@YashasSamaga excellent idea :-)

There was one already I see.... hmmm

but that one is in math se

Hii @YashasSamaga

Which room should we use

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

@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

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?

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

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

@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 :/

@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

@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

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

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 $2e^- + 2H_2O \rightarrow H_2 + 2OH-$

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

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 ...

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

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

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

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

o0

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

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

In my it is 6!

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

(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

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?

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

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

@anonymous

let's discuss here?

it is completely irrelevant now at the h bar

nobody is keeping track of our progress