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5:16 AM
@YashasSamaga excellent idea :-)
 
There was one already I see.... hmmm

 The JEE LaunchPad

We love Physics, Chemistry and Mathematics. And we are prepari...
but that one is in math se
 
 
3 hours later…
7:48 AM
Hii @YashasSamaga
Which room should we use
 
8:01 AM
it doesn't matter
but let's keep the JEE spam out of the h-bar
 
I don't understand whats the problem
 
the conversations go very long and we keep asking countless questions which are related to few of us only
the convos get mixed up
difficult to read
 
Welcome @0celo7
 
8:35 AM
@YashasSamaga Isn't the same applicable for most other conversations in hbar? When Balarka, 0celot and Slereah talk about differential geometry even then very few people understand. And what do you mean by "JEE spam" ? If you don't respect the subjects you are learning then there is no point in learning. If differential geometry, music, cooking can all have intense discussions in the hbar so can we. I am strongly against dividing the room into topic based fragments.
 
Agree
 
8:56 AM
Hii @anonymous
You there
@YashasSamaga
 
The option where you won't lose something as gas will be the answer
 
Yeah
But how to find that
 
You lose Cl2 in NaCl
You lose NO in AgNO3
You lose O2 in CuSO4
You lose O2 in K2SO4
I am still thinking
 
In solution it is given as
 
This wasn't a JEE Main question, right?
 
9:08 AM
K2SO4 is a salt of a strong acid qnd strong base . So the electrolysis of K2SO4 is the electrolysis of water
 
OMG
You won't get O2 in K2SO4
^ from K2SO4
 
@YashasSamaga it came in one of my jee main test paper
 
you get H2 and O2 from water
I'm still confused though
nvm
I just messed this up
I suck at Inorganic chem
I need to start revising
The last time I studied chem was in September
 
I hate inorganic chemistry
 
me too
too much to remember
ask anonymous chem questions
he is studying chem atm
 
9:13 AM
He is not visiting this room
 
@Koolman ask anonymous in the h bar
Yay!
 
???
 
anonymous joined, waiting for his answer
I'm going to start a chem revision from tmrow
 
9:28 AM
@Koolman: write down the reactions at the anode and cathode and it should be obvious which ones don't generate a net change in the H+/OH- ratio.
 
does the ratio change?
 
NaCl
 
NaCl + H20 -> NaOH + HCl
 
@JohnRennie it will depend on petential values (E°)
 
the OH and H stay as OH- and H2O+ ions
in K2SO4, H+ and OH- combine to form H2 and O2?
hmm you get Cl2 gas in NaCl too :/
 
9:32 AM
@Koolman yes, but there are only a few of them to learn. For example take copper sulphate. The copper plates on the electrode i.e. $Cu^+ + e \rightarrow Cu$ so there is no change in pH. At the other electrode $OH^-$ is reduced not $SO_4^-$ so we end up with $H_2SO_4$ being produced.
 
We can solve it easily if we know E° values @YashasSamaga
@JohnRennie so we have to learn them
 
The questions will normally only involve the more common ions so it's easy to remember in what order they lie i.e. which of them is oxidised or reduced.
@Koolman if the question gives you the electrode potentials then that's great. If the question doesn't give you the electrode potentials then yes you have to learn them. But there aren't many to learn.
 
@JohnRennie so what does the solution mean
How does it explains the answer
 
@Koolman are you happy that electrolysis of copper sulphate does change the pH as I explained above?
 
@Koolman I was thinking of net charge in the solution. It is about pH lol
Now everything makes sense
 
9:37 AM
@JohnRennie yeah
 
In NaCl, you get NaOH, in CuSO4, you get H2SO4, in AgNO3, you get Ag(OH)2
 
It would be acidic
 
I need to start studying chemistry
that was really stupid of me to think about charges in this question
 
OK. Silver and copper are pretty similar, and sulphate and nitrate are pretty similar. So silver nitrate behaves in a similar way to copper sulphate. The silver plates out and at the other electrode oxygen is given off. We end up nitric acid. OK so far?
 
Yes
 
9:39 AM
Oh, you don't get Ag(OH)2... -.-
 
@YashasSamaga we will get it
According to electrode potentials
 
Now sodium chloride. At the anode chlorine is reduced and bubbles off $$Cl^- \rightarrow Cl + e$$ So no pH change there. At the cathode water reacts instead of the $Na^+$ so we get $OH^-$ produced. So we end up with NaOH and the pH goes up. OK so far?
 
@JohnRennie chlorine will oxidise
 
Since I wasn't sure I Googled it, and the anode reaction reduces $Cl^-$ to gaseous chlorine.
 
If it get reduced it will take electrons
 
9:46 AM
Oops sorry! Yes, the oxidation state goes from -1 to 0 so it gets oxidised.
Oh well, it was 35 years ago I last studied chemistry :-)
 
Np
 
huh I haven't studied chemistry since 6 months and I've forgotten everything except organic chem
 
So it just remains to look at potassium sulphate. Can you give the half reactions for that?
 
does that get electrolyzed?
before all the water has been electrolyzed
 
At cathode $2e^- + 2H_2O \rightarrow H_2 + 2OH-$
 
9:50 AM
OK ... and at the anode?
 
At anode $2H_2 O \rightarrow O_2 + 4H^+ +4e^-$
 
So the overall reaction is?
 
$2H_2 O \rightarrow O_2 + H_2$
 
And does that change the pH?
($2H_2$ not $H_2$)
 
Nah
@JohnRennie sorry
 
9:55 AM
So there's your answer. Potassium sulphate is the only case where the pH doesn't change :-)
 
Yeah I got that
I want to what they mean by the solution given
 
To answer the question you have to just remember the relative reactivities of the ions involved, but that isn't too hard as there aren't that many ions that will be mentioned in this type of question.
 
Yeah
 
@Koolman I'm not sure what you're asking ...
 
48 mins ago, by Koolman
K2SO4 is a salt of a strong acid qnd strong base . So the electrolysis of K2SO4 is the electrolysis of water
 
9:58 AM
Hmm, I'd say that's not a helpful statement.
 
Yeah i also think that
 
Though I suppose electrolysis of an acid stronger than water normally produces $H_2$ and electrolysis of a base stonger than water normally produces $O_2$.
 
May be
 
The way to approach questions like that is to work out what the half reactions are. Once you've done that the answer should be obvious.
 
Yeah
 
 
4 hours later…
1:42 PM
I tried it as

(number of ways selecting 4 girls out of 5)(arranging them on their seats )(ways we can find four consecutive sear for these girls)(selecting 10 seats from the remaining 12 seats )(arranging students in these 10 seats)

$=(^5 _4)(4!)(4)(^{12}_{10})(10!)$ = $^{11}P_6(6!)(2)$

But the answer is not this
Any idea @YashasSamaga
 
That's what I was doing.
It is lengthy.
3 cases
 
But what mistake I have done
 
(number of ways selecting 4 girls out of 5)(arranging them on their seats )(ways we can find four consecutive sear for these girls)
is not related to
(selecting 10 seats from the remaining 12 seats )(arranging students in these 10 seats)
and the first is wrong too
you don't consider the case where
you pick random 3 girls
and a girl comes in the 12C10 selection
 
Why , I have first arranged the girls , then the remaining
 
you are missing a case
 
1:53 PM
Which case
 
GGGG_ you do 5C4 to pick the 4 girls
you don't count a case where
GGGB_
and the girl takes that place
you have to do it case by case
 
@YashasSamaga it cannot be the case , as four girl should sit togther
 
The problem is you need 4 girls to be together
and you have 5 girls
you are undercounting in your method
 
What
 
oh wait
you are overcounting
not undercounting
I'm confused now
 
1:59 PM
My answer is smaller than that is given
 
I am totally confused
is your answer half of the correct answer?
GGGG_
_GGGG
are two different cases
if you compensate for that, you'll overcount
:/
 
Nope
 
The correct answer is given above
 
btw 46/2 = 23
and you don't have anything greater than 12!
so you can never get the correct answer by multiplying
you need to add
or subtract
 
2:08 PM
No there should be some mistake I have done
 
As 23 is a factor of the answer, it is impossible to get the answer by multiplying numbers.
As I said, you did not consider this case
GGGG_
_GGGG
you can place 4 girls inside a van in two different ways
but now you'll over count
 
@YashasSamaga its not 23
 
The answer has 46
23 is a prime
23*2 = 46
your answer does not have a factor of 23
 
@YashasSamaga i have considered that by multiplying it by 4
 
and it isn't possible to get that factor
you need to add/subtract numbers somewhere
@Koolman Your answer must be greater than the correct answer then
 
2:10 PM
In my it is 6!
 
2:24 PM
It is very easy. Answer is $(66)(10!)(4!)(4)(5)+(2)(55)9!(5)!-(55)(5)(4)!(4)(9)!$
$=110170368000$
 
@anonymous mistake in my method
 
@Koolman
@Koolman You overcounted.
 
What
 
First check if my answer is okay
With the answer key
 
36 mins ago, by Koolman
user image
 
2:26 PM
(You had to remove the cases where the 5th girl is by mistake grouped with the arranged 4 girls)
 
Why
It has asked atleast four girls
 
Does my answer match with (1) ?
Check please
@Koolman
 
your answer is correct @anonymous
 
I don't have a calculator
 
I did the same
idk if mine is correct
but I used the same method which his solution showed
 
2:28 PM
Great
 
I placed 4 girls and then did the case with 5 girls and then removed the overcounting
When the 5th girl is placed by mistake with the other 4 (in case 1)
Just number the seats from 1 to 16 and you're done!
 
Your answer is correct
 
There are only 4 ways in which 4 girls can stay together
 
@anonymous i have also counted that
 
And 2 ways in which 5 girls can stay together
Then remove the overcounting at the end
@YashasSamaga Did you get the answer as in the picture?
 
2:31 PM
I haven't checked
 
If it matches the answer key it should be correct
 
but I was using the same method which was given in his solution
 
Check it on a calculator
 
your numbers match mine exactly
 
Then it should be correct
@YashasSamaga
It was a easy sum
Explain your method to Koolman then
I have a test tomorrow
Gotta go
 
2:33 PM
I got yours method
Want to know my mistake
 
you are overcounting
 
But my answer is smaller than given
 
no it is greater
what nubmers did you use?
let me write ur equation
5C4 * 4! * 2! * 2! * 12C10 * 10!?
in that
you are overcounting
this case
G1G2G3G4_
_G2G3G4G5
note that the first _ can be G5 and the second _G1
you count it twice
 
Oh god
Yeah @YashasSamaga
Got it
Thanks a lot @YashasSamaga
 
I should have used numbers earlier ,-,
instead of GGGG_
_GGGG
 
 
2 hours later…
4:19 PM
@anonymous
let's discuss here?
it is completely irrelevant now at the h bar
nobody is keeping track of our progress
 

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