45M as well.

[Adequace]

I get 10.45M as well.

Okay thanks,


I have a question about redox http://m.imgur.com/1vtP82x

I'm a bit confused for part b and c. Since both reactants are only solutions/ions, there won't be anything becoming a solid, so therefore no redox reactions occurs? Is this correct?

Thanks