Unit 3 Questions MEGATHREAD :)

[b^3]

How do you do the pH questions where it asks how many moles of HCl needs to be added to change the pH of a 1.0 L solution from 12 to 2?

I do c1V1 = c2V2, dunno if there's a better way?

Remember that pH = 1x10^-pH, hence a pH of 2 = 1x10^-2 of H+ ions.

lol I don't get it?

i'm not sure if this is right, so someone check

so ph of 12 means

and V1=1L, now

and V2=1L
and we know that
so
so now we need to work out the number of mol needed right?
so
so the difference is how much we need to add so
im not 100% sure so someone just check to make sure.

[b^3]

ok so we know that since temperature and pressure remains the same the volume and amount (in mol) are related so.
since CO:O2 is 2:1 we need 10mL of O2 for 20mL of CO
so that means that all the CO is used up but there is 80mL-10mL=70mL of O2 remaining
The amoutn of gas produced is this plus the amount of CO2
so v(CO2)=v(CO)=20mL
so add the two and well 70mL + 20mL = 90mL of gas after completion

[ttn]

How do you do the pH questions where it asks how many moles of HCl needs to be added to change the pH of a 1.0 L solution from 12 to 2?

I do c1V1 = c2V2, dunno if there's a better way?

Remember that pH = 1x10^-pH, hence a pH of 2 = 1x10^-2 of H+ ions.

lol I don't get it?

i'm not sure if this is right, so someone check

so ph of 12 means

and V1=1L, now

and V2=1L
and we know that
so
so now we need to work out the number of mol needed right?
so
so the difference is how much we need to add so
im not 100% sure so someone just check to make sure.

That's what I did, but none of the options had that as an answer.

[b^3]

ok i think i got it, so

so im guessing if i know where you got this from then option D)?

[mickeymouse]

ok so we know that since temperature and pressure remains the same the volume and amount (in mol) are related so.
since CO:O2 is 2:1 we need 10mL of O2 for 20mL of CO
so that means that all the CO is used up but there is 80mL-10mL=70mL of O2 remaining
The amoutn of gas produced is this plus the amount of CO2
so v(CO2)=v(CO)=20mL
so add the two and well 70mL + 20mL = 90mL of gas after completion

how do we know that v(CO2)=v(CO)=20mL?

[b^3]

becuase the temperature and pressure is constant you can use the nmol ratios for the voulme.
 i got the solution for the second one but it may take a while to put up.

[mickeymouse]

becuase the temperature and pressure is constant you can use the nmol ratios for the voulme.
 i got the solution for the second one but it may take a while to put up.

oh right thanks. just to clarify if it produced for example... 4CO2 would the volume be 20mL x 4? and if other products were produced could I use this ratio to find their volume? 

[b^3]

can someone help me with these questions, answers are there but i don't know how to work it out:
1. Carbon monoxide reacts with oxygen as described by the chemical equation:
2CO(g) +O2 (g) ---> 2CO2 (g)
When 20mL of CO reacts with 80mL of O2 at the same temperature and pressure, what is the volume of the gas mixture after the reaction has reached completion?
Answer is 90mL

2. 5.00g of a mixture of magnesium and zinc was allowed to completely react with dilute sulfuric acid as described by the chemical equation, where M represents either magnesium or zinc
M (s) +H2SO4 (aq) ---> MSO4 (aq) +H2 (g)
If 0.250g of hyrdrogen gas was produced in the reaction, what would the percentage composition of magnesium in the mixture be?
Answer is 37.6%

thanks!

sorry i only read the OT now!  but if anyone wants to give it a try anyway ?

ok so we know that n(H2)=m/M = .250/2 = .125mol
and aswell as

so since u can sub it in
so

now multiply by 65.41
so


so now do the percentage by mass
%(Mg)%
did that help?

[b^3]

well yes and no, it depends on which is in excess or not, the one that is the limiting reagent is the one to base everything else off, and then you have to remember that the is some gases that didnt react that will add up in the end.

[extcar]

sweeeet thanks, it helped a lot
and you worked it out real quick too!

[b^3]

well ill say this, I have a maths brain and nothing else, all you have to realise is that there is two simultaneous equations to manipulate and its easy from there. i hope i didnt make it too complicated though.

[jane1234]

Is an alkene (C=C bond) considered a functional group?

[luken93]

Is an alkene (C=C bond) considered a functional group?

I think it is, but I'd await confirmation from the higher powers.

Benzene also is...

[b^3]

Is an alkene (C=C bond) considered a functional group?

I'd say yes because it influences the properties of the molecule but i'm not sure. I'd like to find out though. According to this it is so -> http://chemed.chem.purdue.edu/genchem/topicreview/bp/2organic/function.html

[jane1234]

Is an alkene (C=C bond) considered a functional group?

I think it is, but I'd await confirmation from the higher powers.

Benzene also is...

higher powers?

And TSFX said they aren't "strictly" functional groups... but this was a VCAA sample exam so maybe it is