# Thread: HSC 2016 Chemistry Marathon

1. ## Re: HSC Chemistry Marathon 2016

So... what's the next question?

2. ## Re: HSC Chemistry Marathon 2016

Why is citric acid(c6h807) triprotic even though it has 8 hydrogens?

Kind of a weird question but whatever...

3. ## Re: HSC Chemistry Marathon 2016

Originally Posted by Bestintheworld
Use recent references to find out how particle accelerators are used to discover new transuranic elements. To process the sources you find, assess their reliability by comparing the information provided. Look for consistency of information.(5)
THIS, is the next question. But I will address the temporary question.
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Not every hydrogen involved in molecules will exhibit acidic properties. This can be visualised by considering essentially every alkane - there are no acidic hydrogens in methane, ethane, propane, etc.. This is also true, by extension, for alkanols which have the hydroxyl (-OH) functional group.

A hydrogen atom is acidic if it can become ionised in water (or, more truthfully speaking, interact through coordinate covalent bonding with a water molecule to produce the hydronium ion). Consider the hydrogen atom in hydrochloric acid to illustrate this example.

HCl(aq) + H2O(l) -> H3O(+) + Cl(-)

Now, consider the nature of the hydrogen atoms in citric acid.

The carboxylic acid functional group (-COOH) is noted to indeed, have one acidic hydrogen. This occurs as the hydrogen can become ionised to produce the carboxylate functional group.

(-COOH) + H2O(l) -> H3O(+) + (-COO) (-)

The three carboxylic acid functional groups supply all the hydrogen atoms to consider citric acid triprotic. All five other hydrogens serve a purpose identical to that of alkanes and alkanols (note - citric acid does have one hydroxyl group). These atoms do not exhibit acidic properties - they cannot detach to form ions in solution, thus they cannot be used to call citric acid "octoprotic".

[Worthwhile mention: Citric acid can be named 2-hydroxypropane-1,2,3-tricarboxylic acid]
________________
EDIT: EDIT: removed.

4. ## Re: HSC Chemistry Marathon 2016

Originally Posted by leehuan
THIS, is the next question. But I will address the temporary question.
_________________________

Not every hydrogen involved in molecules will exhibit acidic properties. This can be visualised by considering essentially every alkane - there are no acidic hydrogens in methane, ethane, propane, etc.. This is also true, by extension, for alkanols which have the hydroxyl (-OH) functional group.

A hydrogen atom is acidic if it can become ionised in water (or, more truthfully speaking, interact through coordinate covalent bonding with a water molecule to produce the hydronium ion). Consider the hydrogen atom in hydrochloric acid to illustrate this example.

HCl(aq) + H2O(l) -> H3O(+) + Cl(-)

Now, consider the nature of the hydrogen atoms in citric acid.

The carboxylic acid functional group (-COOH) is noted to indeed, have one acidic hydrogen. This occurs as the hydrogen can become ionised to produce the carboxylate functional group.

(-COOH) + H2O(l) -> H3O(+) + (-COO) (-)

The three carboxylic acid functional groups supply all the hydrogen atoms to consider citric acid triprotic. All five other hydrogens serve a purpose identical to that of alkanes and alkanols (note - citric acid does have one hydroxyl group). These atoms do not exhibit acidic properties - they cannot detach to form ions in solution, thus they cannot be used to call citric acid "octoprotic".

[Worthwhile mention: Citric acid can be named 2-hydroxypropane-1,2,3-tricarboxylic acid]
Nice

5. ## Re: HSC Chemistry Marathon 2016

Actually, I feel as though the open ended investigation question is just a bit too tedious as a whole. We'll keep the questions in the marathon to exam style questions.

Time to introduce acidic environment I see.
___________________________
NEXT QUESTION:

A student sets up an enclosed environment to simulate acid rain. The system initially consists of only sulfur dioxide gas and litmus paper. When he sprayed water into this environment, the litmus turned red rapidly.

a) Use relevant equations to interpret the results of the experiment and assess its validity. (4)
b) Evaluate the impacts of increasing levels of sulfur dioxide in the environment today. (4)

6. ## Re: HSC Chemistry Marathon 2016

I don't think anyone has gone that far into Acidic Environment or even started it. I'll post up a POM question and afterwards people who've started it, can answer it.
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Question: Iodine-131 is a radioisotope that is frequently used in nuclear medicine. Amongst other things it is used to detect fluid build-up in the brain. The half-life of iodine-131 is 8 days.
How much of a 0.16g sample of iodine-131 will remain undecayed after a period of 32 days?

7. ## Re: HSC Chemistry Marathon 2016

Originally Posted by chanandlerbong
I don't think anyone has gone that far into Acidic Environment or even started it. I'll post up a POM question and afterwards people who've started it, can answer it.
_________________________________________
Question: Iodine-131 is a radioisotope that is frequently used in nuclear medicine. Amongst other things it is used to detect fluid build-up in the brain. The half-life of iodine-131 is 8 days.
How much of a 0.16g sample of iodine-131 will remain undecayed after a period of 32 days?
Hehe yeah I haven't done enough of Acidic Environment to answer leehuan's question

Is there a faster way of answering this question? We haven't covered these kinds of questions at school

8. ## Re: HSC Chemistry Marathon 2016

Originally Posted by Shuuya
Hehe yeah I haven't done enough of Acidic Environment to answer leehuan's question

Is there a faster way of answering this question? We haven't covered these kinds of questions at school
$\noindent Yes there's a much faster way \ddot{\smile}. Note that the half-life is 8 days, so if the original amount is M_0, then after 8 days, the amount remaining is \frac{1}{2} M_0. After 16 days, the amount remaining is \frac{1}{2^2} M_0. Continuing the pattern, after 32 days, the amount remaining is \frac{1}{2^4}M_0. Since we had M_0 = 0.16 = 2^4 \times 10^{-2} (in grams) the amount remaining after 32 days is \frac{1}{2^4} \times 2^4 \times 10^{-2} = 0.01 grams.$

9. ## Re: HSC Chemistry Marathon 2016

$\noindent In general, if given a half-life, it's best to use exponentials with base 2 rather than base e. (This is the whole point of using a \textit{half-life} essentially.) If a substance undergoes radioactive decay with half-life of \tau (in some units of time), then the formula for the amount M(t) remaining after a time t has passed (in same units of time) is M(t) = M_{0}\cdot 2^{-\frac{t}{\tau}}, where M_0 is the initial amount present (i.e. at t=0). \Big{(}So after a time of n half-lives has passed, that is, when t=n\tau, for some positive real number n, the amount remaining is \frac{1}{2^n}M_0, as expected.\Big{)} This is much nicer than using base e, since we don't need to spend time finding the value of k in M_0 \mathrm{e}^{-kt}.$

10. ## Re: HSC Chemistry Marathon 2016

Originally Posted by chanandlerbong
I don't think anyone has gone that far into Acidic Environment or even started it. I'll post up a POM question and afterwards people who've started it, can answer it.
_________________________________________
Question: Iodine-131 is a radioisotope that is frequently used in nuclear medicine. Amongst other things it is used to detect fluid build-up in the brain. The half-life of iodine-131 is 8 days.
How much of a 0.16g sample of iodine-131 will remain undecayed after a period of 32 days?
manthan seems like he had. Acids is section 3 of The Acidic Environment and I gave a question that only required section 1 and 2 (indicators and acidic oxides in the atmosphere).
___________________
NEXT QUESTION

leehuan needs to perform an experiment to deduce the molar heat of combustion of various alkanols. He was given excess samples of ethanol, propan-1-ol, butan-1-ol and pentan-1-ol in respective spirit burners.
a) leehuan needs to perform this experiment in the laboratory. Suggest a suitable method for him to use. (3)
b) When arriving at the final results, he finds that his values are always lower than the theoretical value. Explain why this is inevitable and justify possible solutions to minimise the difference in results. (4)
c) Write down the equation for complete combustion of pentan-1-ol. (1)

11. ## Re: HSC Chemistry Marathon 2016

Originally Posted by Shuuya
Hehe yeah I haven't done enough of Acidic Environment to answer leehuan's question

Is there a faster way of answering this question? We haven't covered these kinds of questions at school
Try to avoid using exponential decay related concepts altogether in chemistry. Although obviously calculus has been avoided, it's nonetheless a mathematics topic and exponentiation is quite rarely necessary in chemistry. (Only time I've seen it required is for pH where an exponential formula is used to find [H3O+])
(Note: pH is a logarithmic scale)

InteGrand's pattern was more than sufficient.

12. ## Re: HSC Chemistry Marathon 2016

Originally Posted by InteGrand
$\noindent Yes there's a much faster way \ddot{\smile}. Note that the half-life is 8 days, so if the original amount is M_0, then after 8 days, the amount remaining is \frac{1}{2} M_0. After 16 days, the amount remaining is \frac{1}{2^2} M_0. Continuing the pattern, after 32 days, the amount remaining is \frac{1}{2^4}M_0. Since we had M_0 = 0.16 = 2^4 \times 10^{-2} (in grams) the amount remaining after 32 days is \frac{1}{2^4} \times 2^4 \times 10^{-2} = 0.01 grams.$
Thanks, that's so much easier!

13. ## Re: HSC Chemistry Marathon 2016

Originally Posted by leehuan
leehuan needs to perform an experiment to deduce the molar heat of combustion of various alkanols. He was given excess samples of ethanol, propan-1-ol, butan-1-ol and pentan-1-ol in respective spirit burners.
a) leehuan needs to perform this experiment in the laboratory. Suggest a suitable method for him to use. (3)
b) When arriving at the final results, he finds that his values are always lower than the theoretical value. Explain why this is inevitable and justify possible solutions to minimise the difference in results. (4)
c) Write down the equation for complete combustion of pentan-1-ol. (1)
a) Method:
1. Measure the initial mass of the spirit burner
2. Measure 100mL of water into a beaker and suspend it directly over the spirit burner
3. Measure the initial temperature of the water
4. Light the spirit burner
5. Allow the temperature of the water to rise by about 10°C. Record the final temperature reached by the water.
6. Measure the final mass of the spirit burner
7. Repeat steps 1-6 for the remaining alcohols

b) The experimental value will be lower than the theoretical value due to the scope of error in the experimental process. Heat from the combustion of the alcohol is lost to the surroundings, through the inevitable heating of the air. Incomplete combustion may have also occurred, meaning that all of the energy released was not absorbed by the water and was instead used to form carbon on the base of the beaker. In order to minimise the difference in results, a copper calorimeter could be used to allow an increased absorption of heat, as copper is a better conductor of heat than a glass beaker. The distance between the flame and beaker can also be minimised in order to reduced the effect of heat loss to surroundings.

c) C5H11OH(l) + 8O2(g) --> 5CO2(g)+ 6H2O(g)

[deleted]

15. ## Re: HSC Chemistry Marathon 2016

Originally Posted by Shuuya
a) Method:
1. Measure the initial mass of the spirit burner
2. Measure 100mL of water into a beaker and suspend it directly over the spirit burner
3. Measure the initial temperature of the water
4. Light the spirit burner
5. Allow the temperature of the water to rise by about 10°C. Record the final temperature reached by the water.
6. Measure the final mass of the spirit burner
7. Repeat steps 1-6 for the remaining alcohols

b) The experimental value will be lower than the theoretical value due to the scope of error in the experimental process. Heat from the combustion of the alcohol is lost to the surroundings, through the inevitable heating of the air. Incomplete combustion may have also occurred, meaning that all of the energy released was not absorbed by the water and was instead used to form carbon on the base of the beaker. In order to minimise the difference in results, a copper calorimeter could be used to allow an increased absorption of heat, as copper is a better conductor of heat than a glass beaker. The distance between the flame and beaker can also be minimised in order to reduced the effect of heat loss to surroundings.

c) C5H11OH(l) + 8O2(g) --> 5CO2(g)+ 6H2O(g)
______________________________________
I might be mistaken, but should the state of water in the products be liquid and not gas? If not, can anyone explain why its in a gaseous state. There has been a debate going on at my school, that the products of ethanol combustion for water, it should be liquid state, others have said it should be gas?

16. ## Re: HSC Chemistry Marathon 2016

Originally Posted by Shuuya
a) Method:
1. Measure the initial mass of the spirit burner
2. Measure 100mL of water into a beaker and suspend it directly over the spirit burner
3. Measure the initial temperature of the water
4. Light the spirit burner
5. Allow the temperature of the water to rise by about 10°C. Record the final temperature reached by the water.
6. Measure the final mass of the spirit burner
7. Repeat steps 1-6 for the remaining alcohols

b) The experimental value will be lower than the theoretical value due to the scope of error in the experimental process. Heat from the combustion of the alcohol is lost to the surroundings, through the inevitable heating of the air. Incomplete combustion may have also occurred, meaning that all of the energy released was not absorbed by the water and was instead used to form carbon on the base of the beaker. In order to minimise the difference in results, a copper calorimeter could be used to allow an increased absorption of heat, as copper is a better conductor of heat than a glass beaker. The distance between the flame and beaker can also be minimised in order to reduced the effect of heat loss to surroundings.

c) C5H11OH(l) + 8O2(g) --> 5CO2(g)+ 6H2O(g)
a) I'm reluctant to say 3 despite my gut instinct to give it because whenever I wrote this method I always wrote to do the calculation specifically as a last step. But from my understanding of chemistry this is sufficient for that 3, because my understanding of the chemistry course is that such details are typically addressed in the discussion.

b) Awesome. 4/4

Suggestion - Try putting the copper calorimeter INTO the method rather than the beaker. Believe it or not heat lost to the surroundings is the main reason why it's inevitable, because in a system where there's essentially unlimited supply of O2, incomplete combustion will be rare. For the fourth mark, instead of suggesting the calorimeter you could alternatively suggest putting a lid on the calorimeter with a hole for the thermometer and enclose the environment with aluminium foil in an attempt to trap the heat.

c)
Moles of C: LHS - 5, RHS - 5
Moles of O: LHS - 1 + 16 = 17, RHS - 10 + 6 = 16 !!!
Moles of H: LHS - 12, RHS = 6

Oops!

Correct answer: 2 C5H11OH(l) +15 O2(g) = 10 CO2(g) + 12 H2O(g) [Source: WolframAlpha]

17. ## Re: HSC Chemistry Marathon 2016

Originally Posted by chanandlerbong
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I might be mistaken, but should the state of water in the products be liquid and not gas? If not, can anyone explain why its in a gaseous state. There has been a debate going on at my school, that the products of ethanol combustion for water, it should be liquid state, others have said it should be gas?
Immediately after the reaction it's a vapour. You don't get water pouring out of a cigarette or burning ethanol or something.

After I set ethanol on fire at school by the time it was done I had pretty much nothing but a few baby drops of water

18. ## Re: HSC Chemistry Marathon 2016

Originally Posted by chanandlerbong
(A) Tritium (H-3) has been used as a radioactive tracer in studies of water movement. What is a radioactive tracer? [1]
I don't think the HSC expects that specific language for the radioisotope however.

19. ## Re: HSC Chemistry Marathon 2016

Originally Posted by leehuan
a) I'm reluctant to say 3 despite my gut instinct to give it because whenever I wrote this method I always wrote to do the calculation specifically as a last step. But from my understanding of chemistry this is sufficient for that 3, because my understanding of the chemistry course is that such details are typically addressed in the discussion.

b) Awesome. 4/4

Suggestion - Try putting the copper calorimeter INTO the method rather than the beaker. Believe it or not heat lost to the surroundings is the main reason why it's inevitable, because in a system where there's essentially unlimited supply of O2, incomplete combustion will be rare. For the fourth mark, instead of suggesting the calorimeter you could alternatively suggest putting a lid on the calorimeter with a hole for the thermometer and enclose the environment with aluminium foil in an attempt to trap the heat.

c)
Moles of C: LHS - 5, RHS - 5
Moles of O: LHS - 1 + 16 = 17, RHS - 10 + 6 = 16 !!!
Moles of H: LHS - 12, RHS = 6

Oops!

Correct answer: 2 C5H11OH(l) +15 O2(g) = 10 CO2(g) + 12 H2O(g) [Source: WolframAlpha]

So awkward, I always forget about the Oxygen in the alcohol

20. ## Re: HSC Chemistry Marathon 2016

Originally Posted by chanandlerbong
______________________________________
I might be mistaken, but should the state of water in the products be liquid and not gas? If not, can anyone explain why its in a gaseous state. There has been a debate going on at my school, that the products of ethanol combustion for water, it should be liquid state, others have said it should be gas?
Yeah like leehuan said it's water vapour. However, keep in mind that for molar heat of combustion, it's complete combustion with products being liquid water and CO2 gas (Conquering Chemistry p36).

21. ## Re: HSC Chemistry Marathon 2016

Originally Posted by Shuuya
[/B]

So awkward, I always forget about the Oxygen in the alcohol
I think the trend is that for odd number carbons you need two moles whereas for even number you only need 1

22. ## Re: HSC Chemistry Marathon 2016

Originally Posted by leehuan
I don't think the HSC expects that specific language for the radioisotope however.
For fun

23. ## Re: HSC Chemistry Marathon 2016

Actually, you're right Leehuan. The questions too hard, and not required for the HSC.
_____________________________________
NEXT QUESTION:
1.
2L of concentrated (10M) hydrochloric acid was spilled in a laboratory accident. Three substances were considered for use to minimize the damage, solid sodium hydrogen carbonate, powdered limestone (calcium carbonate), and 2M sodium hydroxide solution.
(a) CALCULATE the minimum mass of calcium carbonate needed to neutralize the acid. [3]
(b) ASSESS each of the three substances, sodium hydrogen carbonate, powdered limestone and 2M sodium hydroxide for use in the neutralization of the spilt acid. [4]

24. ## Re: HSC Chemistry Marathon 2016

Originally Posted by chanandlerbong
Actually, you're right Leehuan. The questions too hard, and not required for the HSC.
_____________________________________
NEXT QUESTION:
1.
2L of concentrated (10M) hydrochloric acid was spilled in a laboratory accident. Three substances were considered for use to minimize the damage, solid sodium hydrogen carbonate, powdered limestone (calcium carbonate), and 2M sodium hydroxide solution.
(a) CALCULATE the minimum mass of calcium carbonate needed to neutralize the acid. [3]
(b) ASSESS each of the three substances, sodium hydrogen carbonate, powdered limestone and 2M sodium hydroxide for use in the neutralization of the spilt acid. [4]

25. ## Re: HSC Chemistry Marathon 2016

Originally Posted by chanandlerbong
Actually, you're right Leehuan. The questions too hard, and not required for the HSC.
_____________________________________
NEXT QUESTION:
1.
2L of concentrated (10M) hydrochloric acid was spilled in a laboratory accident. Three substances were considered for use to minimize the damage, solid sodium hydrogen carbonate, powdered limestone (calcium carbonate), and 2M sodium hydroxide solution.
(a) CALCULATE the minimum mass of calcium carbonate needed to neutralize the acid. [3]
(b) ASSESS each of the three substances, sodium hydrogen carbonate, powdered limestone and 2M sodium hydroxide for use in the neutralization of the spilt acid. [4]
Ok so I tried (b) and I'm not sure how to assess the substances... I wrote out the equations and found that NaHCO3 and CaCO3 produce a salt, water as well as CO2 whilst NaOH only produces a salt and water. Does this have anything to do with the suitability?

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