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March 29, 2024, 09:37:06 pm

Author Topic: HSC Chemistry Question Thread  (Read 1040787 times)  Share 

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conic curve

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Re: Chemistry Question Thread
« Reply #555 on: August 03, 2016, 07:03:30 am »
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Assuming that this just means 0.005% of the solution was Mercury (I can't think of another way to answer this question), we can convert it straight to ppm. Parts per million are exactly that; if there were a million 'parts', how many would be Mercury? Well, 0.005% of them. You can plug this straight into a calculator, and find that the answer is 50ppm

Thanks Jake

Need help:

The haber process is an industrial process used to produce ammonia (NH3) from hydrogen and nitrogen gas in the air. The reaction shown below details this process

(it is attached)

Identify three factors that can increase the forward reaction rate and explain how this occurs (6 marks)
« Last Edit: August 03, 2016, 07:21:33 am by conic curve »

HighTide

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Re: Chemistry Question Thread
« Reply #556 on: August 03, 2016, 07:46:22 am »
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Thanks Jake

Need help:

The haber process is an industrial process used to produce ammonia (NH3) from hydrogen and nitrogen gas in the air. The reaction shown below details this process

(it is attached)

Identify three factors that can increase the forward reaction rate and explain how this occurs (6 marks)
You should have a go at these. Read the principles here: http://www.chemguide.co.uk/physical/equilibria/lechatelier.html
Hopefully you have a go at it first, then read the spoiler.
Spoiler
Increase reaction rate by:
1. Increase temperature--> increase kinetic energy = more collisions  at right orientation in a given time
2. Increase pressure --> Decrease volume= less room= more collisions at right orientation in a given time
3. Add a catalyst--> Decrease activation energy = increase proportion of particles which can surpass activation energy = faster 
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conic curve

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Re: Chemistry Question Thread
« Reply #557 on: August 03, 2016, 08:02:36 am »
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You should have a go at these. Read the principles here: http://www.chemguide.co.uk/physical/equilibria/lechatelier.html
Hopefully you have a go at it first, then read the spoiler.
Spoiler
Increase reaction rate by:
1. Increase temperature--> increase kinetic energy = more collisions  at right orientation in a given time
2. Increase pressure --> Decrease volume= less room= more collisions at right orientation in a given time
3. Add a catalyst--> Decrease activation energy = increase proportion of particles which can surpass activation energy = faster 

Le chaletier (if that;s how you spell it) is not part of the year 11 syllabus

In order to compare the reactivity of three different metals, Macy conducted an experiment where in which she placed the metals in different electrolyte solutions and recorded observations that occurred over time.

1. When metal A was placed in a 0.1M nitrate solution of Metal B there was no reaction observed
2. When metal B was placed in a 0.1M nitrate solution of Metal C a precipitate was seen to deposit on the surface of metal A
3. Metal B was then placed in a 0.1M nitrate solution of Metal C. Macy noticed that a precipitate was deposited on the surface of metal B

Arrange the metals in order of decreasing reactivity and explain the reasoning behind your arrangement (4 marks)

RuiAce

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Re: Chemistry Question Thread
« Reply #558 on: August 03, 2016, 08:45:14 am »
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Le chaletier (if that;s how you spell it) is not part of the year 11 syllabus

In order to compare the reactivity of three different metals, Macy conducted an experiment where in which she placed the metals in different electrolyte solutions and recorded observations that occurred over time.

1. When metal A was placed in a 0.1M nitrate solution of Metal B there was no reaction observed
2. When metal B was placed in a 0.1M nitrate solution of Metal C a precipitate was seen to deposit on the surface of metal A
3. Metal B was then placed in a 0.1M nitrate solution of Metal C. Macy noticed that a precipitate was deposited on the surface of metal B

Arrange the metals in order of decreasing reactivity and explain the reasoning behind your arrangement (4 marks)
If I remember correctly whilst you don't explicitly learn LCP in prelim the same rules are applied and you still need to know what he said.

Now, precipitation reactions in this context are intended to demonstrate metal displacement reactions.
The more reactive metal will displace the less reactive metal and become ionised.

E.g. Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
Zinc being more reactive than copper displaces it.

In #1, metal A did not displace metal B. So metal A is not as reactive as metal B
#2 I think you had a typo. You started talking about metal A out of nowhere
In #3, metal B did displace metal C. So metal B is more reactive than metal C

wesadora

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Re: Chemistry Question Thread
« Reply #559 on: August 03, 2016, 05:16:39 pm »
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For anyone that does Shipwrecks...I HAVE A QUESTION.
 I am aware rusting does not occur when pH is above 9, but why does a more acidic environment (lower pH) accelerate rusting?
The explanation given to me was that the reduction reaction O2 + 2H2O + 4e-<--> 4OH- exists in equilibrium - so, when H+ ions (from acid) are present it reacts with the OH- produced, and thus by LCP removing the product (OH) will cause the equilibrium to shift to the right and consequently accelerating rusting.

HOWEVER, my teacher and many other sources say that the reaction at the cathode is not reversible and i'm....confused.
cheers :)
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onepunchboy

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Re: Chemistry Question Thread
« Reply #560 on: August 03, 2016, 05:31:15 pm »
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Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?

wesadora

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Re: Chemistry Question Thread
« Reply #561 on: August 03, 2016, 05:48:35 pm »
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'The pH of 5 x 10-3M H2SO4 is'?

apparently it's 2. I'm getting 2.3 from -log10[H+], but apparently you're supposed to do -log10[2 x 5 x 10-3]. why? :/
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jakesilove

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Re: Chemistry Question Thread
« Reply #562 on: August 03, 2016, 06:07:47 pm »
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For anyone that does Shipwrecks...I HAVE A QUESTION.
 I am aware rusting does not occur when pH is above 9, but why does a more acidic environment (lower pH) accelerate rusting?
The explanation given to me was that the reduction reaction O2 + 2H2O + 4e-<--> 4OH- exists in equilibrium - so, when H+ ions (from acid) are present it reacts with the OH- produced, and thus by LCP removing the product (OH) will cause the equilibrium to shift to the right and consequently accelerating rusting.

HOWEVER, my teacher and many other sources say that the reaction at the cathode is not reversible and i'm....confused.
cheers :)

I'm sorry I can't answer your question in any more detail, because the explanation you were given is how I understood the process to work! Personally, I would just use that LCP explanation in an exam. Hopefully someone doing Shipwrecks now can help us out here!
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jakesilove

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Re: Chemistry Question Thread
« Reply #563 on: August 03, 2016, 06:11:29 pm »
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'The pH of 5 x 10-3M H2SO4 is'?

apparently it's 2. I'm getting 2.3 from -log10[H+], but apparently you're supposed to do -log10[2 x 5 x 10-3]. why? :/

Hey! Sulfuric acid is DIPROTIC. That means it donates 2 hydrogen ions (two protons), instead of just one (like HCl does). This means that, for one more of Sulfuric acid, two moles of hydrogen ions are donated. Therefore, you DOUBLE the concentration before plugging it into the formula. If it was triprotic (eg. Citric acid), you would TRIPLE the concentration! Hope that makes sense!
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jakesilove

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Re: Chemistry Question Thread
« Reply #564 on: August 03, 2016, 06:17:41 pm »
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(Image removed from quote.)

Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?

To be completely honest, whilst I can't really answer your question directly, I can answer the question by elimination. I think that's the best way to deal with this question anyway.

Firstly, by looking to the Polymer, we know that one monomer should be 5 carbon lengths long, and one should be 4 carbon lengths long. This immediately rules out A or D, leaving us with B or C.

Now, let's get of an OH from one monomer, and a H from another monomer. We know that a H2O is condensed out of the monomers somewhere, so this is a fair thing to do. According to B, this would result in one of two things: either BOTH monomers still have an OH group, or one doesn't an an OH group, and one has two. Looking to the Polymer, we can't see a single OH in sight. As such, B doesn't make any sense; we should see some left over.

Well damn. Actually, C has two OH in each monomer. Maybe that means two Water molecules are eliminated? I'm stumped, sorry I'm not great at visualising Chemistry stuff like this. Maybe someone can come help us out?
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wesadora

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Re: Chemistry Question Thread
« Reply #565 on: August 03, 2016, 06:27:58 pm »
+1

Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?

Pretty much, you NEED two OH groups to form the (C-O-C carboxylic bond AND have excess water (H2O) on both sides of the monomer.
By process of elimination, you first need two monomers with 4 and 5 carbons, and thus the answer is either B or C.
The reason the answer is C, is because each monomer given has two OH groups. The simplified structural formula has a [CH2CH3], which on the 5 carbon monomer in question C is the two left-most most carbons leftover, after the OH groups connect.

This is really hard to explain just using words and it would be so much easier for you to visualise if I could physically point things out with my finger but....yeah ._.
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RuiAce

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Re: Chemistry Question Thread
« Reply #566 on: August 03, 2016, 06:34:14 pm »
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(Image removed from quote.)

Can someone explain why the answer is c. Shouldnt the h2o be cancelled out in the middle? Where did the oxygen cone from?
The question's answered so I'm going to give a piece of advice here.

Try to work backwards. I always figured out which two MONOMERS would give the POLYMER in question
« Last Edit: August 03, 2016, 06:41:07 pm by RuiAce »

Loki98

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Re: Chemistry Question Thread
« Reply #567 on: August 04, 2016, 12:55:17 pm »
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How would i do this question?
The molar heat of combustion of 1-propanol is 2021kJmol-1. What mass of 1-propanol must be combusted to boil 300mL of water if the water has an initial temperature of 20°C and only absorbs 75% of the heat released?


RuiAce

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Re: Chemistry Question Thread
« Reply #568 on: August 04, 2016, 01:22:37 pm »
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How would i do this question?
The molar heat of combustion of 1-propanol is 2021kJmol-1. What mass of 1-propanol must be combusted to boil 300mL of water if the water has an initial temperature of 20°C and only absorbs 75% of the heat released?
Because the density of water is just 1 g mL-1, we know that the mass of water present is 300g.

Because only 75% of the heat is released, our enthalpy is

q = 75% * mc∆T = 0.75 * 300g * 4.18 J g-1 K-1 * 20 K = 18810 J

So q = 18.81 kJ

Using the fact that ∆H * n = -q

-2021 kJ mol-1 * n = -18.81 kJ
n = 0.00930727362... mol

Lastly, using the fact that m = n*MM

m = (0.00930727362...mol) * (60.094 g mol-1) = 0.5590320831... g

Then round to 2 s.f.

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Re: Chemistry Question Thread
« Reply #569 on: August 04, 2016, 01:42:01 pm »
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Thx for the help RuiAce :D