HSC 2016 Chemistry Marathon (3 Viewers)

[InteGrand]

Re: HSC Chemistry Marathon 2016

Determine the number of moles of chromium atoms in 140.0g chromium(III) oxide.




I thought I knew how to do this but I can't get the answer given in the solutions.

(Relevant page on Wikipedia: https://en.wikipedia.org/wiki/Chromium(III)_oxide.)

From the Wikipedia link, the molar mass of chromium (III) oxide is 151.9904 g/mol. (You could also have found this from a Periodic Table if you knew that chromium (III) oxide is Cr₂O₃.)

I'm lazy, so let M = 151.9904 g/mol (the molar mass) and let m = 140.0 g (the mass of the sample).

Hence the moles of chromium (III) oxide in the sample n = m/M.

Hence the number of chromium atoms in the sample is 2n = 2m/M, as there are 2 Cr's per Cr₂O₃. The rest is a calculator exercise.

 

[Flop21]

Re: HSC Chemistry Marathon 2016

(Relevant page on Wikipedia: https://en.wikipedia.org/wiki/Chromium(III)_oxide.)

From the Wikipedia link, the molar mass of chromium (III) oxide is 151.9904 g/mol. (You could also have found this from a Periodic Table if you knew that chromium (III) oxide is Cr₂O₃.)

I'm lazy, so let M = 151.9904 g/mol (the molar mass) and let m = 140.0 g (the mass of the sample).

Hence the moles of chromium (III) oxide in the sample n = m/M.

Hence the number of chromium atoms in the sample is 2n = 2m/M, as there are 2 Cr's per Cr₂O₃. The rest is a calculator exercise.

Oh I was looking at the formula for chromuium(III) oxide and saw there was 2 chromium atoms so molar mass of the chromium atoms are 2* 51.996(or 52) then calculating from there. Is this not correct?

Also calculating 1511.9904/140 = 0.92, but answers say 1.842?

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

Oh I was looking at the formula for chromuium(III) oxide and saw there was 2 chromium atoms so molar mass of the chromium atoms are 2* 51.996(or 52) then calculating from there. Is this not correct?

Also calculating 1511.9904/140 = 0.92, but answers say 1.842?

The answer is 2*m/M = 2*140/(151.9904) (explanation given in my previous post).

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

Oh I was looking at the formula for chromuium(III) oxide and saw there was 2 chromium atoms so molar mass of the chromium atoms are 2* 51.996(or 52) then calculating from there. Is this not correct?

Also calculating 1511.9904/140 = 0.92, but answers say 1.842?

What was the method you were using before? I think you were treating molar mass as M' = 2m_Cr, where m_Cr is the molar mass of chromium, and then saying: moles of chromium = m/M' = 2(m_Cr)/(140) ? If so, that method is invalid, which is why you'd have been getting the wrong answer.

 

[Flop21]

Re: HSC Chemistry Marathon 2016

The answer is 2*m/M = 2*140/(151.9904) (explanation given in my previous post).

Oh thanks, I read over your post again more thoroughly and now I understand.

 

[Flop21]

Re: HSC Chemistry Marathon 2016

What was the method you were using before? I think you were treating molar mass as M' = 2m_Cr, where m_Cr is the molar mass of chromium, and then saying: moles of chromium = m/M' = 2(m_Cr)/(140) ? If so, that method is invalid, which is why you'd have been getting the wrong answer.

Yeah I wasn't understanding why you were using the molar mass of the whole chemical, rather than just 2*Cr molar mass. But I understand why you do that now from reading over your post.

 

[Flop21]

Re: HSC Chemistry Marathon 2016

Calculate the mass of copper(I) sulfide [CU2S] which contains 4.6 x 10^23 copper atoms.


Can someone take me through this one. Finding it super difficult.

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

Calculate the mass of copper(I) sulfide [CU2S] which contains 4.6 x 10^23 copper atoms.


Can someone take me through this one. Finding it super difficult.

I'm lazy again, so I'll do it algebraically for now (I'm essentially doing the general case).

Let the required mass be m. Let the molar mass of Cu₂S be M (you can find this value online or calculate it from a Periodic Table). Let N = 4.6 * 10²³. So our goal is to find m in terms of M and N (which we know the values of).

Now, we each mole of copper(I) sulfide contains 2 moles of copper. Hence there'll be N copper atoms exactly when there are N/2 moles of copper(I) sulfide.

So we need N/2 = m/M (using "number of moles = mass/(molar mass)", a standard HSC Chemistry formula). Rearranging, we get m = MN/2. Now just plug in the values for M and N to get the answer.

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

Can someone help me with this one?

A student conducting this experiment weighs out 2.40 g of CoCl2.6H2O.

For this part you will be calculating the amount of sodium carbonate required for the complete precipitation of cobalt(II) carbonate.

Calculate the moles of CoCl2.6H2O used in this reaction.

 

[leehuan]

Re: HSC Chemistry Marathon 2016

Can someone help me with this one?

A student conducting this experiment weighs out 2.40 g of CoCl2.6H2O.

For this part you will be calculating the amount of sodium carbonate required for the complete precipitation of cobalt(II) carbonate.

Calculate the moles of CoCl2.6H2O used in this reaction.

They would have to provide you with the equation in the HSC because the hexahydrate will screw up the students' mind.

Care to show some theory relating to how to generate the equation?

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

They would have to provide you with the equation in the HSC because the hexahydrate will screw up the students' mind.

Care to show some theory relating to how to generate the equation?

This isn't a HSC question, it's a question I'm doing for chemistry at uni.

But here's the equation I'm assuming it meant to be used?

Na2CO3 + CoCl2.6H2O ===> 2NaCl + CoCO3 + 6H2O

 

[leehuan]

Re: HSC Chemistry Marathon 2016

Yeah I figured you were in uni so that's why I asked for the equation. Which makes sense, I think.

Then, the mass of CoCl2.6H2O is 2.40g. This is basically the mass of cobalt(II) chloride hexahydrate (the whole thing)
You can use your university periodic table sheet (or whatever is relevant) to determine the molar mass M.

The formula is then simply n=m/M?

(Sorry, if I actually tried it out it seems so trivial. I mean it doesn't look like we have a limiting reagent here.)

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

Yeah I figured you were in uni so that's why I asked for the equation. Which makes sense, I think.

Then, the mass of CoCl2.6H2O is 2.40g. This is basically the mass of cobalt(II) chloride hexahydrate (the whole thing)
You can use your university periodic table sheet (or whatever is relevant) to determine the molar mass M.

The formula is then simply n=m/M?

(Sorry, if I actually tried it out it seems so trivial. I mean it doesn't look like we have a limiting reagent here.)

Yeah it seems like I just need to use n=m/M but why would the question include "For this part you will be calculating the amount of sodium carbonate required for the complete precipitation of cobalt(II) carbonate." When it's just as simple as converting grams to moles to answer the question.

 

[parad0xica]

Re: HSC Chemistry Marathon 2016

Question: Is pure water acidic, basic or neutral? Justify your claims.

 

[leehuan]

Re: HSC Chemistry Marathon 2016

Yeah it seems like I just need to use n=m/M but why would the question include "For this part you will be calculating the amount of sodium carbonate required for the complete precipitation of cobalt(II) carbonate." When it's just as simple as converting grams to moles to answer the question.

Is it like, a follow-up question?

Question: Is pure water acidic, basic or neutral? Justify your claims.

Obviously not answered according to HSC student standards:

Pure water has a pH of 7, but this just means that [H3O+]=10^-7. According to the equation for self ionisation of water 2 H2O(l) <=> H3O(+) + OH(-), the water constant Kw = 1.0*10^-4 (mol L^-1)^2 = [H3O+][OH-]. Direct substitution of [H3O+] shows that [OH-]=10^-7. If [H3O+]=[OH-], then the substance is neutral.

The self ionisation of water does indeed imply that the concentration of the hydronium ion and the hydroxide ion is not negligible; it is indeed a small number. It is strictly the fact that their concentrations are equal (implying that they cancel out when considering acidity/basicity) that resort in the final answer.

Note that pure water is actually defined to have a pH of 7. Also note that when we measure the concentrations of aqueous solutions, generally we actually refer to the concentration of such substances in water itself.

 

[Nailgun]

Re: HSC Chemistry Marathon 2016

Is it like, a follow-up question?



Obviously not answered according to HSC student standards:

Pure water has a pH of 7, but this just means that [H3O+]=10^-7. According to the equation for self ionisation of water 2 H2O(l) <=> H3O(+) + OH(-), the water constant Kw = 1.0*10^-4 (mol L^-1)^2 = [H3O+][OH-]. Direct substitution of [H3O+] shows that [OH-]=10^-7. If [H3O+]=[OH-], then the substance is neutral.

The self ionisation of water does indeed imply that the concentration of the hydronium ion and the hydroxide ion is not negligible; it is indeed a small number. It is strictly the fact that their concentrations are equal (implying that they cancel out when considering acidity/basicity) that resort in the final answer.

Note that pure water is actually defined to have a pH of 7. Also note that when we measure the concentrations of aqueous solutions, generally we actually refer to the concentration of such substances in water itself.

I used to watch h20 just add water when i was bored
does that count as chemistry

 

[Occupied]

Re: HSC Chemistry Marathon 2016

Calculate the mass of copper(I) sulfide [CU2S] which contains 4.6 x 10^23 copper atoms.


Can someone take me through this one. Finding it super difficult.

yo flop, is the answer 607.9226171 g = 610g (2 sig fig)?

 

[BlueGas]

Re: HSC Chemistry Marathon 2016

Is it like, a follow-up question?

Yeah it is, the questions are in this order:

Identify the net ionic equation for:

1. CoCl2.6H2O is dissolved in water
2. The cobalt chloride solution is added to an aqueous solution of sodium carbonate
3. The cobalt carbonate preciptate is dissolved in excess sulfuric acid
4. The CoSO4.6H2O crystals are formed from solution after evaporation
5. Then it's the calculating the moles question I posted earlier

 

[Flop21]

Re: HSC Chemistry Marathon 2016

yo flop, is the answer 607.9226171 g = 610g (2 sig fig)?

61g apparently

 

[InteGrand]

Re: HSC Chemistry Marathon 2016

yo flop, is the answer 607.9226171 g = 610g (2 sig fig)?

You appear to be off by a factor of 10 compared to Flop21's reported answer. What method did you use? Did you try the method I posted (I have t actually typed numbers into a calculator but the method should be right)?