The h Bar

John Rennie

11:00

OK, shout if you want me to clarify what I mean.

Mew

Check this one out serebii.net/pokedex-sm/792.shtml

user228700

U'll probably hear me shout but ::still processing::

user228700

@JohnRennie: ::shouts:: I think it'll be quicker if u explained it. I'm not sure I'm thinking correctly about the situation...

user228700

@Mew DO NOT DISTRACT!! :-P

John Rennie

A liquid boils, i.e. turns to vapour, if the pressure is lower than its vapour pressure.

user228700

11:04

One moment.

Mew

:P

John Rennie

@Kaumudi ...

user228700

Okay, so when u say

user228700

> "boils, i.e. turns into vapour"

user228700

I have a problem with that :-P

user228700

11:06

A liquid turns into vapour even before it starts boiling...

Mew

me 2

^ what she said

Anonymous

@Kaumudi Do you know that boiling point is the temp at which liquid's vapour pressure =atm pressure ?

Slereah

So there's a user called "Time Traveler"

Anonymous

The boiling point of a liquid is the temperature at which its vapor pressure is equal to the pressure of the gas above it.The normal boiling point of a liquid is the temperature at which its vapor pressure is equal to one atmosphere (760 torr).

Slereah

He's been asking a lot of CTC questions lately

John Rennie

11:07

@Kaumudi if the pressure is kept lower than the vapour pressure then all the liquid will turn to vapour. Yes?

Slereah

You'd think he'd have looked up CTCs before going time traveling

Mew

@Slereah dpo you have some links, sounds interesting

user228700

@S007 'Course I do.

Mew

Give her more credit S007

Anonymous

@Kaumudi So where are you facing a problem in John's statement ?

user228700

11:08

@JohnRennie It's a little weird, thinking about changing the pressure and not the temperature but yeah, I agree.

Slereah

Time traveler

75 3

Mew

thanks slareash

how gracious of him to visit stackexchange of all places

John Rennie

@Kaumudi yes, it is a bit weird thinking about changing the pressure but keeping the temp constant, and in fact that's why this all seems a bit confusing.

Anyhow, does it make sense that if $P < P_b$ then all the liquid (both A and B) turns to vapour?

user228700

@S007 Well, he said turns into vapour = boiling. That was the problem.

Anonymous

@Kaumudi Yes, that is called boiling only !

Anonymous

11:11

Boiling may be induced by pressure change too

user228700

No. That is called evaporation. If it happens so that the saturation vapor pressure is equal to the atmospheric pressure at that temperature, then only boiling occurs.

Mew

indeed

take 70 degree water to everest

and it will boil

John Rennie

I think this is all getting a bit distracting ...

user228700

:-/

John Rennie

@Kaumudi suppose I put some pure B in an evacuated sealed vessel. Then B will evaporate until the pressure inside rises to $P_b$. At that point the vapour and liquid are in equilibrium. Yes?

user228700

11:14

Yep.

user228700

$P_b$ being its saturated vapour pressure?

John Rennie

OK, now I increase the volume of the vessel, so the pressure drops. When this happens more liquid evaporates and the pressure rises back to $P_b$. Yes?

Yes, $P_b$ is the saturated vapour pressure at our (constant) working temperature.

user228700

@JohnRennie Brain not able to digest this completely.

Anonymous

I can't agree with that. Evaporation can occur at all temperatures.
On average, a fraction of the molecules in a glass of water have enough heat energy to escape from the liquid. The reverse also happens — water molecules from the air enter the water in the glass — but as long as the relative humidity of the air in contact is less than 100% (i.e., saturation), the net transfer of water molecules will be to the air. The water in the glass will be cooled by the evaporation until an equilibrium is reached where the air supplies the amount of heat removed by the evaporating water. In an enclose…

John Rennie

@Kaumudi this is a key point, you need to be sure about this.

Anonymous

11:17

But what John was talking about was lowering of the BP itself by reducing pressure

Mew

@S007, you're distracting the main conversation

user228700

@S007: I'll get back to you.

user228700

@JohnRennie Yeah, Ik :-( ::Thinking::

Anonymous

Evaporation is a different thing @Kaumudi Okay, but I wanted to clear that misconception...we will get back to it later

Mew

@Kaumu, if you have a liquid in a jar, the liquid will evaporate until the partial vapour pressure in the jar is Pb

if you make the jar bigger somehow

the pressure in the jar is now less than Pb

so what happens?

user228700

11:19

@S007 I'm pretty sure this is a miscommunication and that both of us are clear regarding this concept. Still, I will get back to you to clear the miscommunication later.

Mew

more liquid evaporates

until the pressure in the jar reaches Pb again

if you expand the jar even more

the pressure drops again since PV = nRT

T and n are constant

so PV = constn

so if V increases, P drops

so if volume of the jar increases, pressure drops, allowing more liquid to evaporate

user228700

@Mew This. My stupid brain was unable to use this point. Atghf.

user228700

Okay, I understand.

Mew

yay

Anonymous

@Kaumudi I don't think it is a miscommunication. Till few months ago I had the same misconception about evaporation and boiling. Anyway I'm stopping here till the conversation ends.

user228700

11:20

Thanks @Mew. @JohnRennie: Re-reading your messages.

Mew

don't be too hard on yourself Kaumu

ur pretty smart

user228700

(Still reading)

Secret

@S007 unrelated question, thus I decided to ask you in a separate room

user228700

@JohnRennie: Okay. I agree with everything u said :-P

Anonymous

@Secret could you give the link to the room ?

Secret

11:27

$Mathematics$ Room for Secret and S007

John Rennie

@Kaumudi OK. So if I keep increasing the volume of my sealed vessel more and more of the liquid evaporates to keep the pressure at $P_b$. Yes?

user228700

Yes.

John Rennie

And if I keep increasing the volume until the last of the liquid has boiled then at that point the pressure starts to fall below $P_b$

So the conclusion is that if the pressure falls below $P_b$ there will be no liquid left. The whole system will just be vapour.

Mew

agreed

becuase there is no more liquid to evaporate and make up the pressure back to Pb

John Rennie

@Kaumudi Yes?

user228700

11:31

Okay. Have to back up and ask a really dumb question. Boiling→saturated vapour pressure of liquid=pressure of surroundings, yeah?

Slereah

@Qmechanic I'm not sure just adding a google search for a term really makes a post much clearer

user228700

So in this case...

John Rennie

I would avoid the word boiling

All we're talking about is the vapour-liquid equilibrium

Whether you describe the liquid-vapour transition as boiling or evaporation doesn't matter.

user228700

Sigh. Okay, one second, reading again. (I'm sorry, pls. bear w/ me...)

Slereah

From Time Traveler's questions I can tell he's trying to understand Hawking's 92 paper

unfortunate since my answer involved things from Visser's paper

Hawking's paper just used Misner space

It was a pretty vague conjecture

user228700

11:39

@JohnRennie Yes.

user228700

(Although, I will torture u another time, about how boiling relates to this, please rest assured :-P)

John Rennie

OK, what we've just proved is that if we hold the temperature constant and keep lowering the pressure then there is a pressure below which all the liquid turns to gas. For a pure liquid that pressure is the saturated vapour pressure.

Now let's consider what happens if we increase the pressure.

Once more start with my sample of pure B in my sealed vessel. So the pressure inside the vessel is $P_b$. Is that OK as a starting point?

user228700

@JohnRennie Okay, I think I finally understand. Phew. Okay.

user228700

@JohnRennie Yes.

John Rennie

OK, if I now compress my vessel the pressure goes up. Some of the vapour now condenses back into liquid so the pressuire falls back to $P_b$. Yes?

user228700

11:43

Yeah...

John Rennie

And if I compress it again the same happens, and again. As I keep compressing the vapour condenses into liquid to keep the pressure at $P_b$.

user228700

Oh God, the point I kept losing was the fact that at a given temperature, the saturation vapour pressure is constant!

John Rennie

But eventually all the vapour has turning to liquid and now the pressure starts to rise above $P_b$.

@Kaumudi aha, I suspect you've made the leap necessary to understand this :-)

user228700

:-) Okay. One moment...

user228700

Okay, will u be mad if I asked a dumb question again? Also, dyou have to go have lunch or s'thing?

user228700

11:50

Ah, it's only 11:50 AM. OK, does that mean that lunch comes a little later?

John Rennie

@Kaumudi I have to go out in about an hour, but I'm around until then. Ask your question - I bet it's not dumb :-)

user228700

Oh, I bet it'll (my question) piss u off :-P I'm asking about how, if we decrease/increase the volume of the container, won't its pressure have to be always equal to the surrounding pressure if it's a sealed container?

koolman

@Kaumudi @Kaumudi can you please tell me which book you are reading

user228700

@koolman Uhh, I will, if u tell me why u want to know so badly...

Mew

surround pressure?

do you mean surrounding the steal container?

John Rennie

11:54

@Kaumudi no because the container walls are rigid.

Mew

sealed*

John Rennie

After all, a diver's air tanks have a pressure higher then the surroundings.

Mew

well no because if it's say a glass container, there is no reason the inside pressure should be releated to the outside pressure at all

koolman

@Kaumudi because i want to also learn from that

Mew

@Kaumudi a balloon will tend to have the same pressure inside and outside

Because it's volue will change if there is a pressure difference (ignoring elastiicity for now)

but a steal container for instance won't move in response to pressurte differenc between inside and outside

so it is certainly possible for a container with strong walls (e.g. steal, glass etc.) to have differeing pressures inside and outside

user228700

11:56

@koolman Oh, well, you're not going to be able to buy it from the market :-| Its the material given by the institute Resonance.

koolman

Ohk

user228700

@JohnRennie This. Okay, okay.

Mew

Remember pressure = F/A thus inward force on wall = internal pressure * area of wall

similarly external force on wall pushing inwards = external pressure * area

If the container can withstand force, there is no reason for these to be equal

John Rennie

@Kaumudi Cool. Where we got to is that for our pure component B if $P < P_b$ then it's all gas and if $P > P_b$ then it's all liquid. Are you happy this is a good starting point?

Mew

But a baloon will expand or contract until inward force = outward force

user228700

11:58

@Mew: Okay, thanks!

user228700

@koolman You're in 11th, right? I can give u my books next year, if u want them.

user228700

@JohnRennie Yep.

koolman

@Kaumudi thank you but i am in 12

Mew

@Kaumudi will u give me the books?

user228700

@koolman Ah, I see. Okay.

Mew

11:59

I would like to sell them on the market

user228700

@Mew xD I think I'll do that meself, sir!

John Rennie

@Kaumudi and of course the same applies if we have pure A. In this case if $P < P_a$ then all the A is vapour and if $P > P_a$ then all the A is liquid.

Anonymous

@koolman You can buy resonance books from their website...dlpd.resonance.ac.in

koolman

@Kaumudi can you send me the screen shots of that

user228700

@JohnRennie Right.

user228700

12:00

@koolman Of what? O.o

koolman

Only this topic

user228700

The whole chapter?!

John Rennie

@Kaumudi now suppose I put a mixture of 50% liquid A and 50% liquid B in my evacuated vessel. Suggestions as to what will happen?

koolman

Not whole chapter just things about phase diagrams

user228700

Dude, these books suck at explaining things properly. @JohnRennie knows exactly how much they suck. In fact, he is bearing the brunt of this "suckiness" :-P

Mew

12:02

@JohnRennie I have a physics question will you help?

Qmechanic

@Slereah : A google search does indirectly (even if unsuccessful) diagnose something about a question. I agree with your comment under the question.

John Rennie

@Mew I think we've nearly sorted K's question. Can it wait a minute or two?

Mew

np

user228700

@JohnRennie Uhh, when we do what? If we change the pressure, u mean?

John Rennie

Well when we start with just liquid in our vessel both liquids evaporate until the vapour partial pressures are $P_a x_a$ and $P_b x_b$. That's what Raoult's law says.

user228700

12:04

@Mew Like I said before, @JohnRennie: Please consider cloning urself for the benefit of all students :-P

user228700

@JohnRennie Yes...

John Rennie

Because $P_b < P_a$ that means more B evaporates than A. So the liquid has a slightly decreased proportion of B and the vapour has an increased proportion of B.

user228700

Riight...

John Rennie

And the total pressure is $P_a x_a + P_b x_b$

user228700

Yes...

Mew

12:06

What is $x_a$

John Rennie

@Mew mole fraction of A in the liquid

user228700

@Mew Mole fraction of component A in the liquid mixture.

Mew

correct

user228700

—.—

John Rennie

@Kaumudi now suppose I expand my vessel as before. The pressure drops as I expand my vessel and both liquids start evaporating to increase the pressure again, just as happened in the pure liquid case.

user228700

12:08

Yes...

John Rennie

As before, more B evaporates than A.

And that means that the composition of the liquid changes.

user228700

Yes...

John Rennie

So if the mole fraction of B was $x_b$ before the expansion it is now $x'_b$ where $x'_b < x_b$. So far so good?

user228700

Yep.

John Rennie

And likewsie the liquid has become enriched in A so $x'_a > x_a$

user228700

12:11

Yes...

John Rennie

the pressure is now (Raoult's law again) $P' = P_a x'_a + P_b x'_b$

user228700

Yes...

John Rennie

But remember that $x'_b \ne x_b$ and $x'_a \ne x_a$ so the pressure has changed

user228700

Yes...

Mew

So Raoult's law only applies momentarily?

it is not an equillibrium result?

John Rennie

12:13

And this is the key difference between a mixture and the pure component. For a mixture the pressure can change while the liquid and vapour coexist.

Mew

oh sorry

I missed the "expand vessel" part

user228700

@JohnRennie Riight...

user228700

@Mew Huh?

Mew

WEll @JohnRennie said the he expands the vessell

and then he applied Raoult's law again. I initially didn't read that bit

John Rennie

@Kaumudi Suppose we start at a very high pressure, well above the vapour pressure of either component, then no vapour will be present because $P > P_a$ and $P > P_b$. Does that now make sense?

Mew

12:16

sup @heather

heather

@Mew, hello

user228700

@JohnRennie Yep!!

Mew

hows pokemon goin

user228700

@heather: Ello :-)

heather

@Kaumudi, hi =)

John Rennie

12:18

@Kaumudi Now we decrease the pressure until it is equal to whichever of A and B has the greater vapour pressure. And at that point if we decrease the pressure any more the more volatile component will start evaporating.

user228700

@JohnRennie Yes...

user228700

@S007: It seems that @koolman has followed ur footsteps and gotten into trouble (likely) at The Periodic Table.

Anonymous

@Kaumudi which footsteps ?

koolman

Which trouble @Kaumudi

John Rennie

And if we decrease the pressure further some of the less volatile component starts evaporating as well. Because the composition of the liquid is changing the total vapour pressure changes as well. So the pressure changes smoothly.

user228700

12:20

@S007 Go check for urself, I'm busy thinking about pressure :-P

Anonymous

@Kaumudi Leave, i'm too lazy

Anonymous

:-P

user228700

@S007 He pinged everyone :-P

user228700

@JohnRennie Ah, yes...

Anonymous

@Kaumudi Oh that? LOL :-D....

heather

12:21

I don't suppose anyone would mind helping me with calculus?

especially integral calculus

John Rennie

@Kaumudi If I keep expanding my vessel the pressure keeps falling until all the more volatile component has evaporated, and at that point if I keep expanding the vessel all the remaining liquid will evaporate.

heather

@0celo7, hello =)

Mew

@heather yeah bring it on

Secret

@heather Ask and see if they can help

user228700

@heather That's correct. Don't suppose such atrocities.

heather

12:23

@Kaumudi, lol, okay =)

Anonymous

@Kaumudi Now, I want to give a lecture on evaporation and boiling, get ready!

user228700

@JohnRennie Yes, OK...

user228700

@S007 Nooo. My brain is effed up, man, leave me alone :-P I'll come back later tonight.

heather

so basically, I've been trying to teach myself calculus, and I don't really understand how to use, i.e., the product rule for integration. Given a derivative, I can probably figure it out, given an integral, I have no chance basically of figuring it out, unless it is really simple, and I'm not sure why.

@Kaumudi, good luck with that exam btw!

Mew

is ur exam tmr Kaumu?\

user228700

12:25

@heather Thanks :-) Half the people here (including me) are writing the exam (next year) so good luck, everyone! :-P

John Rennie

@Kaumudi And that's what your original graph was showing you. At low pressures we have both A and B as vapour. At high pressures we have both A and B as liquid, and in between there is a range of pressures where we have some A and some B but the composition of the liquid and vapour are different.

Mew

@Kaumudi so no more questions after tonight?

@heather do you have some examples

heather

@Kaumudi, oh, wow, that's nuts, studying a year in advance?

user228700

@JohnRennie Ah, I see...

heather

@Mew, for instance, given the integral $\int^{12.5}_0 (50-3t)(9.8) dt$, I'm not sure what I'm doing

John Rennie

12:26

@heather I suspect you're thinking of revising rather than studying. After all, I started studying for my degree finals three years before the exam! :-)

Mew

@heather a constant can be taken out of the integra

user228700

Okay, I'ma have to read everything again but I'm pretty sure I'm on my way to understanding it properly.

user228700

@JohnRennie :-D Thanks, guru!

heather

I started by doing $\int(50-3(12.5))(9.8) - \int (50-3(0))(9.8)$

user228700

@JohnRennie Lol, agreed.

heather

12:27

because it is a definite integral

Mew

so that equals $9.8\int 50 - 3t dt$

Anonymous

@Kaumudi Okay, be ready for the lecture when me meet next time on h bar :-P :-D Bye for now!

heather

@Mew, but I didn't think you could do that, because constants turn into 9.8x (for instance)

Anonymous

user228700

@S007 I wish you'd stop calling it "lecture". I'm still very sure that it's just a matter of miscommunication.

Mew

12:28

@heather you can still do it

heather

@Mew, why?

Mew

i'll explain with simple example

Anonymous

@Kaumudi No, no. Let me preserve my ego :-P Bye for now :)

user228700

Because I understand where you're coming from @S007. Still, alright, alvida, for now :-)

Mew

$/int 2x dx$ = 2x^2/2 = x^2

also

hang on let me get the latex right

Anonymous

12:29

Alvida :-D

Mew

$/int 2x dx = \

John Rennie

@heather you're integrating $\int f(t)dt$ where $f(t) = (50-3t)(9.8)$. Just multiply it out to get $f(t) = 490 - 29.4t$. Then do $\int (490 - 29.4t)dt$

user228700

@Mew Are u kidding?! I'll stick around and pester anyone who wants to be pestered, don't u worry!

Mew

@JohnRennie, that is one way, but you can also take it out of the integral

@heather integral of 2x is x^2

The integral of x = x^2/2

John Rennie

@Mew @heather the point is that the same applies if you have e.g. $f(t) = (50-3t)(9.8t)$

Mew

12:31

thus 2 * integral of x is the same as integral of 2x

my point is $k\int f(t)dt = \int kf(t)dt$

John Rennie

Multiply out to get $\int (490t - 29.4t^2) dt$

Mew

for any constant k

@heather, an integral is a sum of small parts, you can take out any common factor

Imagine integral as an area under a curve, if the curve is multiplied by k, the area is multiplied by k too

Secret

$$\int u(x)v'(x)dx = u(x)v(x) - \int u'(x)v(x) dx$$

There are couple of integrals you want to use the integration by parts, such as if your integrand is a polynomial multiplied by an exponential.
e.g.
$$\int (x^2+x) e^x dx$$

Here you want to set your $u(x)$ to be $(x^2+1)$ and $v'(x)=e^x$, so that after using integration by parts, the power of $u(x)$ will decrease for each time you use it. Therefore after 2 iterations, the integral on the rightmost side of the expression will be just $\int e^x dx$ which is one of the easy cases. Thus

Electrodynamist

Hello, everyone! Can we ensure that the electromagnetic four-potential will always be timelike? Or, what physical conditions decide whether it will be timelike or spacelike?

heather

@S007 @JohnRennie, okay. One thing I missed, though, in the videos I watched: what is $\int cx dx$ where $c$ is some constant?

Mew

12:34

@heather, another example. $\int 2 = 2\int 1 = 2*x $

insert dx behind each integral

John Rennie

$\int cx dx = $c\int x dx = \frac{cx^2}{2}$

Mew

yes you can take the constant out of the integral

so it iss $c\frac{x^2}{2}$

heather

@Mew, wait, no, I thought $\int 2 = 2x+c$

Mew

yes

that si what i said heather

Secret

timeout typo: Previous message of mine should be $2e^x$ not $e^x$

Mew

12:35

I wrote $2*x = 2x$

heather

oh, nevermind I missed the second equals sign

okay, so I factor out the constant, and get $9.8 \int^{12.5}_0 50-3t \, dt$

John Rennie

Strictly speaking $\int 2 dx = 2 \int dx = 2x$

Mew

@JohnRennie, strictly speaking = $2x + c$

Secret

don't forget the C for indefinite integrals

John Rennie

If you're integrating wrt $dx$ then any factor that doesn't involve $x$ can be taken outside the integral.

Mew

12:38

@heather, that's correct so far

heather

so really you can take out 50 too

right?

Mew

no

heather

?

Mew

because 3 doesn't have 50 as a common factor

but you can split the integral up into:

heather

$\int 50\, dx - \int 3t \, dx$

Mew

12:39

$9.8\int 50 dt - 9.8\int 3tdt$

John Rennie

@heather integration is distributive so $int (f(x) + g(x))dx = \int f(x)dx + \int g(x)dx$

Mew

Now you can take out the 50 and the 3 if you wahnt

but the integral of 50 is so simple there is really no point

heather

50x

Mew

ype, or 50t rather in this case

heather

* 9.8

Mew

12:40

yep

so the first term is 9.8*50*t

and the second term can be simplified to $-9.8\times 3 \int t dt$

$-9.8\times 3 \times \frac{t^2}/{2}$

heather

right, that makes sense

Mew

that last bit is meant to be t^2/2

Secret

NB I thought when heather said integration product rule, is referring to the integration by parts. It seems I understood it wrongly

heather

@Secret, I was, but this was one of the earlier problems I had trouble with before the integration product rule.

Mew

@heather, now that you have the answer, you must plug in the limits of integration

heather

12:42

right

Mew

and go the top limit pluggin in mins the bottom limit plugged in

heather

okay, let me work this out real quick

I got 8421.875

Mew

I got

3828.125

heather

wow, that's not good

okay, let me do it again

I redid it and got the same result

Mew

Ok this is what I did

the solution to the integral is 9.8*(50t - 3t^2/2)

we need to evalute this for t = 12.5 and t = 0

when t = 12.5 we get

9.8*(50*12.5 - 3*12.5^2/2) =

3828.125

heather

12:49

I started with $((9.8*50*12.5) - (-9.8*3*\frac{12.5^2}{2})) - ((9.8*50*0)-(-9.8*3*\frac{0^2}{2}))$

Mew

and when t= 0, we get 0

why the double negative?

why -(-9.8 ?

heather

and then why did I put *12.5 in the first parens...

Mew

no everything is correct

except it should be only a single negative in front of the 9.8

heather

oh, wait, I know why

11 mins ago, by Mew

and the second term can be simplified to $-9.8\times 3 \int t dt$

Mew

yeah

I was assuming first term + second term

thus I absorbed the negative in the second term

sorry for the confusion

heather

12:52

yeah, and I wasn't, so I messed it up =) no, you're good, I should have realized it didn't make sense.

Mew

but yeah, just as constants can be taken out of integrals, so can a negative sign

heather

okay, yeah, because a - sign is really -1

Mew

$\int -f(x)$ = -\int f(x)$

precisely

heather

well thank you that helped a lot

could I have an integral to solve?

Mew

np any other questions?

ok

$\int 2x(2x^3 + 4x^7) dx$

heather

12:57

=) here goes

Mew

oh and let's say the limit is from 0 to 3

heather

okay

Mew

Also remember only constants can be removed from the integral

$x$ cannot be

heather

okay

Mew

too hard?

heather

13:06

nope just had to do something for my mom =)

Mew

awwww

heather

i got -17301.6

got to go

Mew

laterz

koolman

@S007 @Secret I have posted an see ithttp://chemistry.stackexchange.com/questions/62882/azeotropic-distillation/628‌87#62887

1

A: Azeotropic distillation

Here an answer for , I know it is bit confusing

Anonymous

13:22

@koolman good answer

Secret

13:35

Yeah that makes sense, so basically the vapor move towards the right and down and the residue move towards the left and up in the phase diagram

thus resulting in the pure A and the vapor B (which is the azeotrope)

koolman

13:59

@S007 this is because of your explanation

@Secret yeah

Sir Cumference

@Mew Hey Smithy, can you do me a favor?

Check over this code

Anonymous

14:21

@koolman :)

Anonymous

Keep asking your doubts whenever you have one :)

Secret

DHMO, MAFIA
[Division by zero]
Definition 1: Given a non annihilating semiring $S$, a left zero term is an expression $0a\neq 0$, while a right zero term is an expression $a0 \neq 0$. It is a two sided zero term (or just zero term) if it is both a left and right zero term. The order of a zero term is the power $n$ of the element $a$ (which can be zero).

Theorem 1: All left zero terms are idempotent in a left distributive semiring, while all right zero terms are idempotent in a right distributive semiring

Mew

@SirCumference wat is it bro

ACuriousMind

14:47

@koolman I don't know how exactly chemistry.SE sees it, but we don't like the posting of images in answers. First, you didn't give any source where you got that from, and second, images are not searchable for the text in them. If you can't be bothered to write an answer in your own words, don't write an answer.

Secret

[Division by zero] What I am suspecting is that these objects are mostly controlled by these idempotents (which has some absorbing properties), thus in a way they are similar to semigroups. This might suggest they can be handled using the framework of semigroup theory

ACuriousMind

@heather Here's a tip (I think @DanielSank also tried to give it to you yesterday, but I'm not sure he said it in a way you understood): When I look at your calculation steps, all I see is a bunch of numbers that means nothing to me, and I would have to dig up the problem itself to get what they signify. Writing e.g. $g$ instead of 9.8 and $t_0,t_1$ instead of what I think are the starting and end times makes your formulae a) shorter b) more readable.

Sir Cumference

@Mew Huh?

ACuriousMind

At the very end, when you have simplified what you wanted to compute as much as possible, you can then plug in all the numbers again

Sir Cumference

Oh crap, wrong room, wrong person

Sorry haha

Room for Secret and S007

Transcript for

$Mathematics$ Room for Secret and S007