HSC Chemistry Question Thread

[Bells_123]

Hey, I was wondering whether anyone could give me a hand on these two calculations and how to do them. This question and the attached one as well please. Cheers 

6. 14 g of coal undergoes complete combustion in 58 g of oxygen gas.
What volume of gas (measured at 25 C and 100 kPa) is present when the reaction is
complete?
(A) 16 L
(B) 29 L
(C) 45 L
(D) 74 L

For the second screenshot question, you have to find the mass of the petrol using the density formula (mass/volume --> make sure you convert 45L to mL!). From this mass you can find moles which you would multiply by the molar heat of combustion (5460) to find how many kilojules is released from that amount of petrol.

You should get something like 1487131 kJ that you divide by 1370 (the heat of combustion) to find the number of moles of ethanol needed to liberate that amount of energy. From this you find the mass of ethanol using the mole formula, and then use the density formula again and rearrange to find the volume (which you convert to litres). So the answer should end up being A.

Hope this helps 

[Bells_123]

So the Zinc electrode is oxidising so is called the reductant, you can see this as the Zinc comes off the electrode and reacts with the chlorine to make the ZnCl2 solution. To answer the question, the Platinum electrode is undergoing reduction.

The zinc is the site of oxidation, but I think the platinum is an inert electrode that doesn't take part in the chemical reaction and is just responsible for electron transfer. So the chlorine that's bubbled into the solution should be the one undergiong reduction as it's reduced to chlorine ions.

[itssona]

hey help with this please
A chemist analysed aspirin tablets. Initially they had to determine the molarity of a NaOH solution. Three 25mL samples of a 0.1034mol/L solution of standardised HCl were titrated with the NaOH solution.
The average volume of the standardised NaOH required for neutralisation was 25.75mL
(a) Calculate the molarity of the NaOH solution

[Bells_123]

hey help with this please
A chemist analysed aspirin tablets. Initially they had to determine the molarity of a NaOH solution. Three 25mL samples of a 0.1034mol/L solution of standardised HCl were titrated with the NaOH solution.
The average volume of the standardised NaOH required for neutralisation was 25.75mL
(a) Calculate the molarity of the NaOH solution

With titration calculations I always like to write a balanced equation first to determine the mole ratios. This neutralisation reaction should be in a 1:1 mole ratio between NaOH and HCl, so you can find the moles of HCl by using the concentration equation (moles/volume in litres) which also equals the moles of the NaOH. Then you can find the molarity/concentration of the NaOH by plugging it in the concentration equation again 

[fkkiwi]

Anyone have an explanation to this?

Which of the following is the best oxidising agent?

A
A+
D2+
D

given A+ + e- --> A = 1.8V and D2+ + 2e- --> D = -1.7V

[Fergus6748]

Anyone have an explanation to this?

Which of the following is the best oxidising agent?

A
A+
D2+
D

given A+ + e- --> A = 1.8V and D2+ + 2e- --> D = -1.7V

Well the oxidising agent is what is being reduced and if you look at the standard potential table, the voltage values increase from negative to positive. The oxidising agent is usually the lower element on the table so I would say A+ (B). Pretty sure that's right

[amelia20181]

how do you do this question from the 2005 paper

[Bells_123]

how do you do this question from the 2005 paper

If it's a condensation polymer it most likely means that a water molecule comes off when the monomers join. By looking at the polymer you can see that 4 monomers are joined together (2 of each type and they alternate). If you circle around the first monomer from the first oxygen to the second oxygen on either side of the ring (counting left to right), you can see it repeats again after the second monomer, so where the monomers join must be at the second oxygen (right side of ring). The answer looks like D because when the monomers join you can see it will leave one oxygen when a water molecule forms.

Hopefully that made sense and helped you

[clovvy]

Volume of water: 200mL, initial temp: 22
When 1.635g of ethanol was burnt, the temp of water was increased to 47. Calculate the energy released, in kJ, when 1.50g of ethanol was burnt
a) 19.2kJ
b) 5.89kJ
c) 5.44kJ
d) 18.8kJ

[Fergus6748]

Volume of water: 200mL, initial temp: 22
When 1.635g of ethanol was burnt, the temp of water was increased to 47. Calculate the energy released, in kJ, when 1.50g of ethanol was burnt
a) 19.2kJ
b) 5.89kJ
c) 5.44kJ
d) 18.8kJ

Hey, so what I understand is that I would find the heat of combustion (mcat) and then divide by the mass of fuel burnt (the original 1.635g) to give the heat of combustion per gram of ethanol:

DeltaH = mcdeltat
            = (200)(4.18)(25)
            = 20900
            = 20900/1.635
            = 12782.87462

Then multiply by the new mass of fuel burnt (1.50g):

 = 12782.87462 x 1.50
 = 19174.31193 J
 = 19.17431193 kJ
 = 19.2 kJ (A)

[Dragomistress]

Hey,
Just curious, how do I know what compound is the most volatile?
For example, which is the most volatile out of propane, propan-1-ol, propanoic acid and methyl propanoate. (It is propane)

[KT Nyunt]

Hey,
Just curious, how do I know what compound is the most volatile?
For example, which is the most volatile out of propane, propan-1-ol, propanoic acid and methyl propanoate. (It is propane)

Since a volatile substance means that it easily becomes a gas. To determine which one of these are the most volatile, I think we have to looks of the intermolecular forces they form. Since propanol, propanoic acid and methyl propanoate all can form some hydrogen bonds, whereas propane can only form weak dispersion forces, it is more likely to be found as a gas than propanol, propanoic acid and methyl propanoate. Hence, propane is the most volatile.

[Dragomistress]

Hello again!
How do I complete this question?
A lawn fertiliser lists the sulfate content as 38.5%(w/w). What mass of barium sulfate precipitate would be expected to form if a 1.50g sample of the fertiliser were analysed by reacting the sample with excess barium nitrate solution? (0.238g, 0.578g, 1.40g or 3.64g)? (Answer is 1.40g)

Thank you!

[jazcstuart]

Hello again!
How do I complete this question?
A lawn fertiliser lists the sulfate content as 38.5%(w/w). What mass of barium sulfate precipitate would be expected to form if a 1.50g sample of the fertiliser were analysed by reacting the sample with excess barium nitrate solution? (0.238g, 0.578g, 1.40g or 3.64g)? (Answer is 1.40g)

Thank you!

Hi, this is how I would do it (I did it on paper first, I attached a picture if that is easier to read):

You know the weight of the sample, therefore you can find the mass of sulfate in the sample using the sulfate content % given:
38.5% x 1.50 = 0.5775g

Now you can work out the moles of sulfate:
n = m/MM
n(SO4) = 0.5775/(32.07+4x16.00)
             =0.0060112418...mol

Using the mole ratio, n(SO4) = n(BaSO4), therefore n(BaSO4) = 0.0060112418...mol
From here you can work out the mass os BaSO4 precipitate using n = m/MM again:
0.0060112418 = m/(137.3+32.07+4x16.00)
m = 0.0060112418 x 233.37
    = 1.402843499...g
    = 1.40 (3 significant figures) (yay)

Hope this makes sense!

[StephTol]

Hey guys!

I'm really confused with this question and I don't understand the solution

Thanks in advance