Hey Syndicate,
thanks for answering but, if i get the energy released as:
H2: 1.43 x 10^5 KJmol^-1 (which is same as 143 000)
C4H10: 4.96 x 10^4 KJ (which is same as 49 600
C8H18: 4.78 x 10^4 KJ (which is same as 47800)
I completely get now how H2 releases the most energy now, followed by butane and octane, but i want to know why its different when i put it in scientific notation? And why did you convert it into MJ instead of leaving it as KJ mol^-1 (think thats the unit i used.. lol) ps- did you convert it from *KJ mol-1* into MJ..? Thanks for replying, its just i did my answer in scientific notation which is what stuffed me up ahah is it preferred to not leave energy released in scientific notation? instead and convert it into KJ mol-1 or MJ (which is more preferred?) lol hope i make sense!
Significant figures are assessed in VCE chemistry, which means you can either use metric system units (ie. MJ, kJ, mJ, kg, g etc...) or scientific notation to put your answer into the right amount of significant figures. You are right saying 1.43 x 10^5 is the same as 143000, however 1.43 x10^5 only has 3 sig figs, whereas 143000 has 6 sig figs (technically 143000 is also 3 sig figs, however VCAA wants us to assume that it is 6 sig figs, just to make things less confusing). The lowest amount of sig figs used to calculate the energy output was 3, hence, you need write you final answer in 3 sig figs.
1 MJ = 1000 kJ, so 143 MJ = 143000 kJ. I also could have written 1.43 x 10^5 kJ which is the same as 143 MJ (it's just my personal preference). So, like I said earlier, it's fine to use anyone of them on the exam (for your SACs, you would be better confirming it with your teacher).
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
Just swap the anions around (unless you have acids or bases involved).
I don't have much time right now, so I'll just do one as an example (will comeback to it later today).
Q3) It's a metal hydrogen carbonate + acid reaction, therefore the products will be CO2, H2O and a salt
full ionic equation: NaHCO3(aq) + HNO3 (aq) -> NaNO3(aq) + CO2(g) + H2O(l)
For the net ionic equation, just remove the elements that are in aqueous state in both sides.
net ionic equation: HCO3-(aq) + H+ (aq) -> CO2(g) + H2O (l)
Challenge: It's a metal + acid reaction, so the final products will be H2 gas and a salt.
full ionic equation: Zn(s) + 2CH3COOH(aq) -> Zn(CH3COO)2 (aq) +H2 (g)
net ionic equation: Zn(s) + 2H+ (aq) -> Zn2+ (aq) + H2(g)