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March 29, 2024, 01:15:25 am

Author Topic: VCE Chemistry Question Thread  (Read 2313181 times)  Share 

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DBA-144

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Re: VCE Chemistry Question Thread
« Reply #7980 on: April 17, 2019, 10:13:00 pm »
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Hey everyone! Got two questions regarding significant figures:

1. If I'm using data from the data book, and the sig figs are lower than the ones in the question, which one should I follow?
(The question I'm on has 3 sig figs as the lowest number, but I'm also required to use the molar mass of hydrogen which is given in the data book as 1.0 which is 2 sig figs)

2. Can someone confirm if this is correct: (this is part of the working out for a larger question)
There's been a temperature change from 18.5 to 13.3 degrees Celsius. So delta T is 5.2 (one decimal place). The smallest number of significant figures given in the question is 3, but 5.2 has two significant figures. I'm now required to leave my answers to two significant figures instead of three because that's the lowest number of s.f in the working out, even if the question had 3 s.f.
Is this correct? This was a question on a practice SAC and I left my answer to 3 s.f., and lost a mark.

Thanks in advance! :) :)

Pretty sure that 2 is correct. You don't know for sure that it's 13.30 or 18.50, so you only know the accuracy of your calculation to 2 sf. Here, you are 'certain' it is that value, but don't know it is, say, 5.20 or 5.2001 or 5.2000001. Hence, you also carry over this uncertainty to any other calculation you make using this value. Therefore, you should be using the fewest number of sf given to you in your working. I am NOT certain about how VCAA or your school does it, this is how I understand it only. 

For 1, I believe that values in the data book are correct to an unlimited degree and hence we would use 3 sf in example 1.

Pretty sure that significant figures are only a mark all up in the exam, so you don't need to stress too much about this, but it's a great idea to be getting this clarified. I have struggled with this stuff for too long haha :P
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-_-zzz

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Re: VCE Chemistry Question Thread
« Reply #7981 on: April 17, 2019, 10:31:36 pm »
+2

Pretty sure that 2 is correct. You don't know for sure that it's 13.30 or 18.50, so you only know the accuracy of your calculation to 2 sf. Here, you are 'certain' it is that value, but don't know it is, say, 5.20 or 5.2001 or 5.2000001. Hence, you also carry over this uncertainty to any other calculation you make using this value. Therefore, you should be using the fewest number of sf given to you in your working. I am NOT certain about how VCAA or your school does it, this is how I understand it only. 

For 1, I believe that values in the data book are correct to an unlimited degree and hence we would use 3 sf in example 1.

Pretty sure that significant figures are only a mark all up in the exam, so you don't need to stress too much about this, but it's a great idea to be getting this clarified. I have struggled with this stuff for too long haha :P

DBA is correct. The values from the data book are assumed to have an 'unlimited' number of sig figs so if you're using a value from the data book that has 2 numbers in it (e.g. 2.5) but the question gives you values with 3 sig figs, then you would round your final answer to 3 sig figs. In regards to your first question, if for example you have 2 numbers which are correct to 3 sig figs such as 24.3 and 19.3 and you are required to subtract 19.3 from 24.3, the value in which you yield can only be rounded accurately to 2 sig figs (5.3). Thus if you were to go and derive an answer by multiplying/dividing with '5.3', then your final answer should be to 2 sig figs despite the fact that the values given to you in the question were rounded to 3 sig figs. Hope that makes sense :)
« Last Edit: April 17, 2019, 10:38:17 pm by -_-zzz »

persistent_insomniac

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Re: VCE Chemistry Question Thread
« Reply #7982 on: April 18, 2019, 12:26:19 pm »
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Are the cathode and anode always the electrodes or is the + or - terminals of the battery? If it is always the electrodes what if they are same such as carbon electrodes?

fiona_atarnotes

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Re: VCE Chemistry Question Thread
« Reply #7983 on: April 18, 2019, 06:09:41 pm »
+5
Are the cathode and anode always the electrodes or is the + or - terminals of the battery? If it is always the electrodes what if they are same such as carbon electrodes?

Hey!
The anodes and cathodes are the electrodes of the battery. For VCE, the cells (electrolytic and galvanic cells) generally have two electrodes where one is an anode and one is a cathode. Non-rechargeable batteries are galvanic cells. Here, the cathode is positively charged and the anode is negatively charged. For electrolytic cells, it's reversed so the cathode is negatively charged and the anode is positively charged. In the case where both electrodes are carbon, since carbon electrodes are inert, that just means the species that is oxidising/reducing is not carbon.

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jasheel

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Re: VCE Chemistry Question Thread
« Reply #7984 on: April 18, 2019, 06:16:40 pm »
+6
Are the cathode and anode always the electrodes or is the + or - terminals of the battery? If it is always the electrodes what if they are same such as carbon electrodes?

Hey, also adding to what Fiona said. If you have a battery attached to your electrolytic cell, the positive terminal of the battery will attach to the positive electrode (the anode), and the negative terminal will attach to the negative electrode (the cathode).

The way you can think about it is the negative side of the battery is 'forcing' negative electrons into the negative cathode (hence reduction). And therefore, the positive side is 'attracting' the negative electrons from the anode (hence oxidation).
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peachxmh

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Re: VCE Chemistry Question Thread
« Reply #7985 on: April 22, 2019, 10:46:06 pm »
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Need help with a question! I cut out irrelevant bits of it as it was a multi-part question so let me know if anything doesn't make sense/there's information missing :p

In acidic solution, ions containing titanium can react according to the half-equation TiO2+ (aq) + 2H+ (aq) + e- ⇌ Ti3+ (aq) + H2O (l) where E° = -0.06 V. In the diagram (attached), A and B are inert graphite electrodes, and all ions have a concentration of 1M. If both A and B electrodes are replaced with iron, what equation will now occur at Electrode B?

I wrote 2TiO2+ (aq) + 4H+ (aq) + Fe (s) -> 2Ti3+ (aq) + 2H2O(l) + Fe2+(aq) but this is wrong - the answer is instead supposed to be Fe(s) + 2H+(aq) -> Fe2+(aq) + H2(g).

I don't understand this - why is 2H+(aq) + 2e- -> H2(g) a half-equation that is considered? Is it because the half-cell containing electrode B has H+ ions in solution and the question states it is an acidic solution? But the half-equation contains H2 which isn't part of the question or in the galvanic cell so shouldn't it be eliminated as an option?
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DBA-144

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Re: VCE Chemistry Question Thread
« Reply #7986 on: April 23, 2019, 07:27:22 am »
+3
I'm note sure about the below, but it might be because the iron is more reactive than the titanium? So the iron is reacting with the titanium rather than the iron. This would also mean that the metal would react with the acid by producing hydrogen gas. This is using the fact that metal + acid --> salt + hydrogen gas. The hydrogen ions come from the acidic solution and the hydrogen gas comes from the reaction between the metal and the acid.

Also, the metal part of it is not that hard to figure out now either. It's a metal and is going to donate electrons, so the metal would just be donating electrons. 

If you don't mind, I have a question of my own- how can we be certain that the iron would form the 2+ ion? I suspect that this is due to the Hydrogen needing to accept 1 electron each, so Iron is donating 2 electrons total?

Just want to say that I am not certain about any of this, just a potential answer. However, I am not sure why else the Titanium would not be reacting like it was before. It's probably a lot more complicated than this but still I hope this helps, sorry if I am wrong.
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persistent_insomniac

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Re: VCE Chemistry Question Thread
« Reply #7987 on: April 23, 2019, 07:28:29 pm »
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Hi!
With electroplating cells does it matter what electrolyte you use? For e.g. A can wanting to be plated with tin - should we only ever use an Sn2+ electrolyte or can we use any solution below Sn2+ on the electrochemical series as Sn2+ will still be the strongest oxidant so regardless will still gain the electrons?
I hope I make sense....

fiona_atarnotes

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Re: VCE Chemistry Question Thread
« Reply #7988 on: April 24, 2019, 07:11:53 pm »
+5
Hi!
With electroplating cells does it matter what electrolyte you use? For e.g. A can wanting to be plated with tin - should we only ever use an Sn2+ electrolyte or can we use any solution below Sn2+ on the electrochemical series as Sn2+ will still be the strongest oxidant so regardless will still gain the electrons?
I hope I make sense....
Hey! So generally, the answer is yes, you should only ever use an electrolyte with the metal ion you're plating. I think if you included other oxidants in the solution that are weaker than Sn2+ would be okay initially. However, it's just that with electroplating, you want the concentration of the metal ion you're plating onto the object to be unchanging. So if you included other oxidants in the solution that are weaker than Sn2+, as the reaction continues, the concentration of Sn2+ ions decreases whereas the concentration of the other weaker oxidants remain the same. Therefore, it would reach a point where the other oxidants reduce preferentially due to the decrease in concentration of the Sn2+ ions despite those other oxidants being weaker oxidising agents.

Hope that makes sense!

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Rameen

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Re: VCE Chemistry Question Thread
« Reply #7989 on: April 25, 2019, 11:59:32 am »
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A sample of calcium chloride, CaCl2, contains 0.030 moles of chloride ions, Cl–. The total mass of the CaCl2 sample is
A.   1.1 g
B.   1.7 g
C.   3.3 g
D.   6.6 g

Hi I need some help with this question. I am not getting the right answer

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #7990 on: April 25, 2019, 12:26:28 pm »
0
A sample of calcium chloride, CaCl2, contains 0.030 moles of chloride ions, Cl–. The total mass of the CaCl2 sample is
A.   1.1 g
B.   1.7 g
C.   3.3 g
D.   6.6 g

Hi I need some help with this question. I am not getting the right answer

What answer are you getting? It would be helpful if you could post your working out. Are you using stoichiometry?

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Re: VCE Chemistry Question Thread
« Reply #7991 on: April 25, 2019, 01:59:53 pm »
+2
A sample of calcium chloride, CaCl2, contains 0.030 moles of chloride ions, Cl–. The total mass of the CaCl2 sample is
A.   1.1 g
B.   1.7 g
C.   3.3 g
D.   6.6 g

Hi I need some help with this question. I am not getting the right answer

All you have to do first if find out how many moles of CaCl2 you have. So given that you have 0.030 moles of Cl-, then the moles of CaCl2 must be half of that because for every 1 mole of CaCl2, you have 2 moles of Cl-. Thus the moles of CaCl2 = 0.015 mol. You can then convert the moles to mass using n = m/M :)

jasheel

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Re: VCE Chemistry Question Thread
« Reply #7992 on: April 27, 2019, 04:02:01 pm »
+4
Need help with a question! I cut out irrelevant bits of it as it was a multi-part question so let me know if anything doesn't make sense/there's information missing :p

In acidic solution, ions containing titanium can react according to the half-equation TiO2+ (aq) + 2H+ (aq) + e- ⇌ Ti3+ (aq) + H2O (l) where E° = -0.06 V. In the diagram (attached), A and B are inert graphite electrodes, and all ions have a concentration of 1M. If both A and B electrodes are replaced with iron, what equation will now occur at Electrode B?

I wrote 2TiO2+ (aq) + 4H+ (aq) + Fe (s) -> 2Ti3+ (aq) + 2H2O(l) + Fe2+(aq) but this is wrong - the answer is instead supposed to be Fe(s) + 2H+(aq) -> Fe2+(aq) + H2(g).

I don't understand this - why is 2H+(aq) + 2e- -> H2(g) a half-equation that is considered? Is it because the half-cell containing electrode B has H+ ions in solution and the question states it is an acidic solution? But the half-equation contains H2 which isn't part of the question or in the galvanic cell so shouldn't it be eliminated as an option?

In a galvanic cell the strongest oxidant will always react with the strongest reductant.

In this case they gave you the E° value of TiO2+ (aq) + 2H+ (aq) + e- ⇌ Ti3+ (aq) + H2O(l), which is -0.06 V. If you place this in the electrochemical series, you'll find that 2H+(aq) is a stronger oxidant than TiO2+. Because of this 2H+(aq) will be reduced in preference (hence: 2H+(aq) + 2e- -> H2(g)).

The reaction is considered because it's part of the half-cell containing electrode B. It doesn't matter that H2 isn't part of the galvanic cell because its a product (if H2 was a reactant than it would be a different matter). It only matters that H2+ is being reduced. The H2 is just going to be released into the air as a gas!
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jasheel

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Re: VCE Chemistry Question Thread
« Reply #7993 on: April 27, 2019, 04:07:45 pm »
+2
If you don't mind, I have a question of my own- how can we be certain that the iron would form the 2+ ion? I suspect that this is due to the Hydrogen needing to accept 1 electron each, so Iron is donating 2 electrons total?
Iron (Fe(s)) on the electrochemical series will lose two electrons, it can't ever lose one electron only! (Fe2+ can lose another electron to form Fe3+ later on however). 
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f0od

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Re: VCE Chemistry Question Thread
« Reply #7994 on: April 29, 2019, 09:08:14 pm »
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In our recent reduction potentials prac, our predicted cell voltage (0.57) was just over the actual cell voltage (0.56). I was wondering what might be the reason for this (slight, but still a) difference aside from it not being in SLC conditions?

Thanks! :)
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