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Author Topic: HSC Chemistry Question Thread  (Read 1040853 times)  Share 

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RuiAce

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Re: Chemistry Question Thread
« Reply #390 on: July 12, 2016, 08:58:49 pm »
0
Hey,
Need Help for this q.
Its worth 5 marks
A sample of lemon juice is to be analysed in the laboratory. A student took 25.00 mL of the juice
and diluted it to 250.00 mL. Exactly 25.00 mL of the diluted lemon juice is titrated with standard
0.1045 mol L-1
 sodium hydroxide solution using phenolphthalein as the indicator. An average titre
 of 24.05 mL of sodium hydroxide was required
Assuming that the lemon juice contained only citric acid (molar mass = 192.1 g/mol), calculate
the concentration in mol L-1 of citric acid in the undiluted lemon juice.

Thank You!
We always start with an equation. Even if you forget the formula for citric acid (2-hydroxypropane-1,2,3-tricarboxylic-acid) you should be well aware of how it is triprotic.

The reaction goes to completion as we have a strong base.
C3H4OH(COOH)3(aq) + 3 NaOH(aq) -> C3H4OH(COO)3Na(aq) + 3 H2O(l)

We may determine the (average) moles of NaOH solution.
n(NaOH)
= CV
= 0.1045 mol L-1 * 0.02500 L
= 2.6125 * 10-3 mol

1 mol of NaOH reacts with 1/3 mol of citric acid.

n(C3H4OH(COOH)3) = n(NaOH)/3 = 8.7083333333... * 10-4 mol

This gives us the moles of citric acid used in the reaction. We know that 25 mL of citric acid was used in the reaction. Hence the concentration of the citric acid is

C2
= n/V
=  (8.7083333333... * 10-4 mol) / (0.025 L)
= 0.03483333... mol L-1

Notice that this is the concentration of citric acid AFTER dilution. To obtain the concentration BEFORE dilution, we use the concentrations formula C1V1 = C2V2

C1 * 25.00 mL = 0.03483333... mol L-1 * 250.00 mL

C1 = 0.34833333... mol L-1

= 3.483 * 10-1 mol L-1 (4 sig. fig.)


(There may be a slight calculation error somewhere but the method should be good.)

anotherworld2b

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Re: Chemistry Question Thread
« Reply #391 on: July 12, 2016, 11:28:45 pm »
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How would you do these questions?
« Last Edit: July 13, 2016, 01:09:35 am by anotherworld2b »

kristine.faelnar

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Re: Chemistry Question Thread
« Reply #392 on: July 12, 2016, 11:32:00 pm »
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Hi guys!!
So I was making my notes for my option topic (Forensics) and I came across this syllabus point and I have no bloody idea what I'm doing. I don't really understand it and it hasn't been gone through much in class.

"Perform a first-hand investigation and gather first-hand information to identify the range of solvents that may be used for chromatography and suggest mixtures that may be separated and identified by the use of these solvents."

Please help!!
Kristine x0x0

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Re: Chemistry Question Thread
« Reply #393 on: July 12, 2016, 11:47:12 pm »
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Hi guys!!
So I was making my notes for my option topic (Forensics) and I came across this syllabus point and I have no bloody idea what I'm doing. I don't really understand it and it hasn't been gone through much in class.

"Perform a first-hand investigation and gather first-hand information to identify the range of solvents that may be used for chromatography and suggest mixtures that may be separated and identified by the use of these solvents."

Please help!!

I believe that is a practical

kristine.faelnar

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Re: Chemistry Question Thread
« Reply #394 on: July 12, 2016, 11:53:10 pm »
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I have never done it before, so I have no idea what to do for that :(
I tried finding notes online as well and I couldn't find anything
Kristine x0x0

RuiAce

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Re: Chemistry Question Thread
« Reply #395 on: July 13, 2016, 07:49:52 am »
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How would you do these questions?
These were...a bit awkward to read... sideways and upside-down photo

Q4: Not really important at all. One thing to note about the ammonium ion is just the coordinate covalent bond from the nitrogen to the fourth hydrogen ion. Yes, the coordinate bond is to a hydrogen ion (aka. a proton), not a hydrogen atom. This hydrogen ion basically gives that extra +1 charge.

Whereas for hydroxide, the hydroxide ion is formed when water undergoes self ionisation. What happens is that one of the hydrogens in H2O essentially "comes off", but it doesn't take its electron with it. The hydrogen and oxygen were initially covalently bonded, but the hydrogen comes off and leaves its electron with oxygen. This causes the -1 charge.

I have no idea how this relates to the fertiliser practical. Maybe the rest of the textbook could help here.

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Re: Chemistry Question Thread
« Reply #396 on: July 13, 2016, 08:49:59 am »
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I have never done it before, so I have no idea what to do for that :(
I tried finding notes online as well and I couldn't find anything

Just wait until your teacher does it in class and then you compose the practical

RuiAce

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Re: Chemistry Question Thread
« Reply #397 on: July 13, 2016, 10:22:34 am »
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I have never done it before, so I have no idea what to do for that :(
I tried finding notes online as well and I couldn't find anything
If you haven't done a practical yet purely because you're not that far into the syllabus yet, you cannot be asked it in the exam. If you do, take it up to the teacher or faculty coordinator.

(Unfortunately I have no clue about forensic chemistry.)

RuiAce

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Re: Chemistry Question Thread
« Reply #398 on: July 13, 2016, 10:42:13 am »
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Did you do Industrial chemistry?
Yes

Posts like these aren't necessary. Please delete it.

(You're free to private message me about this stuff all you like though :) )

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Re: Chemistry Question Thread
« Reply #399 on: July 13, 2016, 11:10:52 am »
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Yes

Posts like these aren't necessary. Please delete it.

(You're free to private message me about this stuff all you like though :) )

Oh I was just asking ahaha  :D

Anyways if anyone here is generous enough could you please explain to me spetator ions, ionic equations, half equations, OILRIG, etc all with examples (if it's too much do 1-2 at a time). I seriously suck at those and I want to continue chemistry in year 12 (I heard it comes up in year 12 BTW and that's the main reason why I'm worried)

Thanks so much  :)

HighTide

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Re: Chemistry Question Thread
« Reply #400 on: July 13, 2016, 11:24:47 am »
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Oh I was just asking ahaha  :D

Anyways if anyone here is generous enough could you please explain to me spetator ions, ionic equations, half equations, OILRIG, etc all with examples (if it's too much do 1-2 at a time). I seriously suck at those and I want to continue chemistry in year 12 (I heard it comes up in year 12 BTW and that's the main reason why I'm worried)

Thanks so much  :)
Alright, although this is a question, you should probably have a go at these first. A simple google search, or using a textbook would give you the answers.

Spectator ions don't participate in the reaction. So if have like a compound like sodium sulfate and barium nitrate, the precipitate would be barium sulphate, the sodium ions and nitrate ions are spectator ions as they are not involved in the main reaction, but they are in the solution.
Ionic equations are kind of self explanatory. If you have a specific question about it, I can elaborate.
Half equations: One for oxidation and another for reduction since both occur spontaneously.
OILRIG is just oxidation is losing electrons. Reduction is gaining electrons.

Alright examples you don't really need at this point. But since you're keen on learning this (it does indeed come up in year 12), you should check out videos from Khan Academy, or other youtube sites because they introduce you to the basics. These basics will suffice for year 11 and year 12. If you have any specific question after that, I'd be happy to answer it.
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Re: Chemistry Question Thread
« Reply #401 on: July 13, 2016, 02:12:49 pm »
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Alright, although this is a question, you should probably have a go at these first. A simple google search, or using a textbook would give you the answers.

Spectator ions don't participate in the reaction. So if have like a compound like sodium sulfate and barium nitrate, the precipitate would be barium sulphate, the sodium ions and nitrate ions are spectator ions as they are not involved in the main reaction, but they are in the solution.
Ionic equations are kind of self explanatory. If you have a specific question about it, I can elaborate.
Half equations: One for oxidation and another for reduction since both occur spontaneously.
OILRIG is just oxidation is losing electrons. Reduction is gaining electrons.

Alright examples you don't really need at this point. But since you're keen on learning this (it does indeed come up in year 12), you should check out videos from Khan Academy, or other youtube sites because they introduce you to the basics. These basics will suffice for year 11 and year 12. If you have any specific question after that, I'd be happy to answer it.

How come spectator ions don't take place in the reaction?

Yeah I know OILRIG, I just tend to get confused with examples. Usually does it require the periodic table?


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Re: Chemistry Question Thread
« Reply #402 on: July 13, 2016, 03:19:10 pm »
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Yeah I know OILRIG, I just tend to get confused with examples. Usually does it require the periodic table?

No it does not, or not to my knowledge, require the periodic table. Basically when looking at an equation you are trying to determine the characteristics of the reactants ad products in determining which has lost (oxidation) and gained (reduction) electrons....HINT: look at the change in states/charges

As a rule of thumb, most of the time that is...the metal e.g. Mg will be oxidised

Hope this is somewhat helpful

Alex
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RuiAce

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Re: Chemistry Question Thread
« Reply #403 on: July 13, 2016, 03:23:19 pm »
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Regarding spectator ions:

Spectator ions are present because they're a part of the species. However, after the reaction is complete, nothing has happened to them. For example:

CuSO4(aq) + Zn(s) -> Cu(s) + ZnSO4(aq)

In the reaction, copper (II) ions turned to copper solid
Zinc solid turned to zinc ions

The sulfate ion did nothing. The species present are copper(II) sulfate and zinc sulfate, but the sulfate itself never got altered. It remained as SO42-.

We call this a spectator ion due to the fact that it is there just to ensure the reaction happens. It doesn't participate in the reaction.

It isn't a condition either because it doesn't HAVE to be there for the reaction to happen.
____________

As to why they don't take part in the reaction? Note that the definition of spectator ion is something THAT does not take part. It's just there because it's there in the mixture.
« Last Edit: July 13, 2016, 03:25:41 pm by RuiAce »

wesadora

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Re: Chemistry Question Thread
« Reply #404 on: July 13, 2016, 08:57:22 pm »
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I'm having trouble with a following two questions in one of my school's past trial exams:

Spoiler
"A student used this equipment [experimental setup for alkanol's heat of combustion prac: i.e. retort stand, clamp, spirit burner, water, thermometer, conical flask] to heat 250g of water. The mass of the spirit burner, which contained ethanol, decreased from 296.52g to 295.95g.

Given the heat of combustion of ethanol is 29.7kJ/g, calculate the maximum possible change in the temperature of the water."


--> this question I'm confused about the kJ/g...seeing ∆H = Q/n for kJ/mol, i just thought of doing ∆H=Q/m as it was in kJ/g (kilojule per gram), and solving for Q, as: 29.7 = Q/0.57 (0.57 is mass of ethanol burned)...yeah i got lost :(

Spoiler
"Ozone concentrations are measured in Dobson units (DU). DU are the standard way to express ozone concentration in the stratosphere. A concentration of one DU means there would be 2.7x1020 ozone molecules in a layer of air that was one square metre in area and 0.01mm thick.

A baseline value of 220 DU is chosen as the starting point for an ozone hole in the stratosphere since total ozone values of less than 200 DU were not found in historic obsevations over Antarctica over 1979.
Which of the following concentrations, in moles per cubic metre (molm'3), is most nearly equivalent to 220 DU?"

A) 0.05 molm-3
B) 10 molm-3
C) 5000 molm-3
D) 10 000 molm-3

--> this one i am just lost in.

« Last Edit: July 13, 2016, 09:01:33 pm by wesadora »
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