HSC 2016 Chemistry Marathon (2 Viewers)

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leehuan

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Re: HSC Chemistry Marathon 2016

Hm, maybe I did rush a bit then.

REPLACEMENT QUESTION (in the meantime):

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)
 

hsccheese

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Re: HSC Chemistry Marathon 2016

Thanks! True that though, so I'll answer a few questions every now and then to promote activity as well as possible. Maybe one every 2-3 days when there isn't a previous answer. Later on I'll just let the thread flow.



Although I didn't use the word "compare" or "contrast", it helps to state that alkanes only have single bonds within the hydrocarbon chain structure, which are typically chemically unreactive. The third mark was a bit hidden, and required you to justify the process - we can say that the double bond is easily opened up through interaction with other molecules, for example, through addition reactions or polymerisation. Alternatively, use the bromine water practical equations (if you have done so).
Hexane: C6H14(l) + HOBr(l) -(UV)-> C6H12BrOH(l) + H2(g)
1-hexene: C6H12(l) + HOBr(l) -> C6H12BrOH(l)
This is acceptable as the necessity of UV to force a substitution reaction is obviously more energy consuming than the simple addition reaction.

Maybe 1/3, but keep trying!
I'm a student from the US, but I learned that the alkene series is generally more reactive then the alkane series because of the following.

Bonding can be perceived as the overlap of electron orbitals. A sigma bond (the only bond in an alkane) is a covalent bond formed by head on overlap of atomic orbitals/hybrid orbitals along the bond axis. They have good orbital overlap, strong bonding due to the shorter bonds and the electron density is symmetric about the internuclear axis.

A double bond is made of a pi bond and a sigma bond (one head on and one overlap). A pi bond is a bond formed by the sideways overlap of p orbitals on adjacent atoms, perpendicular to any sigma bonds between the same atoms. A pi bond has 2 orbital lobes, 1 above and below the planes of the sigma bond and they are weaker than sigma bonds due to the poorer orbital overlap and hence more reactive.

Here's a little diagram of a triple bond with 2 pi bonds. The overlap breaking more easily due to being weaker is what makes alkenes more reactive than alkanes. Note (i'm not saying anything about alkynes, just showing you what the orbitals looks like)

Capture.PNG
 

hsccheese

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Re: HSC Chemistry Marathon 2016

Thread revival.

There should be enough of you that have done this now.
Transuranic elements are produced by 1) Neutron bombardment or 2) Fusion reactions.
More early transuranic elements were produced by bombarding heavy elements with neutrons in a reactor. By bombarding U-238, you get U-239 which beta decays to Np-239 and Pu-239 after. Hard to write nuclear equations without the subscripts/superscripts so I'll write them in word.
U-238 + neutron -> U-239 -> electron + Np-239
Np-239 -> electron + Pu-239
You can produce larger transuranic elements by bombarding heavy nuclei with high velocity positively charged particles like He/C nuclei. You have to do this in a particle/linear accelerator or cyclotron because they are charged particles unlike neutrons.
You can get Cf-246 by bombarding U-238 with a C nuclei in a linear accelerator.
U-238 + C nucleu -> Cf-246 + 4 neutrons
 

leehuan

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Re: HSC Chemistry Marathon 2016

I'm a student from the US, but I learned that the alkene series is generally more reactive then the alkane series because of the following.

Bonding can be perceived as the overlap of electron orbitals. A sigma bond (the only bond in an alkane) is a covalent bond formed by head on overlap of atomic orbitals/hybrid orbitals along the bond axis. They have good orbital overlap, strong bonding due to the shorter bonds and the electron density is symmetric about the internuclear axis.

A double bond is made of a pi bond and a sigma bond (one head on and one overlap). A pi bond is a bond formed by the sideways overlap of p orbitals on adjacent atoms, perpendicular to any sigma bonds between the same atoms. A pi bond has 2 orbital lobes, 1 above and below the planes of the sigma bond and they are weaker than sigma bonds due to the poorer orbital overlap and hence more reactive.

Here's a little diagram of a triple bond with 2 pi bonds. The overlap breaking more easily due to being weaker is what makes alkenes more reactive than alkanes. Note (i'm not saying anything about alkynes, just showing you what the orbitals looks like)

View attachment 32704
I saw this information once. However, the 12th grade chemistry course lacks significant depth in contrast to the full explanation of the reason. But yes, the alkene series is more chemically reactive.
 

leehuan

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Re: HSC Chemistry Marathon 2016

Transuranic elements are produced by 1) Neutron bombardment or 2) Fusion reactions.
More early transuranic elements were produced by bombarding heavy elements with neutrons in a reactor. By bombarding U-238, you get U-239 which beta decays to Np-239 and Pu-239 after. Hard to write nuclear equations without the subscripts/superscripts so I'll write them in word.
U-238 + neutron -> U-239 -> electron + Np-239
Np-239 -> electron + Pu-239
You can produce larger transuranic elements by bombarding heavy nuclei with high velocity positively charged particles like He/C nuclei. You have to do this in a particle/linear accelerator or cyclotron because they are charged particles unlike neutrons.
You can get Cf-246 by bombarding U-238 with a C nuclei in a linear accelerator.
U-238 + C nucleu -> Cf-246 + 4 neutrons
Yep.
The marks were for:
- Two examples correctly identified
- Equations (1mk for both situations)
- A process mentioned
 

Bestintheworld

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Re: HSC Chemistry Marathon 2016

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)
 

leehuan

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Re: HSC Chemistry Marathon 2016

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)
Don't tell them how to watch out for reliability. That could drop the mark allocation to 4.

Science students are expected to know what reliability is.
 

RachelGreen

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I saw this information once. However, the 12th grade chemistry course lacks significant depth in contrast to the full explanation of the reason. But yes, the alkene series is more chemically reactive.
Alkene series being more reactive than the alkane series is due to the high electron density associated with the double bond in alkenes. Isn't this reason a bit more simpler?
 

leehuan

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Re: HSC Chemistry Marathon 2016

Alkene series being more reactive than the alkane series is due to the high electron density associated with the double bond in alkenes. Isn't this reason a bit more simpler?
You don't even need to know why the double bond is more reactive, you just need to know that is. But justification with the high electron density will never lose you any marks though.
 

ze-

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Re: HSC Chemistry Marathon 2016

So... what's the next question?
 

manthan1999

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Re: HSC Chemistry Marathon 2016

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

Kind of a weird question but whatever...
 

leehuan

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Re: HSC Chemistry Marathon 2016

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.
_________________________

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.
 
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manthan1999

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Re: HSC Chemistry Marathon 2016

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
 

leehuan

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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)
 
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chanandlerbong

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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.
_________________________________________
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?
 

Shuuya

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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.
_________________________________________
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 :)

[HR][/HR]

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

InteGrand

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Re: HSC Chemistry Marathon 2016

Hehe yeah I haven't done enough of Acidic Environment to answer leehuan's question :)

[HR][/HR]

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

leehuan

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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.
_________________________________________
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)
 
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