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04:08
Hi @JohnRennie
I have a maths question can I ask it here?
04:48
@sanya You are welcome to ask. I'm not so good at maths questions but someone else may be able to answer.
Hello @JohnRennie !
Hi :-)
How are you feeling rn?
Okay:)
@khaxan It wasn't a great weekend but I feel better this morning.
04:50
Oh good! Get well soon then!!
Thanks :-)
@JohnRennie do you have time for a question?
This question
How far have you got with it?
04:54
If I know "the number of" all the possible functions that can defined from one set to another, how do I find the number of onto functions?
@JohnRennie HSac is weak. It dissociates partially. I found out the equilibrium conc. But then realised that H+ will be eaten up by the Buffer also the Sac- ions. Too many simultaneous changes taking place. I cannot understand how to set up the new equilibrium
@sanya I don't feel there is a direct way to find it, if you know the number of elements in domain and co-domain, there is a principle to find the number of onto functions.
@khaxan It's simpler than you think. The buffer keeps the pH at 3 and the dissociation of the HSaC cannot change this. So [H⁺] stays constant at 10⁻³ mol/litre.
@KavinIshwaran yes the number of elements are given thats how I found all the possible functions, but im lost after that.
@JohnRennie so h+ is 10^-3 at equilibrium. All of it contributed by saccharin?
05:02
No, the H⁺ comes mainly from the buffer.
@sanya There is a direct formula for onto functions that comes from inclusion and exclusion principle
13
Q: Number of surjective functions from A to B

Justin D.Am I on the right track? I am not sure about my reasoning... Number of surjective functions from $A$ to $B$ $$A = \{1,2,3,4\} ; B = \{a,b,c\}$$ We must count the surjective functions, meaning the functions for which for all $b \in B$, $\exists~a \in A$ such that $f(a) = b$, $f$ being one of those...

The equation is Kₐ = [H⁺][SaC⁻]/[HSac]
Yes?
Yes
@KavinIshwaran I haven't been introduced to that yet, my module tells me to find the number of into functions and subtract it from the total no. Of functions. I do not understand why
05:05
Got it!
And the buffer keeps [H⁺] constant at 10⁻³ so we get:
10⁻¹² = 10⁻³ [SaC⁻]/[HSaC]
@khaxan OK :-)
Yes, (Sacharin dissociation is suppressed no?)
Correct!
Is there any strategy you apply in such questions. For eg: questions where common ion effect plays a role and there are simultaneous interactions. How do you find the equilibrium concentrations?
@sanya by inclusion - exclusion, he meant them as a counting principle. For finding a set containing elements satisfying A or B, you find the elements satisfying A and add them to the elements satisfying B and subtract those element which both satisfy A and B.
So addition and product rule?
05:13
Did you see the answer I linked? It contains the formula you wish to know
@sanya n(Aub)=n(a) + n(b) - n(a intersection b)
I dont get why cardinal number of A intersection B =number of onto functions @khaxan
This does give me some idea ill try again
It is just a technique, don't confuse it with the formula!
05:57
Hey @JohnRennie Sir !
Hi :-)
Could you please help me with this ?
1
Q: The Work-Energy Theorem in Non-Inertial Frames: A Critical Analysis and Misconceptions

TonyPhysicsloverIt is evident from these queries and responses that the Work Energy Theorem holds true in non-inertial frames (with work performed by pseudo force included in the equation). Is work-energy theorem valid in non-inertial frames? Work associated with pseudo force Now if we Consider the following s...

06:31
@JohnRennie Hi !
Hi :-)
How is your cold ?
I'm feeling better this morning :-)
Great :-)
I am suffering from viral fever, couldn't able to think much :-)
:-(
Get well soon!
06:37
Thank you :-)
I really hate the way that your head is fuzzy when you're feeling unwell and you just can't concentrate. People were asking me questions and I just couldn't concentrate enough to answer them.
07:02
@JohnRennie hi sir.. How are you ?
Hi :-)
I had a bit of a cold over the weekend but I'm feeling much better now :-)
Are you good ? seems like you got cold. Take care. Get well soon :)
@JohnRennie That's good :)

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