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March 29, 2024, 07:04:42 am

Author Topic: VCE Chemistry Question Thread  (Read 2313380 times)  Share 

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caffinatedloz

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Re: VCE Chemistry Question Thread
« Reply #8430 on: January 21, 2020, 02:54:23 pm »
+1
Will increasing the temperature in HPLC decrease the retention time and increase degree of separation and why?
Hey jnlfs2010! Can you explain what you already understand about this question and what you are stuck on so that we can help you better?

whys

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Re: VCE Chemistry Question Thread
« Reply #8431 on: January 21, 2020, 02:58:07 pm »
+3
Will increasing the temperature in HPLC decrease the retention time and increase degree of separation and why?
Retention time is the time taken for the solute to pass through the chromatography column. Higher column temperatures usually lead to materials eluting faster - so decreased retention time and higher quality performance. There is usually a decrease of retention by 1-2% for each degree celsius on average. This is because the more components exist in the form of a gas. This can cause poor separation, although separation is faster. More components exist as a gas due to higher temperatures giving energy to the liquid particles, causing them to evaporate and form a gas. If the temperature is lowered, then the compound will condense. Temperature doesn't effect HPLC as much as gas chromatography though. But basically, as an evaporated gas, it transfers down much faster than a liquid will when it remains condensed due to cooler temperatures. I'm pretty sure the typic temperature range for HPLC is around 60-70 degrees celsius.

I hope this sort of made sense! It was challenging to word out haha
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Evolio

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Re: VCE Chemistry Question Thread
« Reply #8432 on: January 24, 2020, 07:47:40 am »
0
Hey guys!
Relating to redox reactions and the electrochemical series, I was wondering why Iodide I2(s) and Cu(s) would not cause a reaction?
Because the I2(s) is on the higher left hand side than the Cu(s) which is on the right hand side. So the I2(s) would be  the oxidant and the Cu(s) would be the reductant?
It's a question on my homework sheet and the answers say 'no reaction' which I'm a bit confused about.

sk2000

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Re: VCE Chemistry Question Thread
« Reply #8433 on: January 24, 2020, 09:14:48 am »
+8
Hey guys!
Relating to redox reactions and the electrochemical series, I was wondering why Iodide I2(s) and Cu(s) would not cause a reaction?
Because the I2(s) is on the higher left hand side than the Cu(s) which is on the right hand side. So the I2(s) would be  the oxidant and the Cu(s) would be the reductant?
It's a question on my homework sheet and the answers say 'no reaction' which I'm a bit confused about.

In the context of a galvanic cell, iodine solid is typically not found in any type of redox reaction because it does not conduct electricity. Iodine is found as a covalently bonded diatomic I2 molecule, meaning that there are no free electrons or ions to conduct electricity.

Electrodes in a galvanic cell must be conductive to alllow for the transfer of electrons to complete the redox reactions, i.e for the reductant to actually be able to lose electrons and the oxidant able to gain them.

For an iodine-based galvanic cell to be functional, the iodine would have to be in an aqueous solution as an ionic compound (such as potassium iodide) and using an inert electrode (such as platinum). Potassium iodide is an ionic bond, and the ions dissociate in solution to become K+ and I-; the charged particles able to move freely and thus the solution is electrically conductive. Then, iodide ions in solution become the reductant.
« Last Edit: January 24, 2020, 10:36:42 am by sk2000 »
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Evolio

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Re: VCE Chemistry Question Thread
« Reply #8434 on: January 24, 2020, 09:41:39 am »
+2
Ohhh, I understand now. So, we should take into account the states and whether the charged particles are able to move freely.
Thank you, sk2000!  ;D

Chocolatemilkshake

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Re: VCE Chemistry Question Thread
« Reply #8435 on: January 27, 2020, 03:13:18 pm »
0
Hi all,
I was reviewing unsaturated and saturated fatty acid esters (in biodiesel) and was wondering which ester would be more viscous?

My thinking...
Considering that unsaturated esters (due to double bonds) have a changed shape, they fit together less nicely, hence decreasing the dispersion forces between the molecules. Therefore, the molecules are less attracted to each other and unsaturated biodiesel esters are less viscous.

Is this correct? Thanks  8)
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sk2000

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Re: VCE Chemistry Question Thread
« Reply #8436 on: January 27, 2020, 05:28:20 pm »
+8
Hi all,
I was reviewing unsaturated and saturated fatty acid esters (in biodiesel) and was wondering which ester would be more viscous?

My thinking...
Considering that unsaturated esters (due to double bonds) have a changed shape, they fit together less nicely, hence decreasing the dispersion forces between the molecules. Therefore, the molecules are less attracted to each other and unsaturated biodiesel esters are less viscous.

Is this correct? Thanks  8)

You are correct! It is also important to consider the source of the components of certain biodiesels.
Plant oils often contain C=C double bonds, making them unsaturated, whilst animal fats such as tallow are known to contain only C-C single bonds, making them saturated.

The fatty acid's tail will determine most of its properties, including viscosity. Unsaturated fatty acids contain C=C double bonds, which often have geometric isomers i.e cis and trans formations (although trans formations are uncommon in natural fatty acids; they must be artifically produced). So, the cis arrangement around the double bond will affect the chains 3D shape because each carbon atom on a double bond is only connected to three other atoms (as opposed to the usual four of a saturated fatty acid) which will rearrange the atoms locations in 3D space, as they tend to localise the furthest possible from each other. As a result, the chains will take up more space, and there will be more space separating the neighboring chains because these chains simply cannot get any closer to each other. The increased distance that dispersion forces will have to travel weakens the forces, which will cause unsaturated biodiesel molecules to flow better, i.e have a lower viscosity.

Saturated fatty acids are straight chains, thus they will occupy less 3D space, meaning that neighboring chains can pack closer together, strengthening the dispersion forces as they will not have to travel as far. Thus, biodiesel sourced from saturated fats will not be able to flow as smoothly, i.e they have a higher viscosity.

We may also consider that neighboring biodiesel molecules will be able to form dipole-dipole bonds with the oxygen atoms in each ester group. As dipole-dipole forces are a lot stronger than dispersion forces, this will mean that all biodiesels are more viscous in general than petrodiesels.

We can see that the viscosity of biodiesel quite neatly relates back to it's source. Overlooking structure for a quick justification, animal fats are found as solid at room temperature, whilst plant oils are liquids at room temperature. Thus we can logically deduce that biodiesel molecules sourced from animal fats (saturated) will be more viscous than those sourced from plant oils (unsaturated).

Sorry if this was long, but hope it helps :)
« Last Edit: January 27, 2020, 05:34:40 pm by sk2000 »
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Chocolatemilkshake

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Re: VCE Chemistry Question Thread
« Reply #8437 on: January 27, 2020, 06:13:03 pm »
+3
You are correct! It is also important to consider the source of the components of certain biodiesels.
Plant oils often contain C=C double bonds, making them unsaturated, whilst animal fats such as tallow are known to contain only C-C single bonds, making them saturated.

The fatty acid's tail will determine most of its properties, including viscosity. Unsaturated fatty acids contain C=C double bonds, which often have geometric isomers i.e cis and trans formations (although trans formations are uncommon in natural fatty acids; they must be artifically produced). So, the cis arrangement around the double bond will affect the chains 3D shape because each carbon atom on a double bond is only connected to three other atoms (as opposed to the usual four of a saturated fatty acid) which will rearrange the atoms locations in 3D space, as they tend to localise the furthest possible from each other. As a result, the chains will take up more space, and there will be more space separating the neighboring chains because these chains simply cannot get any closer to each other. The increased distance that dispersion forces will have to travel weakens the forces, which will cause unsaturated biodiesel molecules to flow better, i.e have a lower viscosity.

Saturated fatty acids are straight chains, thus they will occupy less 3D space, meaning that neighboring chains can pack closer together, strengthening the dispersion forces as they will not have to travel as far. Thus, biodiesel sourced from saturated fats will not be able to flow as smoothly, i.e they have a higher viscosity.

We may also consider that neighboring biodiesel molecules will be able to form dipole-dipole bonds with the oxygen atoms in each ester group. As dipole-dipole forces are a lot stronger than dispersion forces, this will mean that all biodiesels are more viscous in general than petrodiesels.

We can see that the viscosity of biodiesel quite neatly relates back to it's source. Overlooking structure for a quick justification, animal fats are found as solid at room temperature, whilst plant oils are liquids at room temperature. Thus we can logically deduce that biodiesel molecules sourced from animal fats (saturated) will be more viscous than those sourced from plant oils (unsaturated).

Sorry if this was long, but hope it helps :)

Wow! Super helpful really tied all my understandings together  ;D
Thanks so much!!
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Katherinet123

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Re: VCE Chemistry Question Thread
« Reply #8438 on: February 01, 2020, 01:13:57 pm »
0
Hi could i please have some help with this question I keep getting the wrong answer!!

1.Calculate the energy released when 5.00 kg of methane (CH4) is burnt in an unlimited supply
of oxygen. The heat of combustion of methane is –890 kJ mol–1. (Give your answer in megajoules, MJ.

Thanks in advanced!!!

ssillyssnakes

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Re: VCE Chemistry Question Thread
« Reply #8439 on: February 01, 2020, 02:19:20 pm »
+8
Hi could i please have some help with this question I keep getting the wrong answer!!

1.Calculate the energy released when 5.00 kg of methane (CH4) is burnt in an unlimited supply
of oxygen. The heat of combustion of methane is –890 kJ mol–1. (Give your answer in megajoules, MJ.

Thanks in advanced!!!

With these questions you have to first convert the mass of the fuel (in this case methane) into mols.
n(CH4) = 5000g/16g/mol = 312.5 mols of methane
Then times the heat of combustion (per mol) by how many mols are present
890kJmol-1 * 312.5 mol = 278125 kJ
Then convert this into whatever value you need and you have your answer!
278125kJ = 278 MJ (3 sig figures)

Hope this helps!
« Last Edit: February 01, 2020, 10:19:13 pm by ssillyssnakes »
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Katherinet123

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Re: VCE Chemistry Question Thread
« Reply #8440 on: February 01, 2020, 03:48:11 pm »
0
thank you so much!

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8441 on: February 09, 2020, 06:52:18 pm »
0
Soooooo fractional distillation vs ...can someone please explain. This is my understanding at the moment

FD: on crude oil because it is useless just as it is separating it into fractions is required to get the hydrocarbons? Is that all that happens?

Fracking: Natural gas is deep within deposits of rocks. This is the process of collecting it. Also does biogas undergoes fracking after the anaerobic decomposition of organic matter?

Thanks :))
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Evolio

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Re: VCE Chemistry Question Thread
« Reply #8442 on: February 09, 2020, 07:45:51 pm »
+12
Hey Coolgalbornin03Lo!  ;D

Yes, you are correct in saying that fractional distillation is used for crude oil. However, we could add a bit more detail to it. Crude oil contains different sizes of hydrocarbons where some have small chains of carbons and some have long chains of carbons. The smaller carbon chains have a lower molar mass. And we know that molar mass corresponds to the strength of the dispersion forces where the larger the molar mass, the stronger the dispersion forces. This means that the smaller carbon chains (hydrocarbons) have weaker dispersion forces with each other while the larger carbon chains have a larger molar mass and so they have stronger dispersion forces with other long carbon chains.
Now, we can also form a relationship between the strength of the dispersion forces and the boiling points of the hydrocarbons.  The stronger the dispersion forces between the hydrocarbons, the more energy, specifically heat energy, requires to break the forces apart, thus meaning that they require a higher boiling point to break the forces apart.
Okay, that was background info.

So, in fractional distillation the crude oil is heated outside the column, where it converts to a gas and moves inside the column. There are different sections of the fractionating column where it is cooler at the higher up sections and hotter at the more bottom sections and the hydrocarbons begin to condense back into liquid form. That is why it is called 'fractional' because there are different sections. The hydrocarbons with the higher boiling point where they need more heat energy required to break the dispersion forces, collate at the top while the smaller hydrocarbons collate at the bottom.
Here's a very useful video to explain fractional distillation: https://www.youtube.com/watch?v=PYMWUz7TC3A

With regards to your second question, I'm pretty sure that fracking is only to extract natural gas and it not used to obtain biogas. Fracking involves disrupting porous rock layers using sand, water and other chemicals which allows the natural gas to be released. I'm not too sure about fracking and how much we need to know however.

I hope this helped! And please correct me if I'm wrong!  :)

This diagram may help you visualise the fractional distillation process and it also shows the sectional location of the different fuels due to their respective boiling points :

« Last Edit: February 09, 2020, 07:53:53 pm by Evolio »

jnlfs2010

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Re: VCE Chemistry Question Thread
« Reply #8443 on: February 12, 2020, 06:39:12 pm »
0
Hey, this is an MC question from the 2018 Chemistry exam. I don't really understand the question and the answer like I don't know how to start to achieve a correct thought process to the answer. What does it mean when air is present, and where is it present?
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SmartWorker

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Re: VCE Chemistry Question Thread
« Reply #8444 on: February 14, 2020, 04:49:27 pm »
0
Hi all,

this is a 3 mark question. Why does the atom when heated only emit a few wavelengths of light.

Thanks
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