HSC 2012-2015 Chemistry Marathon (archive) (87 Viewers)

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HeroicPandas

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Question: Identify the systematic name of citric acid
 
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1.22 g of an unknown gas has a volume of 15.0 L at 100 kPa and 25°C.

Identify the gas.
 
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Calculate the pH of the resulting solution when 50.0mL of 0.15mol/L potassium hydroxide solution and 150mL of 0.10mol/L sulfuric acid are mixed.
 
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mahmoudali said:
can you show how you did it with working out so we can all understand
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Yeah, sorry lol

at 100 kPa and 25oC

1 mol ---- 24.79L
x mol ---- 15.0L

x = 0.6050...moles

n = m/M
M = m/n
= 1.22g/0.6050...moles
= 2.016253..g/mol

Since all the gases occur as diatomic molecules, divide the molar mass by 2.

2.016253..g mol-1 / 2
= 1.008126..
= 1.008

1.008 is the molar mass of hydrogen.

Therefore, the gas is H2.

Note: The gas isn't a noble gas because none of the noble gases have a molar mass of 2.016253 g/mol.
 
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bedpotato said:
Calculate the pH of the resulting solution when 50.0mL of 0.15mol/L potassium hydroxide solution and 150mL of 0.10mol/L sulfuric acid are mixed.
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2KOH + H2SO4 -----> K2SO4 + 2H20

n(KOH) = cV = 0.05L x 0.15 mol/L = 0.0075moles
n(sulfuric acid) = 0.15L x 0.1 mol/L = 0.015moles

since moles of base and acid are in stoichiometric ratio (2:1), there is no limiting reagent; i.e. all moles of the base and acid have reacted to form K2S04 and water

hence pH is 7 as K2SO4 is neutral
 
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2 advantages of refluxing?
 
Deliriously

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leesh95 said:
2 advantages of refluxing?
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1) It prevents loss of volatile reactants+products as the vapors can condense back into liquid form.
2) It's an open system - allows higher temps to be used safely without pressure build up.

Q: Identify two methods for determining the hardness of water. Evaluate the appropriateness of each method for testing the water quality (for hardness) of a local river.
 
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superSAIyan2 said:
2KOH + H2SO4 -----> K2SO4 + 2H20

n(KOH) = cV = 0.05L x 0.15 mol/L = 0.0075moles
n(sulfuric acid) = 0.15L x 0.1 mol/L = 0.015moles

since moles of base and acid are in stoichiometric ratio (2:1), there is no limiting reagent; i.e. all moles of the base and acid have reacted to form K2S04 and water

hence pH is 7 as K2SO4 is neutral
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don't we have to take into account that H2SO4 is diprotic?

H2SO4 ----> 2H+ + SO4-
 
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Deliriously said:
1) It prevents loss of volatile reactants+products as the vapors can condense back into liquid form.
2) It's an open system - allows higher temps to be used safely without pressure build up.

Q: Identify two methods for determining the hardness of water. Evaluate the appropriateness of each method for testing the water quality (for hardness) of a local river.
Click to expand...
Volumetric Analysis and AAS
This technique involves testing the titrant 'ethylene-Diamine-tetra-acetic-acid' to react with Ca+ and Mg+ ions with the help of Erichrome Black indicator. This method is appropriate in the sense that the sample being analyzed is assumed to have high concentrations of the ions. However if the sample contains low concentrations of the ions, then this method would be ineffective. This is where AAS is used due to it's high sensitivity in detecting low concentrations of Ca+ and Mg ions+. That's all I think of, water management is my weakest field.
 
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superSAIyan2 said:
2KOH + H2SO4 -----> K2SO4 + 2H20

n(KOH) = cV = 0.05L x 0.15 mol/L = 0.0075moles
n(sulfuric acid) = 0.15L x 0.1 mol/L = 0.015moles

since moles of base and acid are in stoichiometric ratio (2:1), there is no limiting reagent; i.e. all moles of the base and acid have reacted to form K2S04 and water

hence pH is 7 as K2SO4 is neutral
Click to expand...
Good try, but this is incorrect.

When calculating pH, you have to take into account the amounts of H+ and OH- present, not the actual acids and bases themselves.

The answer's here:

n(KOH) = 0.0075mol
n(OH-) = 0.0075mol (since KOH is a strong base)

n(H2SO4) = 0.0150mol
n(H+) = 2 x 0.0150mol (since H2SO4 is diprotic) = 0.0300mol

0.0075mol of OH- only requires 0.0075mol of H+ to be neutralised. Therefore, H+ is in excess.

n(H+ in excess)= 0.0300 - 0.0075 = 0.0225mol

[H+] = 0.0225mol/200mL = 0.0225mol/0.2L = 0.1125mol/L

Hence, pH = -log[H+] = -log(0.1125) = 0.94887... = 0.95
 
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Ca(OH)2(s) <-> Ca2+(aq)+2OH-(aq)

1)In the above equilibrium reaction, The concentrations of OH-, Ca2+, and Ca(OH)2 were increased, explain what would happen to the equilibrium.

2)The amount of Ca(OH)2 was increased, explain what would happen to the equlibrium
 
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HeroicPandas

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Menomaths said:
Ca(OH)2(s) <-> Ca2+(aq)+2OH-(aq)

1)In the above equilibrium reaction, The concentrations of OH-, Ca2+, and Ca(OH)2 were increased, explain what would happen to the equilibrium.

2)The amount of Ca(OH)2 was increased, explain what would happen to the equlibrium
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1) Changing the amount of solid in an equilibrium has no effect as solids have a constant concentration, so if you increase conc. of calcium ions and hydroxide ions, the system will adjust to minimise the change (Le Chatelier's Principle), hence it will favour the reverse reaction to decrease the rise in conc of Ca2+ and OH- --> so more calcium hydroxide and less OH-/Ca2+

2) Equilibrium does not change
 
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HeroicPandas said:
Good work for question 1 but u made a small mistake in Q2 (a) which costed u the rest of the question
1 (b) i would say potassium acetate ionises to form acetate ions instead of "which comes out from the KCH3COO when put into water"

2 (a) Lead is more reactive

(b) only '3' is correct

(c) He could heat the lead nitrate and then sniff it in lol
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okay thanks for tips, what's the answer for b?
 
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Dementured said:
a) The raw material that is used to produce ethylene, sugar cane, is a renewable material. However, the processes used to create the ethylene from sugar cane result in the emission of CO2, as well as the emissions produced from the transportation of the ethylene. However, the process of photosynthesis requires CO2, so in effect, the growing of sugar cane results in the entire process being 'carbon neutral'.

b) CH2=CH2 (g) -> -CH2-CH2-n (s)

c) Sugar cane (extraction) -> Glucose (fermentation) -> Ethanol (distillation) -> Ethylene

Question 2:

a)
i) Chloroethene
ii) Poly..ethenylbenzene?
iii) Ethenylbenzene

b)
(i) polyvinylchloride: (1) flexible raincoats (due to water resistance), (2) pipes and gutters (water resistant, rigid), (3) cable insulation (through plasticisers)
(ii) polystyrene: (1) foam/insulated cups (addition of air in molten state), (2) protective packaging (hard, brittle, transparent), (3) disposable cutlery (hard, brittle)
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1st thing in bold: polymerisation is the key word
2nd thing in bold: dehydration (bigger)
3rd thing in bold: ethene
 
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Dementured said:
Burette: solution of known concentration
Flask: solution of unknown concentration + indicator
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hmm this is just the recommendation, it will depend on the solution (in the hsc they might put some random test and I guess it will depend on the colour change and which will be easier to detect)
 
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bedpotato said:
Yep, it is!

Which stateent best describes the equivalence point in a titration between a strong acid and a strong base?
(A) The point at which the first sign of a colour change occurs
(B) The point at which equal moles of acid and base have been added together
(C) The point at which equal moles of H+ ions and OH- ions have been added together
(D) The point at which the rate of the forward reaction equals the rate of the reverse reaaction
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it's c?
 
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