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Author Topic: HSC Chemistry Question Thread  (Read 1040642 times)  Share 

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RuiAce

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Re: Chemistry Question Thread
« Reply #465 on: July 22, 2016, 10:13:30 pm »
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Yes I know that but when it comes to answering them, I get stuck

Could you at least give me hints on them?

What does it mean by like 3-methy-2-difluro-1-ene (or something similar to that)?
Should've been more specific about methyl groups and that stuff, yeah.

Firstly, let's clear up a nomenclature: 1-propene and prop-1-ene mean the same thing. The convention is to use the latter (very retarded change they implemented a few years ago, but we use it).

Let's fix your question into something more manageable.
3-methy-1,2-difluro-hept-1-ene

Getting a CFC (or rather HCFC) with a methyl group present is pretty darn rare but we can still do it.

We start by identifying the prefix: hept. This means there are 7 carbons

Note that it's hept-1-ene. Hence, there is a double bond coming off carbon 1

C=C-C-C-C-C-C

We are told that a methyl group comes off carbon number 3. We can add that in.

        C
C=C-C-C-C-C-C

We are told that a fluorine comes off carbon 1, AND carbon 2. We can draw them in wherever appropriate

Cl     C
C=C-C-C-C-C-C
    Cl

Then just chuck all the appropriate hydrogens in there last.

conic curve

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Re: Chemistry Question Thread
« Reply #466 on: July 22, 2016, 10:20:02 pm »
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Should've been more specific about methyl groups and that stuff, yeah.

Firstly, let's clear up a nomenclature: 1-propene and prop-1-ene mean the same thing. The convention is to use the latter (very retarded change they implemented a few years ago, but we use it).

Let's fix your question into something more manageable.
3-methy-1,2-difluro-hept-1-ene

Getting a CFC (or rather HCFC) with a methyl group present is pretty darn rare but we can still do it.

We start by identifying the prefix: hept. This means there are 7 carbons

Note that it's hept-1-ene. Hence, there is a double bond coming off carbon 1

C=C-C-C-C-C-C

We are told that a methyl group comes off carbon number 3. We can add that in.

        C
C=C-C-C-C-C-C

We are told that a fluorine comes off carbon 1, AND carbon 2. We can draw them in wherever appropriate

Cl     C
C=C-C-C-C-C-C
    Cl

Then just chuck all the appropriate hydrogens in there last.

Thanks so much  :D

anotherworld2b

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Re: Chemistry Question Thread
« Reply #467 on: July 23, 2016, 07:07:15 pm »
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I've tried to answer q6 b and c but i got them wrong and i dont know why

RuiAce

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Re: Chemistry Question Thread
« Reply #468 on: July 23, 2016, 07:35:14 pm »
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I've tried to answer q6 b and c but i got them wrong and i dont know why
Your answer to b) is unsimplified. The correct answer should be that without the 2's
H+ + OH- -> H2O(l)

You are safe to cancel out the numbers because this is a net IONIC form we're talking about.
_________________________________

Your working out for c) is wrong in the products. Here are proposed amendments.

1. Seperate CH3COOH(aq) into H+ and CH3COO- for better clarity in the reactants first
2. You wrote acetic acid when it's not there. It's only the acetate ion. Thus, replace the mistaken acetic acid with CH3COO-.

You should get MgO(s) + 2 H+ -> Mg2+ + H2Ol
« Last Edit: July 23, 2016, 07:38:14 pm by RuiAce »

anotherworld2b

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Re: Chemistry Question Thread
« Reply #469 on: July 23, 2016, 11:09:18 pm »
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I got the answer for q6 b but the answer for q6 is different.

I was also wondering how would you do q 8? I dont understand the question itself.

 
Your answer to b) is unsimplified. The correct answer should be that without the 2's
H+ + OH- -> H2O(l)

You are safe to cancel out the numbers because this is a net IONIC form we're talking about.
_________________________________

Your working out for c) is wrong in the products. Here are proposed amendments.

1. Seperate CH3COOH(aq) into H+ and CH3COO- for better clarity in the reactants first
2. You wrote acetic acid when it's not there. It's only the acetate ion. Thus, replace the mistaken acetic acid with CH3COO-.

You should get MgO(s) + 2 H+ -> Mg2+ + H2Ol

RuiAce

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Re: Chemistry Question Thread
« Reply #470 on: July 23, 2016, 11:19:04 pm »
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I got the answer for q6 b but the answer for q6 is different.

I was also wondering how would you do q 8? I dont understand the question itself.
CH3COOH(aq) is acetic acid. I disagree with the answer because the fact it is acetic acid means that it is definitely dissociated.

Unless the question is being pedantic. Their answer isn't "completely" unjustified because some unsplit CH3COOH is in there as well. So I can see where they're coming from. But the question should've made that clearer - I don't like these ambiguous questions.
___________________

Q8 is combining equations. You have to do some maths here.

a) Equation 2 has just one CO in the reactants. But equation 1 has two COs in the products
So we have to multiply equation 2 by two to merge the equations:

2 CO(g) + 2 FeO(s) -> 2 Fe(l) + 2 CO2(g)

Substitute equation 1 in there to get:

C(s) + CO2(g) + 2 FeO(s) -> 2 Fe(l) + 2 CO2(g)

Cancel out the duplicated carbon dioxide:

C(s) + 2 FeO(s) -> 2 Fe(l) + CO2(g)

It's sort of like simultaneous equations: Perform (1) + 2*(2)

I'll let you have a try at the other two first. If you're still stuck, however (especially with 8c - that one is trickier), just post again
« Last Edit: July 23, 2016, 11:21:09 pm by RuiAce »

Aliceyyy98

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Re: Chemistry Question Thread
« Reply #471 on: July 24, 2016, 12:44:05 am »
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Hello!

I have a quick question about how coordinate covalent bonds work? Could someone please explain to me, perhaps using examples that would be great :) thanks in advance!

Cheers

RuiAce

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Re: Chemistry Question Thread
« Reply #472 on: July 24, 2016, 01:09:36 am »
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Hello!

I have a quick question about how coordinate covalent bonds work? Could someone please explain to me, perhaps using examples that would be great :) thanks in advance!

Cheers
The basic idea is that whereas in your normal covalent bonds, two atoms share an electron with each other, in a coordinate covalent bond one atom shares BOTH of the electrons with the other.

E.g. The first one you must know of is ozone. In O2, there's a double bond between the two oxygen molecules (there are two covalent bonds, so both atoms share two electrons with each other). However in O3, that third oxygen gets TWO electrons from one of the other oxygen atoms.

It's more easily demonstrated with a Lewis dot structure diagram.

Aliceyyy98

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Re: Chemistry Question Thread
« Reply #473 on: July 24, 2016, 01:17:19 am »
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Hi thank you for you answer, just to add on, how would you know if a molecule has a covalent coordinate bond or not without a diagram or anything :(


The basic idea is that whereas in your normal covalent bonds, two atoms share an electron with each other, in a coordinate covalent bond one atom shares BOTH of the electrons with the other.

E.g. The first one you must know of is ozone. In O2, there's a double bond between the two oxygen molecules (there are two covalent bonds, so both atoms share two electrons with each other). However in O3, that third oxygen gets TWO electrons from one of the other oxygen atoms.

It's more easily demonstrated with a Lewis dot structure diagram.

RuiAce

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Re: Chemistry Question Thread
« Reply #474 on: July 24, 2016, 01:22:35 am »
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Hi thank you for you answer, just to add on, how would you know if a molecule has a covalent coordinate bond or not without a diagram or anything :(
A pro-tip is to try drawing them all with covalent bonds first. If you can't successfully FORCE them to have a full outer shell (8 electrons) using only normal covalent bonds, there's probably a coordinate covalent bond somewhere.

However, the following three are ones that you just need to know that there is a coordinate covalent bond:
- Ozone O3: From one oxygen to the third oxygen
- Hydronium ion H3O+: From the oxygen to the third hydrogen
- Ammonium ion NH4+: From the nitrogen to the fourth hydrogen
« Last Edit: July 24, 2016, 01:26:00 am by RuiAce »

angiezhang9

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Re: Chemistry Question Thread
« Reply #475 on: July 24, 2016, 10:00:56 am »
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Hey :)

I don't understand this multiple choice question from the 2014 Catholic Trials.

An experiment was performed to determine the volume of carbon dioxide gas released during the fermentation of a glucose solution. The reaction took place over several days in a flask containing yeast that was immersed in a water bath at 28 degrees and stoppered with cotton wool. Which of the following factors would most significantly compromise the validity of this experiment?

A) The evaporation of water
B) The temperature of the water bath
C) Measurement error associated with determining the mass lost
D) Lack of oxygen due to the presence of the cotton wool plug

(The answer is A)

Thanks heaps! :)

RuiAce

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Re: Chemistry Question Thread
« Reply #476 on: July 24, 2016, 10:13:55 am »
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Hey :)

I don't understand this multiple choice question from the 2014 Catholic Trials.

An experiment was performed to determine the volume of carbon dioxide gas released during the fermentation of a glucose solution. The reaction took place over several days in a flask containing yeast that was immersed in a water bath at 28 degrees and stoppered with cotton wool. Which of the following factors would most significantly compromise the validity of this experiment?

A) The evaporation of water
B) The temperature of the water bath
C) Measurement error associated with determining the mass lost
D) Lack of oxygen due to the presence of the cotton wool plug

(The answer is A)

Thanks heaps! :)
One again, CSSA is giving very dodgy questions.

Consider the answers:
D) is obviously wrong - The whole process of fermentation is dependent on anaerobic conditions. By removing oxygen, we are promoting the validity of the experiment.
C) is asking for the wrong thing - A measurement error affects the accuracy of the experiment, not the validity.
B) can be a bit more tricky. Obviously the best conditions for fermentation is at 37°C, but the water bath is at 28°C. This does not promote the fermentation of glucose and is also a candidate for why the experiment is invalid.

However, we can't just use this our basis. To consider the validity, look at the aim.

What we want to determine, is the volume of carbon dioxide released IN fermentation. The fermentation itself? It just has to proceed! Note that at some temperature like 28°C the reaction will still proceed, just that it will be hindered by a slow process.

Therefore B is out. But why is the answer A?

This is my proposal. When you have an aim, the idea is that you keep all other factors fixed and leave behind only one variable. As the water evaporates (especially given how slow fermentation is), the impact of the water bath is altered. Potentially the temperature of the water bath changes, but not only that, the entire glucose solution is no longer under the same, consistent 28°C anymore! It could be under the 10°C or 40°C air! Because we don't fix all our other factors, the experiment's validity is more heavily impacted here. I'll let someone else give a better answer.
« Last Edit: July 24, 2016, 11:41:00 am by RuiAce »

Happy Physics Land

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Re: Chemistry Question Thread
« Reply #477 on: July 24, 2016, 12:02:14 pm »
+2
Hey :)

I don't understand this multiple choice question from the 2014 Catholic Trials.

An experiment was performed to determine the volume of carbon dioxide gas released during the fermentation of a glucose solution. The reaction took place over several days in a flask containing yeast that was immersed in a water bath at 28 degrees and stoppered with cotton wool. Which of the following factors would most significantly compromise the validity of this experiment?

A) The evaporation of water
B) The temperature of the water bath
C) Measurement error associated with determining the mass lost
D) Lack of oxygen due to the presence of the cotton wool plug

(The answer is A)

Thanks heaps! :)

Hey Angie!

Rui has asked me to supplement his answer so please have a read what I wrote below :)

A is the correct option simply because all we are trying to measure is the volume of CO2 produced. If water evaporates (especially since we are carrying out the experiment under 28 degrees here, so a large chance that some water would have evaporated) then we obtain a measurement that includes a combined mass loss of CO2 and H2O. Validity means "are we measuring what we want to measure" and this is clearly not the case if water has also evaporated. Keep in mind, WE ONLY WANT TO MEASURE THE MASS OF CARBON DIOXIDE.

Explaining this one step further. When you conducted this experiment what you would have done is to measure the mass of fermentation mixture before fermentation has taken place, and the mass of fermentation mixture after fermentation has taken place. The loss in mass after fermentation you would assume to be CO2 (You would probably have carried out the experiment with a limewater flask connected to the fermentation mixture as well). But keep in mind that even at room temperature, water evaporation still occurs, meaning that whilst CO2 is being evaporated, water is too! So when you measure the lost weight, thats a combined weight of evaporated CO2 and WATER. So evidently we are not measuring JUST the mass of CO2 and therefore the experiment becomes invalid.

So, how do we improve validity? We can do one of the two things:

1. Keep a carton of water under the same temperature as your fermentation mixture. Make sure the volume of the water is the same as the volume of your fermentation mixture. Measure the loss in the amount of water at the same time when you measure the loss in mass of the fermentation mixture, subtract this from the measured weight loss of the fermentation mixture. Why are we doing this? Because we want to see how much water has evaporated under the same conditions and in an equal amount of time as fermentation takes place. This way we can measure only the mass of CO2 produced.

2. This is an easier method. Simply connect your fermentation mixture to a conical flask filled with limewater. When CO2 is released it reacts with Ca(OH)2 (lime) to form a milky liquid CaCO3. By measuring the increase in the weight of the limewater and divide this weight by the molar mass of CO2 we can work out how many moles of CO2 has been produced and therefore calculate the volume of CO2 produced. Make sure when you do this experiment that everything is enclosed to prevent CO2 from escaping into the air. Why do we not worry about water evaporation in this case? Because it doesnt affect anything! It does not react or enter the limewater and therefore all we are measuring here is the amount of CO2 produced!

A bit long (quite long tbh) for a multiple choice answer, but I thought I would just make everything clearer for you. :)

Best Regards
Happy Physics Land
« Last Edit: July 24, 2016, 12:03:51 pm by Happy Physics Land »
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angiezhang9

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Re: Chemistry Question Thread
« Reply #478 on: July 24, 2016, 02:08:32 pm »
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Hey Angie!

Rui has asked me to supplement his answer so please have a read what I wrote below :)

A is the correct option simply because all we are trying to measure is the volume of CO2 produced. If water evaporates (especially since we are carrying out the experiment under 28 degrees here, so a large chance that some water would have evaporated) then we obtain a measurement that includes a combined mass loss of CO2 and H2O. Validity means "are we measuring what we want to measure" and this is clearly not the case if water has also evaporated. Keep in mind, WE ONLY WANT TO MEASURE THE MASS OF CARBON DIOXIDE.

Explaining this one step further. When you conducted this experiment what you would have done is to measure the mass of fermentation mixture before fermentation has taken place, and the mass of fermentation mixture after fermentation has taken place. The loss in mass after fermentation you would assume to be CO2 (You would probably have carried out the experiment with a limewater flask connected to the fermentation mixture as well). But keep in mind that even at room temperature, water evaporation still occurs, meaning that whilst CO2 is being evaporated, water is too! So when you measure the lost weight, thats a combined weight of evaporated CO2 and WATER. So evidently we are not measuring JUST the mass of CO2 and therefore the experiment becomes invalid.

So, how do we improve validity? We can do one of the two things:

1. Keep a carton of water under the same temperature as your fermentation mixture. Make sure the volume of the water is the same as the volume of your fermentation mixture. Measure the loss in the amount of water at the same time when you measure the loss in mass of the fermentation mixture, subtract this from the measured weight loss of the fermentation mixture. Why are we doing this? Because we want to see how much water has evaporated under the same conditions and in an equal amount of time as fermentation takes place. This way we can measure only the mass of CO2 produced.

2. This is an easier method. Simply connect your fermentation mixture to a conical flask filled with limewater. When CO2 is released it reacts with Ca(OH)2 (lime) to form a milky liquid CaCO3. By measuring the increase in the weight of the limewater and divide this weight by the molar mass of CO2 we can work out how many moles of CO2 has been produced and therefore calculate the volume of CO2 produced. Make sure when you do this experiment that everything is enclosed to prevent CO2 from escaping into the air. Why do we not worry about water evaporation in this case? Because it doesnt affect anything! It does not react or enter the limewater and therefore all we are measuring here is the amount of CO2 produced!

A bit long (quite long tbh) for a multiple choice answer, but I thought I would just make everything clearer for you. :)

Best Regards
Happy Physics Land


Thanks so much for clearing this up. It makes sense now :) RuiAce and Happy Physics Land

Happy Physics Land

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Re: Chemistry Question Thread
« Reply #479 on: July 24, 2016, 03:02:33 pm »
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Thanks so much for clearing this up. It makes sense now :) RuiAce and Happy Physics Land

No worries angie, what we do here the best is to clear things up! :)
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