Multiple Choice (3 Viewers)

[Hayzazz]

Ahh shit I put 15c...

Alright, 18/20 is still OK I guess

 

[khorne]

No, there's 5. I'm not sure about 15, I put B as well. Could someone explain this?

Look at the titration curves for weak acid/strong base. It sets up a buffer system. The thing to consider is, how do you calculate molarity if you do not know how much has reacted? So tboth monoprotic weak and strong ionise fully in a titration, as OH ions are added to shift the equilibirum for the weak one.

 

[lolcakes52]

15 is a, the volume is determined by the number of moles, which is determined by the number of protons in the acid. Since they are the same the volume is also the same.

 

[jamesfirst]

Dude...m8...cool your jets

You have the following:

Cl and Br on the same C at the end and at the start is the same, so thats one. Then on the middle, that makes total of 2. You have Cl on end, br on other end which is another 1 (doesn't matter which side) so thats 3. Then you have Br in middle, cl on one end and Br on one end, cl in middle, that makes 2 more, so 5. It's 5. The rest are repetitions.

15 is not B because weak acids ionise fully in titrations with strong ones.

wtffffffffffffffff I HAVE 6 LOOOOOOOOOOOL WTF IS GOING ON.


I'll draw it and post it see if it's right.

 

[khorne]

wtffffffffffffffff I HAVE 6 LOOOOOOOOOOOL WTF IS GOING ON.


I'll draw it and post it see if it's right.

Its because you are repeating them you dork. Look hard, one will be a rotation of another

 

[CalumGemmell]

Look at the titration curves for weak acid/strong base. It sets up a buffer system. The thing to consider is, how do you calculate molarity if you do not know how much has reacted? So tboth monoprotic weak and strong ionise fully in a titration, as OH ions are added to shift the equilibirum for the weak one.

Dammit, oh well, thanks anyway.

 

[KatherinePierce]

So I got a message from Riproot and this is what his teacher thinks the answers are. Not what he got but what his teacher thinks:

1. A
2. D
3. C
4. D
5. B
6. A
7. B
8. C
9. D "the covalent bond constantly changes place so it isn't always polar or something like that"
10. C
11. D
12. C or D (not sure because the term "electrochemical activity" is on the syllabus)
13. C
14. C (Draw ALL of them)
15. A (They're both monoprotic acids so they react in a 1:1 ratio)
16. B
17. D
18. A? (factor of 5 in working)
19. B
20. B

 

[jamesfirst]

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probably wrong but still please have a look.

 

[utanobeiiby]

yes it is..

 

[lolcakes52]

So I got a message from Riproot and this is what his teacher thinks the answers are. Not what he got but what his teacher thinks:

10 is a, 11 is c, 18 is definitely a

 

[khorne]

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probably wrong but still please have a look.

ummm you do realise your branched ones are the same as straight chained ones? Just because it has a branch doesn't make it different

you take the longest carbon chain when naming. In all of those, it's 3.

 

[jamesfirst]

ummm you do realise your branched ones are the same as straight chained ones? Just because it has a branch doesn't make it different

fk........................


okay


what about 9 ???? Isn't it because it's polar ?

 

[Nympha]

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probably wrong but still please have a look.

There are no methyl chains. When you name carbon compounds, you take the longest C chain - in each case its still 3 so its always gonna be something propane.

 

[khorne]

fk........................


okay


what about 9 ???? Isn't it because it's polar ?

Yes thats what I said

 

[lolcakes52]

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in clockwise starting from top left.
2 is identical to 4( or your picture is wrong), 1 is identical to 5, 2 is identical 6
probably wrong but still please have a look.

 

[jamesfirst]

in clockwise starting from top left.
2 is identical to 4( or your picture is wrong), 1 is identical to 5, 2 is identical 6
probably wrong but still please have a look.

Nah, I got it wrong ... oh well..

 

[KatherinePierce]

Fixed.

So I got a message from Riproot and this is what his teacher thinks the answers are. Not what he got but what his teacher thinks:

1. A
2. D
3. C
4. D
5. B
6. A
7. B
8. C
9. D "the covalent bond constantly changes place so it isn't always polar or something like that"
10. C
11. D
12. C or D (not sure because the term "electrochemical activity" is on the syllabus)
13. C
14. C (Draw ALL of them)
15. A (They're both monoprotic acids so they react in a 1:1 ratio)
16. B
17. D
18. A? (factor of 5 in working)
19. B
20. B

 

[nano2e]

I think 15/20, same as what I got in trials so pretty happy. Q20 is so gay though, i thought it meant 400g was used in each reaction, not 400g split between them -.- It isnt worded properly ffs.

 

[Aindan]

I think 15/20, same as what I got in trials so pretty happy. Q20 is so gay though, i thought it meant 400g was used in each reaction, not 400g split between them -.- It isnt worded properly ffs.

but it was worded properly, you just misread =/

 

[Nympha]

I think 15/20, same as what I got in trials so pretty happy. Q20 is so gay though, i thought it meant 400g was used in each reaction, not 400g split between them -.- It isnt worded properly ffs.

Hahaha I did the same for q20, so annoyed at myself. Didn't read it properly