Otago HSFY chat - archive

[chococroissant]

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[TD="align: left"]because the pHka, then the concentration and therefore the moles of CH3COOH and CH3COO- would have to be equal

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so half of the CH3COOH would react with the NaOH to make CH3COO-


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[TD="align: left"]You would have half the CH3COOH left while the other half reacting to form CH3COO-

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[TD="align: left"]CH3COOH= CH3COO-

Answer is 0.4g.

because if it was 0.8g of NaOH, then it would react in a 1:1 mole ratio with CH3COOH. You wouldn't get any CH3COO- then. [/TD]
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[frootloop]

Well thats all nice and well to say that, you are missing the whole point of education then. Sad but true.

I don't mean to be rude, but you haven't done HSFY, so forgive me if I think your opinions about it carry little weight....

 

[miss_universe]

I don't suppose they do carry much weight. But HSFY is not a concept which is special to Otago.

 

[frootloop]

HSFY is not a concept which is special to Otago.

Huh? Other than Auckland and Otago, name me a place in Australasia which uses this kind of thing as the main selection tool for entry into the health science professional courses.

 

[League]

I don't suppose they do carry much weight. But HSFY is not a concept which is special to Otago.

I can respect that. OLY1 exists in the same realms, though students do attempt to understand the topics (except BIOSCI 106 ). Maybe to some extent, those who only have a graduate entry option may treat their entire degree with the same respect as HSFY.

It won't be appropriate for you nor froot neither I to say that you should or should not learn your course and just work on your marks. I think it's a matter of preference of the individual.

 

[frootloop]

I can respect that. OLY1 exists in the same realms, though students do attempt to understand the topics (except BIOSCI 106 ). Maybe to some extent, those who only have a graduate entry option may treat their entire degree with the same respect as HSFY.

It won't be appropriate for you nor froot neither I to say that you should or should not learn your course and just work on your marks. I think it's a matter of preference of the individual.

It appears I've been unclear. This all started with someone saying something along the lines of 'You don't need to learn anything which isn't on the lecture slides', which, in an HSFY context, I agree with whole-heartedly (The lecture slides have a crapton of information on them as is, and you need to know upwards of 90% of it, in order to get into your chosen course, so naturally the lecture slide material is your first priority).
Sure, in like any other course, going and learning stuff that may or may not be directly relevant is a good thing. But in HSFY, most students simply don't have the time/capacity to learn all the stuff they 'need' as well as all the extra interesting tidbits you don't 'need'. So, in that respect, HSFY isn't really about your 'learning experience' and whatnot.
That does *not* mean that you shouldn't try to understand the concepts and whatnot in HSFY, because how else are you going to do well in an exam? Good luck explaining something you don't understand

 

[Sam]

It appears I've been unclear. This all started with someone saying something along the lines of 'You don't need to learn anything which isn't on the lecture slides', which, in an HSFY context, I agree with whole-heartedly (The lecture slides have a crapton of information on them as is, and you need to know upwards of 90% of it, in order to get into your chosen course, so naturally the lecture slide material is your first priority).
Sure, in like any other course, going and learning stuff that may or may not be directly relevant is a good thing. But in HSFY, most students simply don't have the time/capacity to learn all the stuff they 'need' as well as all the extra interesting tidbits you don't 'need'. So, in that respect, HSFY isn't really about your 'learning experience' and whatnot.
That does *not* mean that you shouldn't try to understand the concepts and whatnot in HSFY, because how else are you going to do well in an exam? Good luck explaining something you don't understand

The perfect testament to this is the high number of people who can get 90%+ on PHSI191 and still know nothing about what any of it means, haha. Cheat sheets win!

 

[frootloop]

[offtopic]

The perfect testament to this is the high number of people who can get 90%+ on PHSI191 and still know nothing about what any of it means, haha. Cheat sheets win!

I'll give you that, I understood approx. 5% of PHSI191, and yet got 89% in it... Slight discrepancy between understanding and marks there [/offtopic]

 

[League]

'You don't need to learn anything which isn't on the lecture slides', which, in an HSFY context, I agree with whole-heartedly (The lecture slides have a crapton of information on them as is, and you need to know upwards of 90% of it, in order to get into your chosen course, so naturally the lecture slide material is your first priority).

That goes without saying.

Good luck explaining something you don't understand

Refer to BIOSCI106. No one understands it in their first year. N-5-N-10-Methylenetetrahydrofolate. Seriously?

 

[juna]

Which one of the following indicators would be most suitable for the titration of
0.10 mol L
-1
HBr with 0.10 mol L
-1
KOH?
A) methyl orange (pKa = 3.4)
B) alizarin yellow (pKa = 11.2)
C) thymol blue (pKa = 1.7)
D) bromothymol blue (pKa = 7.1)

Wouldn't methyl orange be the right answer? But apparently no- the right answer is D

 

[Cathay]

You want the indicator to change colours when the titration reaches its end point. An indicator changes colours when pH is equal to its pKa.

HBr is a strong acid, and KOH is a strong base, so their conjugates are of negligible acidity/basicity, meaning that at the end point, the solution would be neutral. This means you'll want an indicator with pKa as close to 7 as possible, so in this case that'll be bromothymol blue (pKa 7.1).

 

[Profiterole]

HBr =Strong acid
KOH=Strong base

Hence when added together=neutralisation reaction-hence pH would be around 7

 

[juna]

omgosh, that's such a good point! I knew that- it is rather interesting though, that in my year 13 chem book, it says that the best indicator to use would be methyl orange......

 

[Cathay]

it is rather interesting though, that in my year 13 chem book, it says that the best indicator to use would be methyl orange......

For this EXACT scenario?

Personally I'd take onboard the principles to determine the best indicator, and be very cautious of blanket statements like that.

 

[juna]

A suggested mechanism for the decomposition of ozone O3(g) is
step 1 (fast equilibrium) O3(g) ⇄ O2(g) + O(g)
step 2 (slow) O3(g) + O(g) → 2O2(g)
The rate law for this mechanism is:
(A) rate = k[O2]2
(B) rate = k[O3]2/[O2]
(C) rate = k[O3]
(D) rate = k[O2]

I got k=(O2)/(O3)^2, but the right answer is B- any ideas as to why?

Also, the rate law for the order of 2- if the concentration of (NO2)^2 is tripled, then the rate will increase by 9 times? So for rate law order of 2, the rate increases by power of 2? Because in our notes, it says that rate increases by *4?

 

[Profiterole]

A suggested mechanism for the decomposition of ozone O3(g) is
step 1 (fast equilibrium) O3(g) ⇄ O2(g) + O(g)
step 2 (slow) O3(g) + O(g) → 2O2(g)
The rate law for this mechanism is:
(A) rate = k[O2]2
(B) rate = k[O3]2/[O2]
(C) rate = k[O3]
(D) rate = k[O2]

I got k=(O2)/(O3)^2, but the right answer is B- any ideas as to why?

Also, the rate law for the order of 2- if the concentration of (NO2)^2 is tripled, then the rate will increase by 9 times? So for rate law order of 2, the rate increases by power of 2? Because in our notes, it says that rate increases by *4?

How to solve these types of questions:
1. write rate law for slow step
2. write k expression for fast step.
3. determine intermediate and make it the subject
4. sub 3. into 1.

so for this:
rate law= k[O][O3]
k=[O2][O]/[O3]
if we rearrange:
k[O3]/[O2] =[O]

hence if we sub:
r=(kk[O3][O3])/[O2]

hence the rate law is k[O3]^2/[O2]

 

[frootloop]

omgosh, that's such a good point! I knew that- it is rather interesting though, that in my year 13 chem book, it says that the best indicator to use would be methyl orange......

If only this was year 13 chemistry
But seriously, your year 13 books are unlikely to be all that helpful in HSFY, the stuff is presented differently, and the lecturers often have different opinions on stuff than your year 13 books, and, especially in chemistry, your year 13 (particularly NCEA) books are likely to be a bit over-simplified.

 

[noblelady777]

Ok, this is a question from the 2009 physics paper.
A uniform cube, each edge of which is 1m long, of solid polystyrene is sitting on a flat floor. What minimum height will you need to lift one edge of the base to get the cube to tip over (in m)?
A. 0.3
B 0.7
C 1.0
D 1.4
E 2.0

Is this about the centre of mass or something related to that? I'm guessing that it's something about lifting it to a height greater than the centre of mass or something? Can someone please explain this to me? Thanks.

 

[Profiterole]

Ok, this is a question from the 2009 physics paper.
A uniform cube, each edge of which is 1m long, of solid polystyrene is sitting on a flat floor. What minimum height will you need to lift one edge of the base to get the cube to tip over (in m)?
A. 0.3
B 0.7
C 1.0
D 1.4
E 2.0

Is this about the centre of mass or something related to that? I'm guessing that it's something about lifting it to a height greater than the centre of mass or something? Can someone please explain this to me? Thanks.

if you draw a diagram, the minimum angle required for the box to tip over is 45 degrees,
and since you know the length of 1 side (1m), you can use Cos 45(degrees)=height needed to lift edge/length of hypotenuse (which is the length of 1 side=1m)
therefore: answer= cos45=~~0.7m (option B)

 

[Kception]

Another way to look at it (basically the same idea as jono) is by pythagorus. When you turn the box around to the point where it is about to fall over it, in essence should look like a diamond. If you draw a line vertically through this diamond and cut it in half - you will have the minimum height required to tip the box over.

Mathematically speaking ((1^2 + 1^2)^0.5) / 2

Hope it helps