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HSC 2012-2015 Chemistry Marathon (archive) (1 Viewer)

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Crisium

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re: HSC Chemistry Marathon Archive

Next question:

Distinguish between the end point and the equivalence point in an acid-base titration.
 

BlueGas

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Next question:

Distinguish between the end point and the equivalence point in an acid-base titration.
That's easy, does the HSC even ask those questions?

Endpoint: The point at which the indicator in the solution changes colour
Equivalence point: The point at which the reaction is complete
 

Crisium

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That's easy, does the HSC even ask those questions?

Endpoint: The point at which the indicator in the solution changes colour
Equivalence point: The point at which the reaction is complete
Yeah sometimes in multiple choice for sympathy marks
 

Drsoccerball

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re: HSC Chemistry Marathon Archive

Independent Trial Paper 2011

A student measures the pH of a can of soft drink, immediately after it is opened, then at 15 minute intervals as shown in the table.
Time 0 , 15 , 30 , 45 , 60 , 75 , 90 , 105 , 120
pH 4.5 , 5.9 , 6.5 , 6.7 , 7.1 , 7.3 , 7.4 , 7.4 , 7.5


Compare the initial hydrogen ion concentration with that after 2 hours (2 MARKS)

Explain the changing pH of the soft drink in terms of Le Chatelier’s principle, and the properties of carbon dioxide. (3 MARKS)
1. At the beginning the hydrogen ion concentration is greater than at the end as there is less H2CO3 in solution



2. Once opened the can looses Carbon Dioxide such that it is equal to the atmospheric carbon dioxide thus the mass of carbon dioxide is reduced from the solution. Le Chateliers principle states that if there is a change in pressure, concentration or temperature the system(in equilibrium) Shifts to counteract this change .As the carbon dioxide concentration is reduced this causes the equilibrium to shift to the left to counter act this change in concentration. This makes the concentration of H2CO3 less thus becoming less acidic.
 

Drsoccerball

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Actually...i found the answer to part 1 only.

IT says, the difference in pH from start to end is 3. Hence, the initial hydrogen ion concentration is 1000 times more than that after 2 hours.
Which is what i wrote... 10^(-4.5)/10^(-7.5) =1000 ...
 

Ununoctium

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Here's my question.

Question 1: (6 marks)

Fuel X
Heat of combustion: 4,800kJ/mol
G/mol: 68.3
Non-renewable

Fuel Y
Heat of combustion: 8,100kJ/mol
G/mol: 108.9
Non-renewable

Fuel H:
Heat of combustion: 7,108kJ/mol
G/mol: 95.1
Non-renewable

Fuel Z:
Heat of combustion: 1,998kJ/mol
G/mol: 31.0
Renewable


a) Calculate which fuel above provides the most heat per kilogram when it undergoes complete combustion. (2 marks)

b) Assess which fuel would be of better use for motor vehicles, with consideration of the environmental and societal impact of the fuel. (4 marks)
 

Kaido

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Its holidays, nothing new >.>
 

Drsoccerball

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Don't worry after trials we'll all be posting questions because that's when we have time to do any subject we wish to work on :D
I need inspiration to do English and Chemistry :(
 

HeroicPandas

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It's been a while...

Suppose that you have a closed system that is under equilibrium (all gases) and the equation is: X(g) + 4Y(g) <-> 8Z(g). If I pump in Argon gas at constant pressure, what does it do to the system? Is equilibrium disturbed or undisturbed? Justify your answer
 

Thunderstorm

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re: HSC Chemistry Marathon Archive

It's been a while...

Suppose that you have a closed system that is under equilibrium (all gases) and the equation is: X(g) + 4Y(g) <-> 8Z(g). If I pump in Argon gas at constant pressure, what does it do to the system? Is equilibrium disturbed or undisturbed? Justify your answer
Disturbed. Le Chatelier's principle states the equilibrium will shift to the left, due to an increase in pressure. The equilibrium attempts to 'balance' itself by becoming equimolar hence shifts left.
 

Fiction

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re: HSC Chemistry Marathon Archive

It's been a while...

Suppose that you have a closed system that is under equilibrium (all gases) and the equation is: X(g) + 4Y(g) <-> 8Z(g). If I pump in Argon gas at constant pressure, what does it do to the system? Is equilibrium disturbed or undisturbed? Justify your answer
undisturbed. Argon gas is an inert gas therefore will not affect the equilibrium as it doesn't affect partial pressure. No changes will happen to the system.
 

Drsoccerball

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undisturbed. Argon gas is an inert gas therefore will not affect the equilibrium as it doesn't affect partial pressure. No changes will happen to the system.
but doesnt it still increase the pressure?
 

Librah

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Read the question properly before answering.

Suppose that you have a closed system that is under equilibrium (all gases) and the equation is: X(g) + 4Y(g) <-> 8Z(g). If I pump in Argon gas at constant pressure, what does it do to the system? Is equilibrium disturbed or undisturbed? Justify your answer
.
 
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Librah

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no, because the initial premise of it was that neither pressure increased or decreased. so therefore, because it's at constant pressure, there was no imposed change and hence the equilibrium remains the same

thus, it has to be undisturbed like fiction said before
If the pressure remains the same, the volume needs to increase if your adding another gas, so the equilibrium will actually shift to compensate for the increase in volume.
 

hawkrider

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If the pressure remains the same, the volume needs to increase if your adding another gas, so the equilibrium will actually shift to compensate for the increase in volume.
oh, does it?

so it comes back to the inverse relationship of volume and pressure yes?

but is thunderstorm's answer correct? (cos it does sound right)
 
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Librah

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oh, does it?

so it comes back to the inverse relationship of volume and pressure yes?

but is thunderstorm's answer correct?
Relate volume/pressure for ideal gases.

Thunderstorm is partially correct. Try to find his mistake.
 

Fiction

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If the pressure remains the same, the volume needs to increase if your adding another gas, so the equilibrium will actually shift to compensate for the increase in volume.
adfalsdjfk

Can someone please point out the difference between the situation as illustrated by HeroricPanda and one where inert gas doesn't affect equilibrium? I'm not seeing the differences. Why does 'constant pressure' rather than just 'pressure' affect anything? Thanks.
 
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