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August 23, 2019, 09:36:50 pm

Author Topic: Chemistry Question Thread  (Read 493911 times)  Share 

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david.wang28

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Re: Chemistry Question Thread
« Reply #3750 on: May 31, 2019, 04:11:53 pm »
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Hello,
I am stuck on two questions, 2.1 and 2.2. Can anyone help me out please? Thanks :)
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mirakhiralla

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Re: Chemistry Question Thread
« Reply #3751 on: June 09, 2019, 08:39:52 pm »
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Dumb q but,
do different types of detergents behave differently in hard water?

horse9996

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Re: Chemistry Question Thread
« Reply #3752 on: June 10, 2019, 02:36:23 pm »
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Dumb q but,
do different types of detergents behave differently in hard water?

No. All detergents behave the same with a head and a hydrocarbon tail which allows it to interact with water and grease.
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annabeljxde

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Re: Chemistry Question Thread
« Reply #3753 on: June 12, 2019, 08:32:15 pm »
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Hi!

Could anyone briefly explain why longer hydrocarbons have a tendency to burn in incomplete combustion?
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r1ckworthy

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Re: Chemistry Question Thread
« Reply #3754 on: June 12, 2019, 11:48:07 pm »
+2
Hi!

Could anyone briefly explain why longer hydrocarbons have a tendency to burn in incomplete combustion?
Edit: There is a limited amount of oxygen (20% in atmosphere) in the air, which will not react with ALL the hydrocarbons in the long hydrocarbon chain. I'll keep my answer up as it is kind of relevant, but don't rely too much on it. Incomplete combustion will always occur when there is an inefficient amount of oxygen. I thought it was too simple at first, but just confirmed it with my teacher ;D

Okay I'm gonna take a shot at explaining this. I got my information from this page.

Combustion is essentially the process of gaseous oxygen reacting with the surface of the hydrocarbon. So if a hydrocarbon is easily volatile (turn into gas), the oxygen will react with the surface (which turns into gas) and then with the next molecule (which turns into gas) and then the next molecule (which turns into gas) and so on with ease. This occurs for smaller hydrocarbons that are easy volatile.

For bigger hydrocarbons however, their volatility decreases (accumulation of dispersion forces increases strength/ bonding), and so it will be much harder for oxygen to "grab" the surface molecule and then the next molecule and so on. Oxygen will react much more readily with gaseous hydrocarbons and get used up. But because of the increased strength of the longer hydrocarbon, the oxygen will react only with the surface molecules, resulting in incomplete combustion.

Super dodgy explanation, but that is kind of the whole premise. I'll probably edit my answer tomorrow and add some more stuff to make it more clear.

« Last Edit: June 13, 2019, 08:00:55 am by r1ckworthy »
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annabeljxde

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Re: Chemistry Question Thread
« Reply #3755 on: June 13, 2019, 05:12:28 pm »
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Edit: There is a limited amount of oxygen (20% in atmosphere) in the air, which will not react with ALL the hydrocarbons in the long hydrocarbon chain. I'll keep my answer up as it is kind of relevant, but don't rely too much on it. Incomplete combustion will always occur when there is an inefficient amount of oxygen. I thought it was too simple at first, but just confirmed it with my teacher ;D

Okay I'm gonna take a shot at explaining this. I got my information from this page.

Combustion is essentially the process of gaseous oxygen reacting with the surface of the hydrocarbon. So if a hydrocarbon is easily volatile (turn into gas), the oxygen will react with the surface (which turns into gas) and then with the next molecule (which turns into gas) and then the next molecule (which turns into gas) and so on with ease. This occurs for smaller hydrocarbons that are easy volatile.

For bigger hydrocarbons however, their volatility decreases (accumulation of dispersion forces increases strength/ bonding), and so it will be much harder for oxygen to "grab" the surface molecule and then the next molecule and so on. Oxygen will react much more readily with gaseous hydrocarbons and get used up. But because of the increased strength of the longer hydrocarbon, the oxygen will react only with the surface molecules, resulting in incomplete combustion.

Super dodgy explanation, but that is kind of the whole premise. I'll probably edit my answer tomorrow and add some more stuff to make it more clear.

This is such an amazing explanation, I can't thank you enough!!! Seriously thank you thank you! ヽ(‿)ノ

Ahhh, I see... that explains why my teacher told us to justify why assuming complete combustion for longer hydrocarbons is invalid..

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Re: Chemistry Question Thread
« Reply #3756 on: June 13, 2019, 05:14:59 pm »
0
URGENT!

Hi everyone!

Can anyone please explain to me how you would find the amount of CO2 produced from complete combustion in terms of energy production (i.e. in kJ/mol). My teacher told me to use the energy output of the hydrocarbon (e.g. 5470kJ/mol) to work this out but I'm a little stuck..
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fun_jirachi

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Re: Chemistry Question Thread
« Reply #3757 on: June 13, 2019, 09:41:10 pm »
+3
Say for the combustion of propane:
C3H8(aq) + 5O2(g) --> 3CO2(g) + 4H2O(g) enthalpy = -2220.0 kJ mol-1
For every mole of propane burnt, you 'expel' -2220.0 kJ of energy.
For every mole of CO2 produced, you get -740.0 kJ of energy.

So say some experiment or whatever 'expelled' 14712 kJ of energy. Then, 14712/740 moles of CO2 would be produced, and hence you have the amount of CO2 produced in the chemical reaction. You need the secondary value, otherwise, you have nothing to work off :)

Hope this helps :)
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emmawazza

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Re: Chemistry Question Thread
« Reply #3758 on: July 24, 2019, 04:02:21 pm »
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Hey Hey!
Just have a question about Module 7.

In addition reactions of alkynes, do you express it as two equations, or one?
e.g. for halogenation, do you do equations:
(1) alkyne -> di-halogenated alkene
(2) di-halogenated alkene -> tetrahalogenated alkane

or jump straight from alkyne -> alkane in 1 equation?
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fun_jirachi

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Re: Chemistry Question Thread
« Reply #3759 on: July 24, 2019, 05:16:34 pm »
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Hey Hey!
Just have a question about Module 7.

In addition reactions of alkynes, do you express it as two equations, or one?
e.g. for halogenation, do you do equations:
(1) alkyne -> di-halogenated alkene
(2) di-halogenated alkene -> tetrahalogenated alkane

or jump straight from alkyne -> alkane in 1 equation?


Hey there!

I think that you should definitely show the steps (just to show the marker where you're going, even though it's pretty obvious). I'd say it's a similar thing to skipping steps in maths; do it at your own peril/up to your own discretion. In this case, it shouldn't be too much work to add in one extra line, just to show your superior understanding :)

Hope this helps :)
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classof2019

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Re: Chemistry Question Thread
« Reply #3760 on: July 30, 2019, 10:17:32 pm »
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Can someone please explain why the answer here is B and not A or D?  Isn't a B/L reaction one which involves the transfer of a proton?

fun_jirachi

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Re: Chemistry Question Thread
« Reply #3761 on: July 31, 2019, 07:21:33 pm »
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Upon searching for the question online (for anyone else searching for the question, it's from the NESA Sample Chemistry Questions), the reason the answer is B is because you've got the question wrong! The question should actually be 'Which of the following is NOT a Bronsted-Lowry reaction?' in which case the answer is in fact B.
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classof2019

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Re: Chemistry Question Thread
« Reply #3762 on: July 31, 2019, 07:37:31 pm »
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Upon searching for the question online (for anyone else searching for the question, it's from the NESA Sample Chemistry Questions), the reason the answer is B is because you've got the question wrong! The question should actually be 'Which of the following is NOT a Bronsted-Lowry reaction?' in which case the answer is in fact B.

Ha, very strange! The question was screenshotted directly from the NESA document entitled "Chemistry Additional Sample Examination Questions". They must have made a typo!

Thank you for clarifying.

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Re: Chemistry Question Thread
« Reply #3763 on: August 06, 2019, 02:41:48 pm »
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Hi!
Could someone help with this attached question?
Thanks

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Re: Chemistry Question Thread
« Reply #3764 on: August 06, 2019, 02:49:01 pm »
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Hi!
Could someone help with this attached question?
Thanks
Use the Henderson-hasselbalch equation .  It is given like this :

pH = pKA + log base 10 of (conjugate base concentration/ conjugate acid concentration).
Since we have the pH, manipulate the equation to find the concentration of acetic acid and ion.
Correct me if I am wrong.