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Author Topic: VCE Chemistry Question Thread  (Read 2313755 times)  Share 

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Shadowxo

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Re: VCE Chemistry Question Thread
« Reply #6000 on: January 23, 2017, 09:58:52 pm »
+1
Hi,

Shadowxo has basically covered it all  ;)

In addition to Shadowxo's answer,
All fuels undergo a combustion reaction to produce energy, which is a type of endothermic reactions, meaning the energy is lost to the surroundings. A negative Hc indicates that energy is lost (the product(s) will have a lower amount of energy than the reactants).

(Btw I answered the same question at the top of this page  :P )

In addition to Syndicate
Combustion reactions are exothermic*, as they release energy, the energy is lost to the surroundings, and they have a negative ∆H. (minor mistake :P) (exothermic = hot  = releasing energy)
And oops, didn't read up :P
« Last Edit: January 23, 2017, 10:00:28 pm by Shadowxo »
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Syndicate

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Re: VCE Chemistry Question Thread
« Reply #6001 on: January 23, 2017, 10:02:35 pm »
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In addition to Syndicate
Combustion reactions are exothermic*, as they release energy, the energy is lost to the surroundings, and they have a negative ∆H. (minor mistake :P) (exothermic = hot  = releasing energy)
And oops, didn't read up :P

Woops  :P

Thanks for the fix  ;D
« Last Edit: January 23, 2017, 10:05:02 pm by Syndicate »
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rpapa

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Re: VCE Chemistry Question Thread
« Reply #6002 on: January 23, 2017, 10:38:40 pm »
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Hi, I was wondering if any one could help me with some Stoichiometry homework questions for Unit 3 on this thread? the question is
"Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide (Li3N solid + 3H2O liquid -> NH3 gas + 3LiOH"
a) What mass of water is needed to react with 98.7g of LiN3? (154g)
b) When the above reaction takes place, how many molecules of NH3 are produced? (1.71 x 10^24 molecules)
c) Calculate the number of grams of Li3N that must be added to an excess of water to produce 45.0L of NH3 (at STP).

I am not quite sure how to answer c). I got a huge number through using n=m/M which is nowhere near the answer (69.7g).
Could I please have some help with this question?
« Last Edit: January 23, 2017, 11:34:36 pm by rpapa »
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Shadowxo

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Re: VCE Chemistry Question Thread
« Reply #6003 on: January 23, 2017, 11:10:44 pm »
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First of all you wrote LiN3 instead of Li3N :P

First of all, at STP the volume of 1 mol of gas is 22.4
To get 45.0L of NH3, the number of mol NH3 would be 45/22.4 = 2.01mol ( it's a shortcut, if it's SLC divide by 24.5 iirc)
n(NH3) produced = n (Li3N) reacted
n = m/M, m (Li3N) = n*M = 2.01*(6.9*3+14) = 69.7g
« Last Edit: January 23, 2017, 11:14:27 pm by Shadowxo »
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rpapa

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Re: VCE Chemistry Question Thread
« Reply #6004 on: January 23, 2017, 11:31:51 pm »
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First of all you wrote LiN3 instead of Li3N :P

First of all, at STP the volume of 1 mol of gas is 22.4
To get 45.0L of NH3, the number of mol NH3 would be 45/22.4 = 2.01mol ( it's a shortcut, if it's SLC divide by 24.5 iirc)
n(NH3) produced = n (Li3N) reacted
n = m/M, m (Li3N) = n*M = 2.01*(6.9*3+14) = 69.7g

oops 😁
thanks so much for answering shadowxo! that makes so much more sense now  :)
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Shadowxo

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Re: VCE Chemistry Question Thread
« Reply #6005 on: January 23, 2017, 11:40:28 pm »
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oops 😁
thanks so much for answering shadowxo! that makes so much more sense now  :)

No worries, any other questions feel free to ask :)
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Gogo14

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Re: VCE Chemistry Question Thread
« Reply #6006 on: January 24, 2017, 01:25:05 pm »
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In my textbook, the heat of combustion is a negative number.
Howrver on websites, it says that heat of combustion is positive, but it is the delta H that is negative.
Which one is right?
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Jakeybaby

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Re: VCE Chemistry Question Thread
« Reply #6007 on: January 24, 2017, 01:38:20 pm »
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In my textbook, the heat of combustion is a negative number.
Howrver on websites, it says that heat of combustion is positive, but it is the delta H that is negative.
Which one is right?
The heat of combustion is the total value of heat released, it doesn't really make sense for this value to be negative as combustion reactions are exothermic, they release heat. Therefore the heat of combustion should always be positive, as it gives you the calorific value as to how much heat is released. i.e. For methane, the heat of combustion is 890kJ mol^-1.

As it is a combustion reaction and we have already established that they are all exothermic reactions, the deltaH value will always be negative for such combustion reactions. The deltaH for methane combustion is -890kJ.mol^-1
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #6008 on: January 24, 2017, 02:09:27 pm »
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In my textbook, the heat of combustion is a negative number.
Howrver on websites, it says that heat of combustion is positive, but it is the delta H that is negative.
Which one is right?

Jakeybaby gave a nice answer, but tbh: this is just semantics. What the website is trying to say is that the amount of heat produced is positive, but deltaH is not a measure of heat produced, it's a measure of change in heat of the solution. Meanwhile, the book is ignoring the concept of heat being produced, and is using "heat of combustion" to mean deltaH. Neither is wrong, they're just using different words to say the same thing, and it's not something that VCAA are ever going to pull you up on.

hodang

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Re: VCE Chemistry Question Thread
« Reply #6009 on: January 25, 2017, 11:22:04 am »
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Hey guys, for Q12 (a) the answer is 4.781 x 10^4 , anyone reassure me that THATS NOT rounded to 3 sig figures? Isnt 4.78 x 10^4 rounded to 3 sig figures?

For Q12 (b) the answer says h2, C4H10,C8H18 - as the order for most energy produced per kG to the least
Anyone can explain this? How can you convert KJ - KG , if i know that ill be able to work out the answer . Thanks heaps :)

Syndicate

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Re: VCE Chemistry Question Thread
« Reply #6010 on: January 25, 2017, 11:42:47 am »
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Hey guys, for Q12 (a) the answer is 4.781 x 10^4 , anyone reassure me that THATS NOT rounded to 3 sig figures? Isnt 4.78 x 10^4 rounded to 3 sig figures?

For Q12 (b) the answer says h2, C4H10,C8H18 - as the order for most energy produced per kG to the least
Anyone can explain this? How can you convert KJ - KG , if i know that ill be able to work out the answer . Thanks heaps :)

Q12a) Yes, you are correct. 4.78 x 10^4 has three sig figs.

Q12b) You can't convert kJ to kg (if that's what you mean?). You can have kJ kg^-1 though, which is helpful, as it indicates the amount of energy is produced in kJ per kg.

You can use the results you got from Q12a to list the fuels in the order specified.
(I had to look up the table in my book)

The Hc listed on table 2.3.1 has the units kJ mol^-1. So from this information, you can calculate the amount of mol in 1 kg of fuel, and multiply it by Hc to get the amount of energy released by that fuel.

n(H2) = 1000/2 = 500 mol
n(C4H10) = 1000/58 = 17.24 mol
n(C8H18) = 1000/114 = 8.77 mol

H2 : energy released = 500 x 286 = 143 MJ
C4H10 : energy released = 17.24 x 2886 = 49.8 MJ
C8H18 : energy released = 8.77 x 5450 = 47.8 MJ

Therefore 1kg of H2 produces the most energy, then C4H10, and finally C8H18.
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hodang

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Re: VCE Chemistry Question Thread
« Reply #6011 on: January 25, 2017, 12:36:04 pm »
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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!

hodang

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Re: VCE Chemistry Question Thread
« Reply #6012 on: January 25, 2017, 12:42:08 pm »
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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 :)

Syndicate

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Re: VCE Chemistry Question Thread
« Reply #6013 on: January 25, 2017, 02:46:09 pm »
+1
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)
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hodang

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Re: VCE Chemistry Question Thread
« Reply #6014 on: January 25, 2017, 07:17:23 pm »
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Thanks Syndicate, youve been a tremendous help. When your free can you also include the steps like Mg2+ (s)  + 2H+ (aq) that step- the breaking things up one