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12:03 AM
Hello could someone help me with an econ problem?
I think I did the math wrong but idk where
 
 
7 hours later…
7:31 AM
@JohnRennie sir hi
Is it right to say that in a bottle there are $N_A$ H20 molecules
 
@Sarabsrimt hi :-)
@Sarabsrimt what mass of water is in the bottle?
 
18gms
 
Then yes, the bottle contains $N_A$ water molecules
 
Then , mass of each water molecule = 18gms / $N_A$
 
Yes
 
7:33 AM
Ohk. Thanks sir.
Just wanted to confirm
Hope you have a great day
Can we say atomic mass means mass of $N_A$ molecules or atoms . For example atomic of Na means mass of $N_A$ atoms
which is 23 grams
 
The atomic mass is the mass of one atom, so for sodium it is $23/N_A$ g
 
Ohk
 
The molecular mass is the mass of one molecule so for water it is $18/N_A$
 
Ohh
Molar mass for total ?
 
The molar mass is the mass of $N_A$ molecules. So the molar mass of water is 18g.
 
7:40 AM
Ohk. In my book it says totally opposite.
 
Really?
 
Like for atomic mass it is written wrong.
Yeah
Many users on stack exchange told me that my book uses old knowledge
Throw that book. Haha
Also , for molar mass . Specific examples is given. Atomic mass of Na = 23gms and atomic mass of cl = 35.5gms . So Molar mass is 58.5 gms
is this right
 
You mean the molar mass of sodium chloride is 58.5g?
If so then yes that is correct.
 
Yes .
It says molar mass is of a salt and not element
 
I'm going to make a coffee. I'll be a few minutes. Back soon!
 
7:43 AM
But now I know it’s wrong
ok sure
@JohnRennie Coffee is great
@JohnRennie if molar mass is mass of $N_A$ molecules , then what is for atoms ?
Or is the definition same
For this Q , the reaction will be
C + 2O —> CO2
Now , isn’t it that no of O atoms are 2 mole since the balancing coefficient is 2
@JohnRennie
If yes , then 4 mole of Carbon atoms and 2 mole of oxygen atoms are present
but require the is that 4 mole of C atoms require 8 mole of O atoms.
 
7:58 AM
2 moles of oxygen atoms is the same mass as 1 mole of O₂ molecules
 
Then , we don’t have that much of O atoms
@JohnRennie k
What we do is that we say 2 mole of oxygen atoms require 1 mole of carbon atoms
therefore CO2 needed is 1 mole
is this correct sir
This Q is by my sir. His answer is different
 
30 secs ago, by Sarabsri mt
therefore CO2 needed is 1 mole
CO2 needed?
 
CO2 produced
not needed. Sorry for that sir
 
You have 16g of oxygen. Yes?
 
Yes
 
8:01 AM
That's one mole of oxygen atoms, or alternatively ¹⁄₂ a mole of O₂ molecules. Yes?
 
Yes
 
So how much CO₂ can be produced from 16g of oxygen?
 
1/2 mole
since O2 is there
C + $O_2$ —> CO2
 
Correct
 
4 mole of carbon and 1/2 mole of O2 is present
C requires 1/2 mole of oxygen only
1/2 mole of carbon requires 1/2 mole of O2 and gives 1/2 mole of CO2 ?
Is this right sir
 
8:05 AM
Yes. So if you react 48g, i.e. 4 moles, of C with ¹⁄₂ mole O₂ you will get ¹⁄₂ mole of CO₂ and there will be 3¹⁄₂ moles of carbon left unreacted.
 
Yes.
4 mole so f carbon also don’t react irght
1/2 mole of carbon only reacts
?
 
Yes
 
Ok. Sorry for little Twist here.
If I take the reaction as C+ 2O —> CO2
Which is in real taken by my sir
since 2O is present.
4 mole of carbon needs 8 mole of oxygen atoms
?
 
Yes
 
1 mole of carbon needs 2 mole of oxygen atoms
But we don’t have 2 mole of O atoms
So , 1/2 mole of C needs 1 mole of O atoms
which gives 1/2 mole of CO2
 
8:08 AM
Yes
 
Ok.
Sor , by saying 1 mole of O atoms
We have in real and which is needed 2O atoms
So , every O atom gives 1 mole or 1/2 mole
It’s like , I’m asking if 2O means 32 gms.
O + O
16 + 16 gms
 
Yes
 
Ok. Then , if we have 1 mole of O atoms.
that means 2O is 2 mole of O atoms
every O + O = 1 mole + 1 mole
 
Yes
 
So , do we reduce this to 1/2 mole + 1/2 mole.
8 GMs + 8gms contribution by each O atom
since total we had is 16gms
Form the Question
 
8:12 AM
If you have ¹⁄₂ a mole of carbon then ¹⁄₂ mole carbon reacts with ¹⁄₂ + ¹⁄₂ = 1 mole of oxygen atoms.
 
Yes. So 8 + 8 gms
@JohnRennie wonderful. I think I got .
Also , If I say I have 16gms of oxygen
does it mean 1 molecule or 1 atom
 
It depends. Unless it tells you whether it's oxygen atoms or oxygen molecules you can't say.
 
Ok
like in my Q. It didn’t say
O2 = 16gms was written specifically
But not oxygen atoms or molecule .
Ok I got it too much better now.
Thank you a lot sir
 
:-)
 
Have a great day sir. May god bless you and give you great health
See you soon
 
psa
8:24 AM
@JohnRennie I'm having trouble getting something into a form that I can solve for. I think I've gotten really close, but I may have done some incorrect approximations or something along the way.
Do you have a second?
 
@psa hi :-)
 
psa
hey
 
Yes, what's the problem?
 
psa
let me upload the question and my work, one sec
I know that I'm really close because I know what the answer should be
Sorry, it's a bit long, but hopefully you can just scan through it.
I'm pretty sure what I should be getting in the end is $\ddot\alpha = -\frac{m+M}{M}\frac{g}{\ell}\alpha$ (this is what I got when I used Lagrangian mechanics)
which is very close to that final answer, but obviously it's not quite that
 
The x position of the centre of mass cannot change because none of the forces acting have a non-zero x component. Yes?
 
psa
8:32 AM
yeah
 
So I would put the COM at x = 0. Then at any time the position of the top end will be $x(t)$ and the bottom end $-x(t)$.
The angle $\alpha$ is then given by $\sin\alpha = \frac{x(t)}{\ell/2}$
The torque is $\tau = x(t)mg$, or $\tau = \tfrac12 \ell \sin\alpha ~ mg$
Oh wait ...
Sorry I hadn't realised it was two masses on a weightless rod. I assumed it was a uniform rod of mass $m$.
 
psa
ah no worries, yeah it's just a mass $M$ at B and $m$ at A
 
That just means the position of the COM isn't at the centre of the rod. The basic idea is the same.
Let me draw a diagram ...
@psa that's what I had in mind.
The COM always lies on the y axis so $a$ is the distance from the top end of the rod to the COM.
The distance $a$ remains constant as the rod oscilaltes.
 
psa
alright I'll give that setup a try and see if it's not as much of a mess
all that should be necessary to solve is conservation of momentum in $\mathbf{\hat x}$ and conservation of energy
 
There will be a tension in the rod as well. So the total normal force on $m$ won't be $mg$.
 
psa
8:47 AM
I'm pretty sure my energy and momentum previously were correct with the origin at O, but for some reason I just wasn't getting the right answer in the end.
or at least I assume it's incorrect
 
@JohnRennie Hi :-)
 
@Wolgwang hi :-)
 
in JEE Chemistry Club, 7 mins ago, by Wolgwang
For example, yesterday my coach completed the "vectors" chapter but due to some reasons I have not tried all the questions of the module. This coming Sunday, he would start the next chapter (id est Kinematics ), and I don't think I would be able to solve all the questions till then.
After the inception of that chapter, I would be doing the questions of that chapter, so when would I practise the questions of the previous chapter?
Any inputs here
 
You mean what do you do about completing all the problems from the previous chapter?
 
@JohnRennie No
@JohnRennie I haven't.
 
8:58 AM
I'm not sure what you are asking ...
 
OK lemme make it more clear
I should have completed all the questions of the previous chapter, but I haven't.
And when the next chapter starts, I won't be able to do so.
 
So what are you asking me? How to complete the questions from the previous chapter?
 
@JohnRennie Yes.
 
I don't think there is a magic solution.
Look through the remaining questions and try to prioritise them i.e. see which ones will teach you most. Do those questions first, then any questions you don't have time for you can keep to use for revision nearer the exam.
 
@JohnRennie I wasn't expecting that :-)
@JohnRennie Thanks :-)
 
9:10 AM
@JohnRennie hi :)
 
@AshishAhuja hi :-)
 
How far have you got with this?
 
Could you help me with finding the velocity gradient in the above question? I believe I can find the torque using integration myself, but I'm struggling with the gradient.
 
Incidentally, when a question says the angles are small it usually means you can use the approximation sinθ = θ
 
9:12 AM
yup
 
The reason there is a gradient is that when a fluid is in contact with a surface it is stationary with respect to that surface.
 
ohh I think I got it.
 
OK :-)
 
Let me give it a try...
yup it worked. Now I'll try the integration...
 
:-)
I think I answered a question on this. I used to do this sort of thing for a living when I worked as a scientist. Commercial viscometers generally use the cone and plate geometry.
Aha!
0
Q: how the form factor (stress - torque) is derived?

Thomson1I am working on an experiment of rheology and I need to calculate shear stress in order to calculate the viscosity. After some research I found that for the type of viscometer I will be using (cone-plate), the stress is calculated by dividing the torque given by the apparatus by a form factor equ...

 
9:27 AM
@JohnRennie that seems interesting!
 
@AshishAhuja it was good fun :-)
 
@JohnRennie Atomic mass = mass of one atom. So , does atomic weight of Na = 23gms
 
I'm not a chemist, so my advice on this may not be 100% reliable. But I would say the molar mass of Na is 23g.
 
Ohk. Yes sir. Thank you.
 
You will often read statements like the atomic weight of sodium is 23, but this means 23 atomic mass units - often abbreviated to AMU.
One AMU is the mass of a hydrogen atom.
 
9:36 AM
Ok.
 
Hmm I'm making some mistake in the integration. I'm getting the right answer but the $\theta$ in the denominator is missing. I'll type out what I'm doing, it's probably something dumb I'm missing...
 
Sir , AMU or g / mol.
 
AMU
 
The reason they give both is that the number *23* is both:
1. the atomic weight in AMU
2. the molar mass in grams
 
9:38 AM
Ohk sir
Getting it.
 
I'm considering a ring at a "height" $x$ from the bottom. The radius of this ring is then $\frac{x}{\theta}$ and the thickness ("height") is $dx$. Surface area $$\frac{2\pi \times x dx}{\theta}$$
 
Do you mean a ring of radius $x$?
Like I used in the answer I linked above?
 
I saw your answer, and the approach I tried first was different, and that's what I'm typing here.
I mean a ring of height $dx$, and then I'll integrate from $0$ to $R \times \theta$
Since $tan \theta = \theta$
Or does this make no sense at all?
 
9:43 AM
Oh, OK, so it is basically a ring of radius $r$ but you're using $h = r\theta$ as the variable to integrate.
 
Yes. To me naturally integrating over height made more sense, but you can express it like that as well.
 
This seems like doing the calculation the hard way ...
 
Hmm...
It's just two more steps, I'll type it out, if that's okay?
 
OK ...
 
Continuing from
5 mins ago, by Ashish Ahuja
I'm considering a ring at a "height" $x$ from the bottom. The radius of this ring is then $\frac{x}{\theta}$ and the thickness ("height") is $dx$. Surface area $$\frac{2\pi \times x dx}{\theta}$$
$$dF = \frac{\eta 2 \pi x dx \omega}{\theta \times \theta}$$
Now using $dT$ for torque:
$$dT = dF \times \frac{x}{\theta} = \frac{\eta 2 \pi x^2 dx \omega}{\theta ^3}$$
Integrating over the limits of
0 to $R \theta$
we get $$T = \frac{\eta 2 \pi R^3 \omega}{3}$$
now I've made a mistake somewhere since the $\theta$ is missing...
I'll go try the method given in your answer I guess.
 
9:52 AM
@AshishAhuja Let's start with the relations $r = x/\theta$ so $dr = dx/\theta$. We'll need these in a moment. OK so far?
 
ok. Here $h$ is what I was using $x$ for right?
 
Oops, yes.
 
Ohh then I think I got my mistake.
 
:-)
 
I believe I'm going wrong in the surface area calculation.
 
9:54 AM
Yes. And you're going wrong because you picked such a hard way to do the calculation!
 
oh well sometimes the hard way is what comes naturally to mind :D
 
:-)
 
I had actually thought about integrating in the way you've done in the answer when I first got the question in a test, but what confused me is since we're considering a flat annulus, how do we justify the surface area = $2 \pi r dr$?
 
When we calculate the torque we will be taking the cross product of $\mathbf r$ with $d\mathbf F$. Yes?
 
yes
ohh got it
 
9:59 AM
And this means all that matters is the component of $d\mathbf F$ normal to $r$ i.e. the vertical component.
And that's going to end up being proportional to the vertical component of the vector area.
 
Yup. That clears it all up, thanks a lot.
 
:-)
 
@JohnRennie quick Q
Free sir ?
My confusion is with amount of O2 reacted.
2 mole of Na to react is given. That is the need and that has been done.
So , 2 mole of Na reacts with 1/2 mole of O2
Why does my teacher even think to write this ( red circle ).
 
@Sarabsrimt I guess they just thought it would make it clearer ...
 
Ohk. We can’t we say 1 mole of Na react with 1/4th mole of O2 right
Since requirement is 2 mole of Na
 
10:08 AM
Well the mole ratio is 4:1
So that could be one mole Na and ¹⁄₄ mole O₂
Or 2 moles Na and ¹⁄₂ mole O₂
 
Ohk ohk.
 
Or 4 moles Na and 1 mole O₂
 
Ohk
we can say this as ratio
 
In this case you have 2 moles Na, so it's 2 moles Na reacting with ¹⁄₂ mole O₂
 
but asked in Q is 2mole
ohk got it
 
10:10 AM
@Sarabsrimt are you trying to solve it using equivalents?
 
Yes.
 
Ah ok, so first you'd write down the oxidation states of the reactants and products since this is a redox reaction.
Na and O2 both have a zero oxidation state on the reactant side.
On the product side Na has an oxidation state of +1 while oxygen has a state of -2
 
Ohk
 
Now we find the n factor for Na and O2. Should I finish the whole thing?
 
Umm. I know this part actually and have written it too. Eq of O2 = 1/2 mole * 4
My Q was related to amount of O2 would react with Na
Yeah, thanks though.
@JohnRennie Thanks a lot. I got it now.
Pretty interesting chapter.
 
10:15 AM
@Sarabsrimt ah ok for that either you could write a balanced chemical reaction (which is much easier than equivalence for this case) or you'd equate Equivalence of Na = Equivalence of O2
 
@AshishAhuja yeah, I got it now.
 
@Sarabsrimt :-)
 
For second reaction
Al + ...
Mole ratio is 4:3
 
Same thing. In any reaction the number of equivalence is the same for all reactants and products.
 
That is because of mole ratio right
 
10:20 AM
No.
 
What is Al reacted was 4 equivalents
 
Then equivalents of O2 will also be 4. You can try it out by actually calculating the moles.
 
Hmm ok.
But answer changes here. @AshishAhuja
if I took eq of Al as 4
 
Give me a second, let me try it out on paper myself...
 
Ohk. Sure
 
10:24 AM
@Sarabsrimt uhh you're getting number of moles of O2 = 1, right?
 
Ohh ohh. My mistake.
yes O2 moles = 1
 
Ok. And then n factor (or valency factor) = 4, so we get number of equivalents of O2 = 4 * 1 = number of equivalents of Al.
 
Yes. But I’m 1st Q
no fo eq of O2 = 2
Not 4
 
5 mins ago, by Sarabsri mt
if I took eq of Al as 4
Aren't we talking about this?
 
Yes
right
I thought you meant to say that equivalent of O2 would be 2 alwasy
@AshishAhuja here
Did you mean to say equivalence only for 1 reaction.
 
10:29 AM
Ohh no. I meant to say that the number of equivalents would be the same for all reactants and products for a given reaction, that's the basis of equivalent concept, and that's the reason n factor is defined the way it is.
For example for the reaction A + B -> C + D
 
Ohk
 
no. of equiv. of A = no. of equiv. of B = no. of equiv. of C = no. of equiv. of D
 
Ohk. Got it.
Thanks a lot.
@AshishAhuja ✌️
 
N factor of Na2O = 2 right
Na2 = 2 and O = -2
 
10:32 AM
It matters. If you're talking about it in the reaction above, then no. As a standalone salt, yes.
definition of N factor is reaction dependent.
 
Ohk.I ask for the reaction since that is what I need for so living equivalents
 
For that reaction, first we write the oxidation states.
(reactant) Na = 0, O = 0
(product) Na = +1, O = -2
Now, we need to equate equivalence of O2 = equivalence of Na
 
Ohk
 
for Na, number of electrons involved when one molecule of Na underwent reaction would be 1 (change in oxidation state)
similarly for O2 it would be 4 (two for each oxygen atom)
so nfactor (Na) = 1, nfactor(o2) = 4
 
Yes
 
10:36 AM
now n factor(Na) * moles of Na = n factor(O2) * moles of O2 = number of equivalence; and now you'd get the answer.
 
One thing before this.
$Na_2$ = 2 Na in terms of no of moles right.
2 moles of both
?
 
Uhh Na2 does not exist.
 
Ohk. But if it did . Take H2 for example.
2H and H2
Both have 2 moles
 
You could say so, yes. But you need to be specific that it's "2 moles of H atoms".
 
Ohk.
Our reaction is Na2O right
@AshishAhuja here. 2 * 1 for Na
How do we know moles for O ?
 
10:40 AM
@Sarabsrimt yes, 2 because 2 moles of Na and 1 because of the n factor.
 
For no of moles of O
 
you mean O2?
 
Yes
O2
 
Ok.
We equate the number of equivalence.
10 mins ago, by Ashish Ahuja
no. of equiv. of A = no. of equiv. of B = no. of equiv. of C = no. of equiv. of D
We have number of equivalence of Na = 2 * 1 as you said above.
 
2 * 1 = x * 4
 
10:41 AM
yes
 
x = 1/2 mole of O2
?
Oh great.
Btw. This is how my teacher told me about n factor
Same way. But I didn’t understand it well enough.
Noe I did more
Ohk. Thanks a lot @AshishAhuja
Only problem was that my teacher said even - 6 we can write
As n factor.
See you later then.
 
n factor is more or less a trick which used to be used by chemists at a time when chemical equations weren't known. So I guess you should take the mod value of n factor. Saying n factor = -6 and getting negative moles of O2 does not make much physical sense.
 
Yeah . Right
 
Bye.
 
 
2 hours later…
12:46 PM
waiting for 7th March ?
No of electrons = summation of atomic number * no of moles * $N_A$ is the formula in my textbook.

Q is to find no of electrons in 5gm of CaCO3 .

No of electrons = 5/100 * $N_A$ * 50 is the answer.

My quick confusion is that if I ask no of electrons in CaCO3 = 1 mole * 50(atomic number ) as no of electrons . Why $N_A$ was not multiplied.

Also , the derivation of the formula . I am not getting how to solve it.
Any idea how we can derive this formula.
 
1:34 PM
I solved it.
 
 
2 hours later…
3:41 PM
Can we say mass of an atom = mass of nucleus ( proton + neutron ) + mass of electron? I’m sure it is not enough but what else do we need ?
 
 
2 hours later…
Sid
5:50 PM
@Sarabsrimt at your level, I think that's the simplest way to understand it.
In fact, you can even think of it as simply the mass of nucleus itself. The protons and neutrons are massive compared to the electrons.
 
@Sid Mass of nucleon means mass of one proton or neutron.

$$\text{No of nucleons} = \frac{\text{mass of one atom} }{\text{Mass of one nucleon}}$$

Now, I have confusion in this formula that why is it division. Mass of an atom = mass of electron + mass of proton + neutron. [![enter image description here][1]][1]Now, I am not exactly sure if this is right. Please correct me here.

It’s like in a bag there are 4 apples and 3 mango and 2 orange. Total = 9 fruits.

Then, total number of oranges are not equal to 9 / 2.
Can you please help in this
 
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