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Author Topic: HSC Chemistry Question Thread  (Read 1040660 times)  Share 

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jakesilove

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Re: Chemistry Question Thread
« Reply #660 on: August 21, 2016, 10:54:36 pm »
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Ahh I see now but how do we know how many moles of lead there are for every lead chloride. Is it because in PbCl2 there is only one mole of lead in the molecule?
Where as, say if the precipitate was for whatever reason Pb3(PO4)2, there would be 1/3 moles for every mole of Pb3(PO4)2?

Almost! For every mole of Pb3(PO4)2, there are actually 3 mole-s of Pb! The little number is the number of 'moles' within the compound. So, for PbCl2, there is two moles of Cl and one mole of Pb. Makes sense?
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cajama

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Re: Chemistry Question Thread
« Reply #661 on: August 21, 2016, 11:03:17 pm »
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Almost! For every mole of Pb3(PO4)2, there are actually 3 mole-s of Pb! The little number is the number of 'moles' within the compound. So, for PbCl2, there is two moles of Cl and one mole of Pb. Makes sense?

Oh yeah I get it completely now. Thank you so much!

conic curve

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Re: Chemistry Question Thread
« Reply #662 on: August 24, 2016, 05:25:00 am »
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Does the pressure a substance exerts on the container increase as it goes from solid to liquid or gas? Or is it the other way around?

RuiAce

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Re: Chemistry Question Thread
« Reply #663 on: August 24, 2016, 08:21:54 am »
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Does the pressure a substance exerts on the container increase as it goes from solid to liquid or gas? Or is it the other way around?
That way is correct.

Pressure only ever relates to the quantity of gaseous substances present in a (generally speaking, closed) system. Liquids and solids are not counted, so if gas is converted into liquid/solid we go the other way and decrease pressure.

massive

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Re: Chemistry Question Thread
« Reply #664 on: August 24, 2016, 04:46:31 pm »
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hey guys, how do you do this :S (answer is d btw)

RuiAce

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Re: Chemistry Question Thread
« Reply #665 on: August 24, 2016, 05:39:22 pm »
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hey guys, how do you do this :S (answer is d btw)
Because our system is constantly at 25oC, 100kPa, observe that VM = 24.79 L mol-1 is constant.

Since n = V/VM, because the molar volume is fixed the reaction that produces the greatest moles of gas will also produce the largest volume.

Write out the relevant chemical equation for each and every one of your reaction to determine the relevant moles. Keep in mind that as HCl is the excess reagent, it does not matter how many moles of it we use.

A) CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)
nCaCO3 = 50/100.9 = 0.49554... mol
nCO2 = nCaCO3 = 0.4955... mol

B) Zn(s) + 2 HCl(aq) -> ZnCl2(aq) + H2(g)
nZn = 50/65.38 = 0.7647... mol
nH2 = nZn = 0.7647... mol

C) Na2CO3(s) + 2 HCl(aq) -> 2 NaCl(aq) + CO2(g) + H2O(l)
nNa2CO3 = 50/105.99 = 0.4717... mol
nCO2 = nNa2CO3 = 0.4717... mol

D) 2 Na(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2(g)
nNa = 50/22.99 = 2.1748... mol
nH2 = 1/2 nNa = 1.0874... mol

By comparison, the greatest moles of gas produced occurs in reaction D.

Note that the purpose of writing all the equations is to identify the mole ratio. The gas produced itself doesn't matter because we are talking about VOLUME, not mass.

massive

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Re: Chemistry Question Thread
« Reply #666 on: August 24, 2016, 05:46:37 pm »
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OOOHH i get it now thanks!

Theres also one other thing i don't get- its about equilibrium. I understand the whole lcp stuff about temperature and pressure but i don't get everything about concentration, i.e. when you add things to the system.

For q9 i know that the answer is b, i was just wondering if part c and d have any effect on the equilibrium.

Also for 10, this is what i was talking about the concentration stuff, how do you do it

massive

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Re: Chemistry Question Thread
« Reply #667 on: August 24, 2016, 06:21:20 pm »
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Hey Jake why do you use 50ml for the volume? they also say that 25ml of sodium chloride was used, don't you add the volumes together :S

RuiAce

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Re: Chemistry Question Thread
« Reply #668 on: August 24, 2016, 07:03:13 pm »
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OOOHH i get it now thanks!

Theres also one other thing i don't get- its about equilibrium. I understand the whole lcp stuff about temperature and pressure but i don't get everything about concentration, i.e. when you add things to the system.

For q9 i know that the answer is b, i was just wondering if part c and d have any effect on the equilibrium.

Also for 10, this is what i was talking about the concentration stuff, how do you do it
With Q9, C and D would end up shifting the equilibrium to the left because you introduce more product, not reactant.

So.

If there is an excess of either reactant or product, then the equilibrium has been disturbed (or it was never at equilibrium). Therefore, according to LCP, the system wants to get rid of that EXCESS.

Conversely, if there is a shortage of either reactant or product, then LCP predicts that the system wants to bring in more of what's lacked to address the shortage. This is how equilibrium is (re-)established.

So for Q10, where the effect is not immediately obvious, you must analyse what happens.

You introduced hydrochloric acid. Obviously it's not going to react with water or the ammonium ion here. Does it react with ammonia?
No. The HCl will have a tendency to react with that hydroxide ion, which is clearly a base.

HCl + OH- -> Cl- + H2O

So since we take out OH- through the introduction of the acid, the equilibrium will shift to the right.

Now, immediately I can tell this question is beyond the HSC's scope, however it's not unreasonable. This is because of unnecessarily introducing option C: Equation is driven to completion.

A few drops of HCl is probably insufficient to do so. In practicality? Yes, this is possible if you had an unlimited amount of HCl. This is because you end up introducing so much acid that all the base ends up neutralised, and you have excess acid instead.

So for this question? It should be A.

massive

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Re: Chemistry Question Thread
« Reply #669 on: August 24, 2016, 07:15:32 pm »
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So for Q10, where the effect is not immediately obvious, you must analyse what happens.

You introduced hydrochloric acid. Obviously it's not going to react with water or the ammonium ion here. Does it react with ammonia?
No. The HCl will have a tendency to react with that hydroxide ion, which is clearly a base.

HCl + OH- -> Cl- + H2O

So since we take out OH- through the introduction of the acid, the equilibrium will shift to the right.


wait so are we just meant to learn this? is this part of module 2? and yeah the answer is A

RuiAce

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Re: Chemistry Question Thread
« Reply #670 on: August 24, 2016, 07:28:53 pm »
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wait so are we just meant to learn this? is this part of module 2? and yeah the answer is A
You were already taught it. It's just a process of deduction. And yes of course it counts under The Acidic Environment
« Last Edit: August 24, 2016, 07:31:36 pm by RuiAce »

massive

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Re: Chemistry Question Thread
« Reply #671 on: August 24, 2016, 11:15:30 pm »
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Guys how do you do part b for the question attached. I was just confused because there is no equation, so how do you use mole ratio comparison like you do if you want to figure out the percent of sulfate in fertilisers.

jakesilove

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Re: Chemistry Question Thread
« Reply #672 on: August 25, 2016, 10:57:20 am »
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Guys how do you do part b for the question attached. I was just confused because there is no equation, so how do you use mole ratio comparison like you do if you want to figure out the percent of sulfate in fertilisers.

You don't need any sort of equation; what you're given is plenty. First, using the molar mass of the substance and the mass you've been given, find the moles. Then, recall that for a substance



that this means that for every one mole of the substance, there are n moles of X and m moles of Y. So, for one mole of the substance in this question, there will be 2 moles of phosphorus. I'll let you do the maths
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anotherworld2b

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Re: Chemistry Question Thread
« Reply #673 on: August 25, 2016, 11:14:33 am »
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I was wondering would my answer to this question be correct?

PF5 abd PH3 have different shapes because of the structure and cimposition of molecukes each one contains. This determines which intermolecular forces are present,  and these forces determine the physical properties of the material

I also wanted to ask what is the difference between a dipole, polar bond and net dipole?
« Last Edit: August 26, 2016, 09:11:55 am by anotherworld2b »

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Re: Chemistry Question Thread
« Reply #674 on: August 26, 2016, 06:05:19 pm »
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I was wondering would my answer to this question be correct?

PF5 abd PH3 have different shapes because of the structure and cimposition of molecukes each one contains. This determines which intermolecular forces are present,  and these forces determine the physical properties of the material

I also wanted to ask what is the difference between a dipole, polar bond and net dipole?

Hey Anotherworld2b!

You are certainly right! PF5 and PH3 do have different shapes because of the structure and composition of the molecules! But this would be a little too general as an answer. When you are asked questions of a similar kind, what you need to include in your answer in "VSEPR principles" and "electronegativity". So according to VESPR rules, molecules are arranged in particular 3D shapes that aim to keep the valence electrons on elements as far apart as possible (since electrons repel each other) in order to maintain a stable structure. The main differences between PF5 and PH3 here are the number of elements involved and the contrast between fluorine's electronegativity and hydrogen's electronegativity. Because there are 5 fluorine elements, the molecule would take on a bitrigonal pyramidal structure with bond angles of 120 degrees and 90 degrees in order to fit all these fluorine atoms in a stable arrangement (more closely packed). In PH3, because there is only 3 hydrogen atoms, it takes on a trigonal pyramidal shape of 120 degrees between all hydrogen atoms (more loosely packed). Because fluorine has a high electronegativity, or a high affinity towards electrons, they must be kept considerably far away from each other to prevent the decomposition of the structure as fluorine tries to gain other electrons in the molecule. These factors all influence the shape and composition of PF5 and PH3.
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