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Author Topic: HSC Chemistry Question Thread  (Read 1040649 times)  Share 

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anotherworld2b

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Re: Chemistry Question Thread
« Reply #645 on: August 19, 2016, 01:05:18 am »
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That being said, this is an example of an ionic equation.

Then how would you do this question? Im confused on what anions and cations to join together?  :-\

RuiAce

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Re: Chemistry Question Thread
« Reply #646 on: August 19, 2016, 08:03:13 am »
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Then how would you do this question? Im confused on what anions and cations to join together?  :-\
This requires you to know what's called your solubility rules.

You need to identify which combination of ions will not dissolve in water. That is, those that are insoluble in water.

Take the first one. According to your solubility rules, Na+, NH4+ and K+ are always soluble. Leaving behind the option of Mg2+

CH3COO- and NO3- are both soluble always. This leaves Cl- and CO32-

But MgCl2 is soluble. In fact, CO32- is ALWAYS INSOLUBLE unless paired with a group 1 metal ion or the ammonium ion.

Hence the precipitate is MgCO3

So we can build our equation by nothing more than looking at what we have:
MgCl2(aq) + (NH4)2CO3(aq) -> 2 NH4Cl(aq) + MgCO3(s)
« Last Edit: August 19, 2016, 08:16:42 am by RuiAce »

anotherworld2b

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Re: Chemistry Question Thread
« Reply #647 on: August 20, 2016, 02:08:08 am »
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Thank you very much rui ace  :D
That link is really helpful
This requires you to know what's called your solubility rules.

You need to identify which combination of ions will not dissolve in water. That is, those that are insoluble in water.

Take the first one. According to your solubility rules, Na+, NH4+ and K+ are always soluble. Leaving behind the option of Mg2+

CH3COO- and NO3- are both soluble always. This leaves Cl- and CO32-

But MgCl2 is soluble. In fact, CO32- is ALWAYS INSOLUBLE unless paired with a group 1 metal ion or the ammonium ion.

Hence the precipitate is MgCO3

So we can build our equation by nothing more than looking at what we have:
MgCl2(aq) + (NH4)2CO3(aq) -> 2 NH4Cl(aq) + MgCO3(s)

onepunchboy

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Re: Chemistry Question Thread
« Reply #648 on: August 20, 2016, 12:59:21 pm »
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What are the tests for the products?
Halp

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Re: Chemistry Question Thread
« Reply #649 on: August 20, 2016, 01:21:22 pm »
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(Image removed from quote.)

What are the tests for the products?
Halp
In the first-hand investigation, the most convenient way to demonstrate electrolysis is by means of a Hoffman Voltameter.

We need to keep in mind the half equations that occur during the electrolysis.

For concentrated NaCl solution, the following half equations occur:
Anode: 2 Cl- -> Cl2(g) + 2e-
Cathode: 2 H2O(l) + 2e- -> H2(g) + 2 OH-(aq)

The presence of the chlorine gas at the anode can be tested using litmus paper. The chlorine gas will bleach the litmus, causing it to turn pale.
The presence of the hydrogen gas at the cathode can be tested using the pop test. Set the hydrogen on fire and anticipate a pop.
The presence of the hydroxide ions at the cathode can be tested using an appropriate indicator. Pour an appropriate indicator into a portion of the liquid produced at the cathode, and observe colour changes.

For dilute NaCl solution, we have the electrolysis of water. Whilst the cathode equation is the same, the anode equation is changed.
Anode: 2 H2O(l) -> O2(g) + 4 H+ + 4e-

The presence of the hydrogen ions at the anode can also be tested using an appropriate indicator.
The presence of the abundance of oxygen can be tested using the burning splint test. Raise a burning splint over the region with excess oxygen. The splint lies on fire itself.

Further info on the burning splint test can be researched on Google.

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Re: Chemistry Question Thread
« Reply #650 on: August 20, 2016, 01:42:51 pm »
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Thanks rui !, just another one

I cant think of any other reasons for 6 marks other than conc. Nacl requires less voltage and produces more favourable products..
Halp

anotherworld2b

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Re: Chemistry Question Thread
« Reply #651 on: August 20, 2016, 02:48:07 pm »
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I am able to do simple precipiation reactions with 2 solutions but how you do this question?

anotherworld2b

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Re: Chemistry Question Thread
« Reply #652 on: August 20, 2016, 07:30:33 pm »
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I have a prac test that is about using precipiation reactions to identify 4 unknown solutions. We will be given the names of the solutions but we need to allocate the right name to the  correct solution. I was wondeing what would be the best way to approach doing this prac?

Happy Physics Land

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Re: Chemistry Question Thread
« Reply #653 on: August 20, 2016, 11:02:01 pm »
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Thanks rui !, just another one (Image removed from quote.)

I cant think of any other reasons for 6 marks other than conc. Nacl requires less voltage and produces more favourable products..
Halp

Hey Onepunchboy!

The chlor-alkali industry favours the use of concentrated NaCl(aq) over molten NaCl(l) because NaOH is a favourable product that can only be produced using concentrated brine and it has a major role in industries that involve the manufacturing of soaps, papers and synthetic fibres. In the electrolysis of Molten NaCl, the products are Na(l) and Cl(g). Whilst Na(l) does have some importance in the manufacture of indigo, pharmaceuticals and petrol additives, the use of NaOH in industry is much more diverse and dominant. In addition, like what you have correctly justified, it takes an energy input of 4.07V for the electrolysis of molten NaCl to take place (2Cl- ——> Cl2(g) + 2e-, Eox= -1.36V. Na+ + e- ---> Na, Ered = -2.71V) whilst it only takes an energy input of 2.19V for the electrolysis of aqueous NaCl. Consequently, chlor-alkali companies can effective reduce energy consumptions, reduce the cost involved and maximise their profits.
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Happy Physics Land

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Re: Chemistry Question Thread
« Reply #654 on: August 20, 2016, 11:09:57 pm »
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I am able to do simple precipiation reactions with 2 solutions but how you do this question?

Hey Anotherworld2b!

This question requires some familiarity with the precipitation rules and observation of the compounds. If we look carefully, amongst all the compounds, potassium, sodium, ammonium and nitrate compounds would all be soluble according to our solubility rules. Sulphide is not included as a part of our solubility rules and in fact, sulphide is unable to form a precipitate compound with any of those cations. So this leaves us with lead and chloride. If you are experienced with these sorts of questions and familiar with solubility rules, you should immediately recognise that lead (II) chloride would produce a white coloured precipitate. So the ionic equation for the formation of this precipitate would be as follow:

Pb2+(aq) + 2Cl-(aq) -----> PbCl2(s)
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Happy Physics Land

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Re: Chemistry Question Thread
« Reply #655 on: August 20, 2016, 11:27:38 pm »
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I have a prac test that is about using precipiation reactions to identify 4 unknown solutions. We will be given the names of the solutions but we need to allocate the right name to the  correct solution. I was wondeing what would be the best way to approach doing this prac?

Hey Anotherworld2b!

You definitely asked the right person, because I did the same prac exam as you in term 1 with 5 unknown solutions. I scored 25/25 in that exam and this one simple table trick I will show you below has really helped me! So essentially how it works is you construct two tables. You will be given on the day what all the four chemicals are, except you dont really know what each individual chemical is. So on one table, you will be constructing a table like what I have shown below, with your different unknown chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your experimental data.



On another table, you will be constructing a table like what I have also shown below, with your different KNOWN chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your theoretical data obtained through your solubility rules.



So now what we do is we match the outcomes. For instance, whichever chemical in the first table that is able to form 2 precipitates should be matched to whichever chemical in the second table that is also able to form 2 precipitates. This way we can identify that the two chemicals are the same and hence identify the unknown chemical as I have shown below.



But now, sometimes you can get situations where you have 2 chemicals from the first table that can form equal amounts of precipitates. Teachers shouldnt give you this sort of situations but if this situation does happen, normally it would not matter which one you would identify as the correct chemical. But, there can be special cases, like what I have shown below, that can help you to determine the correct chemical when two particular chemicals can form equal amounts of precipitates.



I hope my guide helps, and I certainly really enjoyed writing this explanation and I really hope you can achieve a high mark by following what I suggested above!

Best Regards
Happy Physics Land
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anotherworld2b

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Re: Chemistry Question Thread
« Reply #656 on: August 21, 2016, 01:02:41 am »
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thank you very much for your guide Happy Physics Land link  ;D
I had no idea how to even start preparing for this practical but now I feel I have gathered a wealth of confidence :D
I have some questions to ask if that's okay to make sure I thoroughly understand

So basically for the first table you physically add the chemicals together and observe whether there is a precipitate or not.
eg. A+B = NR. Where A is one unknown and B is another unknown chemical. Then repeat for the other parts of the table?
Then for the parts of table where its AA, BB and ect do you leave that empty because your testing for example chemical A with chemical A? Which is done for the second table as well?
Would making a table of some kind to write down the appearance of the reaction such as the colour of the precipitate be necessary as well? How would you know if a precipitate is slightly soluble?
I am also a bit confused about the last part about identifying the chemical when there are two particular chemicals can form equal amounts of precipitates.
I was wondering what other things did you have to do in your prac exam apart from identifying the solutions?
I apologise for asking so many questions and would really like to say that I really appreciate your help

Hey Anotherworld2b!

You definitely asked the right person, because I did the same prac exam as you in term 1 with 5 unknown solutions. I scored 25/25 in that exam and this one simple table trick I will show you below has really helped me! So essentially how it works is you construct two tables. You will be given on the day what all the four chemicals are, except you dont really know what each individual chemical is. So on one table, you will be constructing a table like what I have shown below, with your different unknown chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your experimental data.

(Image removed from quote.)

On another table, you will be constructing a table like what I have also shown below, with your different KNOWN chemicals and the outcomes of the reactions (i.e. whether there are precipitates or not). This is your theoretical data obtained through your solubility rules.

(Image removed from quote.)

So now what we do is we match the outcomes. For instance, whichever chemical in the first table that is able to form 2 precipitates should be matched to whichever chemical in the second table that is also able to form 2 precipitates. This way we can identify that the two chemicals are the same and hence identify the unknown chemical as I have shown below.

(Image removed from quote.)

But now, sometimes you can get situations where you have 2 chemicals from the first table that can form equal amounts of precipitates. Teachers shouldnt give you this sort of situations but if this situation does happen, normally it would not matter which one you would identify as the correct chemical. But, there can be special cases, like what I have shown below, that can help you to determine the correct chemical when two particular chemicals can form equal amounts of precipitates.

(Image removed from quote.)

I hope my guide helps, and I certainly really enjoyed writing this explanation and I really hope you can achieve a high mark by following what I suggested above!

Best Regards
Happy Physics Land

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Re: Chemistry Question Thread
« Reply #657 on: August 21, 2016, 09:33:07 pm »
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Hi I feel like these kinds of questions should be simple since its a multiple choice q. but I still struggle to answer them. Please help! (Question 19 - 2015 HSC Chem. Exam | Screenshots of the question attached)

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Re: Chemistry Question Thread
« Reply #658 on: August 21, 2016, 09:54:38 pm »
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Hi I feel like these kinds of questions should be simple since its a multiple choice q. but I still struggle to answer them. Please help! (Question 19 - 2015 HSC Chem. Exam | Screenshots of the question attached)

Hey!

Firstly, we know that the molar mass of Lead Chloride is 278.1g/mol. Therefore, we have


Since there is one mole of Lead in every mole of Lead Chloride, we have equal as many moles of Lead! We can therefore calculate the concentration in the original sample



And that's that! Somewhere there my order of magnitude is a bit off, but that doesn't matter; clearly, the answer is C. Hope this helps!
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Re: Chemistry Question Thread
« Reply #659 on: August 21, 2016, 10:49:16 pm »
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Hey!

Firstly, we know that the molar mass of Lead Chloride is 278.1g/mol. Therefore, we have


Since there is one mole of Lead in every mole of Lead Chloride, we have equal as many moles of Lead! We can therefore calculate the concentration in the original sample



And that's that! Somewhere there my order of magnitude is a bit off, but that doesn't matter; clearly, the answer is C. Hope this helps!

Ahh I see now but how do we know how many moles of lead there are for every lead chloride. Is it because in PbCl2 there is only one mole of lead in the molecule?
Where as, say if the precipitate was for whatever reason Pb3(PO4)2, there would be 1/3 moles for every mole of Pb3(PO4)2?