2007 Questions (1 Viewer)

Aaron.Judd

Member
Joined
Aug 20, 2008
Messages
372
Gender
Male
HSC
2008
Q13: Why is the answer A?

Q22 a) How do you do this?

Q22 b) ii) Whats this answer?

Q 27 a/b) How do you work this out, especially the calculation?
 

JasonNg1025

Member
Joined
Oct 14, 2007
Messages
295
Gender
Male
HSC
2010
13) Either by reasoning that the other ones are inaccurate (bad way of doing), or reasoning:

N2 (g) + 2H2 (g) <=> 2NH3 (g) + heat (exothermic)

There are clearly more moles of gas on the R.H.S. Decreasing in pressure, therefore, shifts equilibrium to the left. Since the forward reaction releases heat, the reverse reaction absorbs heat, hence the answer is A, absorbing heat.

22) a) Look at your other thread

22) b) ii) The sulfur reduction policy should lead to lower levels of SO2, as:

S + O2 --> SO2.

SO2 forms acid rain:

SO2 + H2O --> H2SO3

2SO2 + O2 --> 2SO3

SO3 + H2O --> H2SO4

Acid rain reduces pH of water bodies, killing less tolerant aquatic life. It reduces plant growth and even kills some plants. Acid rain also corrodes the protective layer on leaves, making them more vulnerable and less tolerant. Limestone buildings and monuments can also be corroded.

The reduce in SO2 also increases general air quality. Therefore, the sulfur reduction policy will be good for the environment.

27) a) Basically (elaborate on this), use scales to weigh water sample, add AgNO3 in excess to fully precipitate out Cl-, pour the reacted water sample through the sintered glass filter. The residue should be AgCl precipitate. Dry, and weigh. Use this to calculate Cl- concentration.

27) b) Precipitate is 3.65g, => moles AgCl = 0.0255.

AgCl --> Ag+ + Cl- => moles Cl- = 0.0255.

Mass Cl- = 0.903g

This is in 50mL (we assume 50g) of water.

Concentration in ppm = 0.903 / 50 x 1000000ppm

This is 18052.67 ppm
 
Last edited:

Aaron.Judd

Member
Joined
Aug 20, 2008
Messages
372
Gender
Male
HSC
2008
With 13
Initally there is 4 mols of gas on the left to 2 mols of gas on the right. So if you reduce the entire system pressure, I.e. half it. Then you have 2 mols on left to 1 mol of gas on the right. Hence shouldn't the reaction still be reactants to products (i.e. to the right)?
 

JasonNg1025

Member
Joined
Oct 14, 2007
Messages
295
Gender
Male
HSC
2010
Aaron.Judd said:
With 13
Initally there is 4 mols of gas on the left to 2 mols of gas on the right. So if you reduce the entire system pressure, I.e. half it. Then you have 2 mols on left to 1 mol of gas on the right. Hence shouldn't the reaction still be reactants to products (i.e. to the right)?
When you reduce the pressure on the system, that's what initially happens. Then the system wants to make up for the pressure it lost, hence it will favour the side with more moles. Hope that makes sense...

Aaron.Judd said:
What about 24 b) ii)
Conditions are different - the theoretical value uses standard conditions (25o C and 100kPa I think), which may differ it a little. Also, in practice, most of the time you get incomplete combustion, making your empirical value much lower.
 

danz90

Active Member
Joined
Feb 2, 2007
Messages
1,467
Location
Sydney
Gender
Male
HSC
2008
Did you get like 2.3L for 22(a) ??
 
Last edited:

Marinatos

Member
Joined
Aug 29, 2007
Messages
77
Location
Newcastle
Gender
Female
HSC
2008
Since you worked out how to do it, how do you do Q22. I'm so brain dead right now.

edit: nver mind. I did eventually get it.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top