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March 28, 2024, 07:10:08 pm

Author Topic: VCE Chemistry Question Thread  (Read 2312952 times)  Share 

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Reus

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Re: VCE Chemistry Question Thread
« Reply #1065 on: June 29, 2014, 02:47:07 pm »
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Are you talking about redox reactions?
If so, I like to look more-so at the loss/gain of electrons!!
No no, uh I'm talking about acid and bases...
I remember there was a specific name for it, but in Units ½
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soNasty

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Re: VCE Chemistry Question Thread
« Reply #1066 on: June 29, 2014, 03:07:57 pm »
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Acid base reactions with diprotric/monoprotic/ acids?
I'm not sure whether or not they're called amphiprotic too

Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #1067 on: June 29, 2014, 03:18:47 pm »
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Acid base reactions with diprotric/monoprotic/ acids?
I'm not sure whether or not they're called amphiprotic too

Yeah I remember the names of species acting in that manner, but not actual reactions.

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #1068 on: June 29, 2014, 05:07:01 pm »
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Amphiprotic = can act as acid or base
So conjugate bases of polyprotic acids are amphiprotic. Try H2CO3's conjugate base of HCO3-, for instance
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vintagea

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Re: VCE Chemistry Question Thread
« Reply #1069 on: June 29, 2014, 10:28:14 pm »
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hello
can someone please explain this? using the electrochemical series
Tin metal is added to solutions of tin(II) chloride to prevent oxidation of the tin(II) ions by oxygen in air.

thank you

#J.Procrastinator

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Re: VCE Chemistry Question Thread
« Reply #1070 on: June 30, 2014, 02:00:17 pm »
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How would I work this out?

The formation of hydrogen iodide from its elements is represented by the equation:
H2(g) + I2(g) → 2HI(g)
This endothermic reaction has an activation energy of 167 kJ mol–1 and the heat of
reaction, ΔH, is +28 kJ mol–1. What is the activation energy for the reverse reaction,
the decomposition of two mole of hydrogen iodide?

With the reverse equation, wouldn't the activation energy still be the same, with ΔH just being -28 kJ mol-1?

The answer is 139 kJ mol-1
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #1071 on: June 30, 2014, 03:16:46 pm »
+1
Let's consider the following two energy profile diagrams:

Endothermic
Exothermic

Notice how the endothermic reaction has to overcome the energy of deltaH AS WELL AS a further little bit? Then, for the exothermic reaction, it only needs to get over that little bit?

So, let's call that little excess bit q, as just some random bit of energy. From the energy profile diagram, we can see that:



And from this, we get:



Since you know the activation energy for the endothermic reaction, to find the activation energy, you either use the above formula to find the activation energy for the reverse, or exothermic, reaction. However, I do suggest logicing it from energy profile diagrams rather than trying to memorise this formula.

#J.Procrastinator

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Re: VCE Chemistry Question Thread
« Reply #1072 on: June 30, 2014, 04:23:07 pm »
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Let's consider the following two energy profile diagrams:

Endothermic
Exothermic

Notice how the endothermic reaction has to overcome the energy of deltaH AS WELL AS a further little bit? Then, for the exothermic reaction, it only needs to get over that little bit?

So, let's call that little excess bit K, as just some random bit of energy. From the energy profile diagram, we can see that:



And from this, we get:



Since you know the activation energy for the endothermic reaction, to find the activation energy, you either use the above formula to find the activation energy for the reverse, or exothermic, reaction. However, I do suggest logicing it from energy profile diagrams rather than trying to memorise this formula.

Oh I get it now, thank you very much! :)

Wait what does the q in the expression represent?
« Last Edit: June 30, 2014, 04:28:39 pm by #J.Procrastinator »
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #1073 on: June 30, 2014, 04:39:53 pm »
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Whoops - sorry, q is the excess bit (which I called k in my original post). Amended.

Lizzy7

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Re: VCE Chemistry Question Thread
« Reply #1074 on: June 30, 2014, 05:11:31 pm »
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Hi everyone, if anyone could please give me explanations for the following... I've tried researching but I haven't figured it out :p

So regarding organic reaction pathways, why is it that I can't directly substitute an 'H' for a 'OH' (hydroxyl) functional group? or an 'H' for a 'NH2' (amine)?

Like it has to be a chlorine from a chloroalkane?

E.g. Chloroalkane + NaOH -> alkanol + NaCl 

#J.Procrastinator

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Re: VCE Chemistry Question Thread
« Reply #1075 on: June 30, 2014, 06:15:54 pm »
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Whoops - sorry, q is the excess bit (which I called k in my original post). Amended.

Ahaha I figured, all goods. :)
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brightsky

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Re: VCE Chemistry Question Thread
« Reply #1076 on: June 30, 2014, 08:23:21 pm »
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Hi everyone, if anyone could please give me explanations for the following... I've tried researching but I haven't figured it out :p

So regarding organic reaction pathways, why is it that I can't directly substitute an 'H' for a 'OH' (hydroxyl) functional group? or an 'H' for a 'NH2' (amine)?

Like it has to be a chlorine from a chloroalkane?

E.g. Chloroalkane + NaOH -> alkanol + NaCl 

H- is a fairly poor leaving group. Cl-, however, is a fairly good one. hence, a nucleophile will readily displace Cl- but not H-. so to form an alcohol, you need to first convert alkane to haloalkane via free radical halogenation, and then convert haloalkane to alcohol in a nucleophilic substitution reaction.
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Lizzy7

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Re: VCE Chemistry Question Thread
« Reply #1077 on: June 30, 2014, 10:16:51 pm »
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H- is a fairly poor leaving group. Cl-, however, is a fairly good one. hence, a nucleophile will readily displace Cl- but not H-. so to form an alcohol, you need to first convert alkane to haloalkane via free radical halogenation, and then convert haloalkane to alcohol in a nucleophilic substitution reaction.


Thank you brightsky!! :D

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #1078 on: June 30, 2014, 10:23:18 pm »
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By 'better leaving group', he means more stable. Chloride we know is quite stable (in water at least) because Cl- has no acid-base properties. H-, however, is a VERY small negative charge that is a ridiculously powerful base (the conjugate base of hydrogen gas and how strong an acid is H2?). Therefore H- is almost never formed.

Brightsky, there is one reaction I can imagine in which H- DOES act as a leaving group :P
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Re: VCE Chemistry Question Thread
« Reply #1079 on: July 01, 2014, 07:51:05 pm »
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Just to clarify something:

In NMR, is chemical shift proportional or inversely proportional to frequency of the radio waves?

I thought that shielded atoms experience the magnetic field of the NMR machine to a lesser extent, and hence require less energy for resonance to be induced (ie. chemical shift is proportional to frequency), but TSFX says the opposite, that is shielded atoms require more energy for resonance to be induced. Does the discrepancy arises due to there apparently being 2 kinds of NMR machines (ones where the magnetic field varies and ones where the radio wave frequency varies)?
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