VCE Chemistry Question Thread

[hodang]

Hey guys, sorry for bothering you guys, i just need help with a few q's from ch: 2 review (i posted pics of the questions)

Q)13 - For q13 (b) i got 78.4 KJ whereas the answers got 78.5 KJ , is this okay? aswell for (c) i got 62.7 KJ whereas the answers say 62.8 KJ and lastly for (e) i got 788 KJ whereas the answers got 789 KJ , not sure its okay or not but just checking in although im sure its fine but just incase

Completely need help with Q14,15,17, 18 (i know how to do the equation, not sure how to find deltaH though, and lastly for Q19

If anyone could help me, would be so great and helpful

God bless

[Shadowxo]

For 13 b,c,d your answers are right, they might have used a slightly higher value for c instead of 4.18. Your answers are fine
For 14, just use E = m*c*∆T and sub in to find ∆T (make sure you use E in J not kJ)
15 find the E used to heat the water, m*c*∆T
then divide that by the grams to find E/g
For 17a, the answer is 1 mol CO and 0.5 mol O2, as the ∆H is negative meaning 1 mol CO2 has less energy than 1 mol CO and 0.5 mol O2
for 17b, i) you just double the ∆H value as it's like the equation is happening twice
ii)You take the value from i) and *-1, as instead of energy being released, energy is being gained
18, You know the equation, as ethane is being burnt and it's complete combustion, 2C2H6(g) +7O2(g) -> 4CO2(g) + 6H2O(g)
Then do E = m*c*∆T = 100*4.18*5.52
n(ethane) = m/M = 0.0450/30
∆H = E/n
∆H = 100*4.18*5.52/(0.0450/30)
Then find ∆H for each mol of ethane,
each mol of equation uses 2 mol ethane so double your previous ∆H value to find the ∆H for the equation
19, can't see the whole question but we know the energy change, E=mc∆T and the grams of wood for that amount of energy, should be able to go from there

[sweetiepi]

Hey guys, sorry for bothering you guys, i just need help with a few q's from ch: 2 review (i posted pics of the questions)

Q)13 - For q13 (b) i got 78.4 KJ whereas the answers got 78.5 KJ , is this okay? aswell for (c) i got 62.7 KJ whereas the answers say 62.8 KJ and lastly for (e) i got 788 KJ whereas the answers got 789 KJ , not sure its okay or not but just checking in although im sure its fine but just incase

Completely need help with Q14,15,17, 18 (i know how to do the equation, not sure how to find deltaH though, and lastly for Q19

If anyone could help me, would be so great and helpful

God bless

For 13 b,c,d your answers are right, they might have used a slightly higher value for c instead of 4.18. Your answers are fine

In addition to what Shadowxo said with regards to Q13 b, c and d, your values may also be a result of being round down throughout your calculations. Hence the difference in decimal placing. To make sure you get the answers correct to 1 decimal place, you shouldn't round the answers until the final calculation for that question, as you can possibly get an incorrect number as your final result if you truncate before then.

[Shadowxo]

In addition to what Shadowxo said with regards to Q13 b, c and d, your values may also be a result of being round down throughout your calculations. Hence the difference in decimal placing. To make sure you get the answers correct to 1 decimal place, you shouldn't round the answers until the final calculation for that question, as you can possibly get an incorrect number as your final result if you truncate before then.

I did the equations myself without rounding so that wasn't the issue, I think it was just the book, likely using a different (higher, maybe not rounded) c value for water

[deStudent]

http://m.imgur.com/aCyXZJL For the HCl, how do you it's H+ becomes a spectator ion?

[Syndicate]

Can anyone help me in writing the ionic equations (totally forgot how) for: (working out would be great and step process)

Na2O (s) + H2So4 (aq) -->

Ba(Oh)2 (aq) + H3Po4 (aq) -->

NaHCO3 (aq) + HNO3 (aq) -->

SO3 (G) + Ca(OH)2 (aq) -->

Challenge:
Zn (s) + CH3COOH (aq) -->

Thanks heaps

Sorry for the late reply.

full: Na2O (s) + H2So4 (aq) -> Na2SO4 (aq) + H2O(l)

net: Na2O (s) + 2H+ (aq) + SO4 2- (aq) -> 2Na+ (aq) + SO4 2- (aq) +H2O(l)
SO4 2- is in aqueous state in both sides of the equation, therefore SO4 is a spectator ion, and the final net ionic equation will look like:
Na2O(s) + 2H+ (aq) -> 2Na+ (aq) + H2O(l)

full: 3Ba(OH)2 (aq) + 2H3Po4 (aq) ->Ba3(PO4)2 (s) + 6H2O (l)
There are no spectator ions.
net: same as the full equation


full: SO3 (g) + 3Ca(OH)2 (aq) -> H2O (l) + CaSO4 (s)
net: same as the full equation

[lzxnl]

One thing to keep in mind when writing ionic equations is that solids must be written in their entire form.
If you have Fe(OH)₃(s) + 3 HCl(aq) -> FeCl₃(aq) + 3H₂O(l), the ionic equation includes the iron because it hasn't dissociated on the left hand side.

[helloimpat]

Could someone help me with this question? in particular part c

[sweetcheeks]

Could someone help me with this question? in particular part c

A) this is quite straightforward,  a simple Q/F= n.e- . Q = Amps x time (seconds). 180000A x (1 hour x 60 minutes x 60 seconds) is equal to 6.5 x 10⁸. 6.5x10⁸ /96500 (charge of one mole electrons) gives an answer of 6.7X10³ moles of electrons. The half equation is Al³⁺ + 3e- --> Al. For every 3 mole of electrons, one mole of aluminium is produced. 6.7X10³/3 = 2.2X10³ moles of aluminium produced. To find the mass of aluminium, multiple moles by molecular weight; 2.2X10³ x 27.0g/mol = 6.0X10⁴ grams aluminium produced in 1 hour

B) this involves working backwards. 10.0 kilograms is equal to 10.0X10³ grams of aluminium. First we need to find the moles of aluminium. This is simply n=m/M 10000grams/27.0g/mol, giving us 370 moles of aluminium. The next step is to find the moles of electrons required. Looking at the half-equation again Al³⁺ + 3e- --> Al we can see that for every one mole of aluminium produced we require 3 moles of electrons. moles Al x 3 (370x3) yields 1.11x10³ moles of electrons. 1.11x10³F x 96500 will give us Q. Q is equal to 1.07x10⁸. Q being a product of Current x time and us knowing the current allows us to rearrange the equation to be Q/current = time (seconds). 1.07x10⁸/180000A = 596 seconds


C is a bit difficult, as it requires you to have a bit of knowledge about aluminium production. Aluminium is refined from the ground as Al2O3 and heated until it is molten. In the electrolytic cell, sacrificial carbon anodes are used, which produce carbon dioxide as the cell operates. The redox equation for the cell is 2Al2O3 + 3C --> 3CO2 + 2Al. We need to determine the volume of CO2 at SLC (24.5L/mole) produced for the entire day (24 hours). The first step will be to determine the amount of CO2 produced. We know how many moles of aluminium is produced in 1 hour, multiplying by 24 will gives us the total for the day. 2.2X10³ * 24 = 5.4X10⁴ moles of aluminium produced. Using stoichiometry we can see that for every 2 moles of aluminium produced 3 moles of CO2 is produced. 3/2* 5.4X10⁴ yields 8.1X10⁴ moles of CO2. The formula V(L)/24.5L/mol=n tells us that one mole of CO2 at SLC will occupy 24.5L. We can rearrange it to be n x 24.5L/mol = VCO2. 8.1X10⁴mol x 24.5L/Mol = 1.97X10⁶ litres of CO2 produced in one day.


n.b Throughout the calculations I gave rounded off answers but I used the actual numbers in the calculations, including that for part C. (transferring moles of aluminium from A). There may be a mistake somewhere, tomorrow I will redo the calculations and see.

[arushi]

someone plz suggest me a good chemistry guide. should i go for addison wesley chemistry?

[deStudent]

Small problem here http://m.imgur.com/AapEIsR
I got A and C, but the answer says it's only C. Is the book wrong here since my answer agrees with the electrochemical series?
Thanks

[Shadowxo]

Small problem here http://m.imgur.com/AapEIsR
I got A and C, but the answer says it's only C. Is the book wrong here since my answer agrees with the electrochemical series?
Thanks

After looking at it quickly, if it's to react with H2S and not H2O it must fall between 0.14 and 1.23 on the left hand side, and only C does that.
Cl2 reacts with water according to the electrochemical series so it's not a valid answer

[tasmia]

Hey guys!
Can someone please help me with two questions?

1. A 10.0 mL sample of nitric acid was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M potassium hydroxide solution. What was the concentration of the original nitric acid?

2. If 3.0 g of NaOH is added to 500 mL of 0.10 M hydrochloric acid, will the resulting solution be acidic or basic? What will the pH be

[sweetcheeks]

Hey guys!
Can someone please help me with two questions?

1. A 10.0 mL sample of nitric acid was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M potassium hydroxide solution. What was the concentration of the original nitric acid?

2. If 3.0 g of NaOH is added to 500 mL of 0.10 M hydrochloric acid, will the resulting solution be acidic or basic? What will the pH be

When solving questions like Q1 I highly recommend drawing a flowchart of the processes that are going on. It allows you to visual what is happening and outline the steps. If you like I can draw one.

We start with a 10.0mL solution of HNO3 of an unknown concentration. It is diluted by factor 10, so that the concentration is 1/10 of the stock. Whilst the concentration has changed the number of moles in the sample has not. Next a 25.0mL sample is taken, it will have the same concentration as the diluted sample but only 1/4 of the moles. A titration was then performed on this sample.

Starting at the titration. The reaction occurring is HNO3_(aq) + KOH_(aq) --> KNO3_(aq) + H2O_(l). 50.0mL of the 0.60M KOH solution neutralised the HNO3 sample. Knowing the concentration and volume, the number of moles of KOH used can be deteremined using C=n/V, rearranged to CxV=n. 0.60M x 0.0500L = 0.030mol of KOH. Using stoichiometry we can determine the amount of HNO3 that reacted. It is a 1:1 ratio HNO3:KOH so we have 0.030mol of HNO3 in the 25.0mL sample.

Above, we found that 25.0mL was taken from the 100.0mL solution. This means that there is 4x the amount of moles in the 100.0mL than the 25.0mL sample. This means that there is 0.12mol of HNO3 in the diluted sample. We can now determine the concentration of the diluted sample. C=n/V becomes 0.12mol/0.1000L giving us a concentration of 1.2M. We need to determine the concentration of the undiluted sample. We can use CiVi=CfVf, we know; the initial volume (10.0mL), the final volume (100.0mL) and the final concentration (1.2M). We can now rearrange to solve for Ci. CfVf/Vi becomes 1.2Mx100.0mL/10.0mL giving us an initial concentration of 12M.

[tasmia]

When solving questions like Q1 I highly recommend drawing a flowchart of the processes that are going on. It allows you to visual what is happening and outline the steps. If you like I can draw one.

Thanks a lot! Seems so easy now that you explained it. If it's not too much trouble, could you please draw the diagram? I wanna learn how to do that myself.