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aga
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They're do-able, but f*king tricky tho.
Mostly it is just following the Rules and applying Logic and some maths, so i guess knowing the Rules, Some Logic & Maths is kinda vital.
Following those Rules and not jumping to easy-path conclusions is also pretty useful !
If we're Lucky (maybe if we ask nicely enough, offer money, sexual services, drugs etc) bloggers might show us all the Answers with a step-by-step
walk-through showing how they can/should be worked out.
@<b>NEMO-Chemistry</b> & <b>Neme</b>
Which bits are the really hard parts for you ?
Stoichimetry ? (aka Balancing Equations)
Algebra ? (aka Squigly Maths-With-No-Numbers)
Molarity ? (aka how much stuff in that solvent)
If you shout up, you could get some help.
For me, it's just thinking clearly that is quite hard !
[Edited on 11-9-2016 by aga]
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Neme
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Well they are not exactly "really hard" as you said, I just did not see that type of problems yet. I will have a look into them more closely later.
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NEMO-Chemistry
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Quote: Originally posted by aga  | They're do-able, but f*king tricky tho.
Mostly it is just following the Rules and applying Logic and some maths, so i guess knowing the Rules, Some Logic & Maths is kinda vital.
Following those Rules and not jumping to easy-path conclusions is also pretty useful !
If we're Lucky (maybe if we ask nicely enough, offer money, sexual services, drugs etc) bloggers might show us all the Answers with a step-by-step
walk-through showing how they can/should be worked out.
@<b>NEMO-Chemistry</b> & <b>Neme</b>
Which bits are the really hard parts for you ?
Stoichimetry ? (aka Balancing Equations)
Algebra ? (aka Squigly Maths-With-No-Numbers)
Molarity ? (aka how much stuff in that solvent)
If you shout up, you could get some help.
For me, it's just thinking clearly that is quite hard !
[Edited on 11-9-2016 by aga] |
I guess its perspective, a few short weeks ago i couldnt balance an equation, now i can do a little bit more than that.
So maybe its simply a question of keep studying and not let it phase you, i am working on the questions in sequence, so its likely things i have no
idea of now will become clearer as i progress.
Maths is my main sticking point, but even that is starting to improve slowly. I think it was more of a knee jerk reaction reading the question and
comparing it to the first one.
I have a clear week from Tuesday (school work wise), as i am almost upto date with my course work. This will give me more time to study this.
I will shout when i get totally stuck  , i havnt finished the others yet as my
course work needed getting upto date.
I guess there is little point in asking questions that dont stretch you anyway, no one learns anything of value if they already know the answer!
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aga
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Aw crap. Does that mean i'm not a noob anymore ?
Feck. I think i am 'cos i know there's a HUGE pile of chem that i have no idea about.
Is there a useful Definition of Noob anywhere ?
Whatever.
If you have a 'sticking' point, U2U me and i'll see if i can help.
Edit:
That's an offer open any Noob of course.
[Edited on 11-9-2016 by aga]
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blogfast25
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As it stands (a few more participants need to get in their answers on #7 and #8 but there's plenty time) the leader board is tight, with so many good
answers.
Depending on the incoming answers a tiebreak question may be needed in 10 days or so. Maybe something with Calculus Lite, mwwhahaha... 
[Edited on 11-9-2016 by blogfast25]
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aga
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I'd suggest that if the Noob can do Calculus, then it's not a Maths Noob question.
$$p_{ain} = \int_0^{\infty} { \frac {aga}{calculusrack}} $$
*Shudder* Healing takes time ...
[Edited on 11-9-2016 by aga]
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blogfast25
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And as the Devil makes idle hands, here's a non-chemistry, Non-Competition question, just for fun.
A square metal plate has one side held at 100 C and the three other sides held at 0 C. Top and bottom of the plate are perfectly insulated (no heat
losses of any kind).
Question:
After thermal equilibrium has been achieved, what is the temperature at the centre of the square, Tc?
[Edited on 11-9-2016 by blogfast25]
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blogfast25
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Your integral lacks a differential. 
Also, you have no idea how light "Lite" can become, in the limit!
[Edited on 11-9-2016 by blogfast25]
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aga
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Should random answers go via U2U ?
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blogfast25
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Yes, please. Forgot to mention that. 
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NEMO-Chemistry
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Ok i might have misunderstood!! I have been working on them and thought we were to submit them in one go??
When is the closing date and do i submit all together or as they are done??
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aga
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Work out question NC#x, send the answer to blogfast25 via U2U using NC#x as the Subject (change 'x' to the actual number).
i.e. send your answers one-by-one. You get feedback !
What you get back is either 'well done' or 'are you sure ?' depending.
(there are probably Other answers, also depending)
Last one i got was an 'are you sure ?'
No, i wasn't.
Looked at it again and was definitely Not Sure when i tried it again, as the first answer i worked out was totally Bull crap.
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NEMO-Chemistry
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Shit my answers must be crap, i didnt get feedback lol. How long we got? I will get these done this week.
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blogfast25
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I have only received one relevant U2U from you and marked it. Check your outbox please.
| Quote: Originally posted by NEMO-Chemistry | Ok i might have misunderstood!! I have been working on them and thought we were to submit them in one go??
When is the closing date and do i submit all together or as they are done??
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#1, #2 and #3 are closed. The rest stays open till the end of the month. I need one U2U PER ANSWER.
[Edited on 11-9-2016 by blogfast25]
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ficolas
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#3 too? yay missef another deadline
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NEMO-Chemistry
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Well that just adds to our noob credentials lol. Head down and study!!
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blogfast25
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Non-competition question solution:
The question what is the temperature at the centre of the disc? was answered correctly as "25 C" by aga and Neme. Neither provided much
rationale for their answer. So here's proof!
As stated in the original problem there are no radiative or convective losses, so the temperature distribution is described entirely by Fourier's heat
equation, here in 2 dimensions and Cartesian coordinates:
$$\frac{\partial u}{\partial t}=k\Bigg(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\Bigg)$$
u is temperature, k is the heat conductivity of the disc's material. Or using the usual shorthand:
$$u_t=k(u_{xx}+u_{yy})$$
Because we're in steady state, so ut=0:
$$u_t=0 \implies u_{xx}+u_{yy}=0$$
Which is essentially Laplace's equation. Now we need to add the boundary conditions:
$$u(0,y)=0,u(L,y)=0$$
$$u(x,0)=0,u(x,L)=u_0$$
(Here u0=100 C)
We can solve this by separation of variables from the following Ansatz:
$$u(x,y)=X(x)Y(y)$$
The solution is fairly straightforward and the process is described here. This gives us:
$$u(x,y)=\displaystyle \sum_{n=1}^{+\infty}A_n\sinh\Big(\frac{n\pi y}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$
$$n=1,2,3,...$$
An is determined from one of the boundary conditions:
$$u(x,L)=u_0=\displaystyle \sum_{n=1}^{+\infty}A_n\sinh(n\pi)\sin\Big(\frac{n\pi x}{L}\Big)$$
Using Fourier series analysis we get:
$$A_n=\frac{2u_0}{n\pi\sinh(n\pi)}\big(1-(-1)^n\big)$$
So that Tc is given by:
$$T_c=u(L/2,L/2)=\frac{2u_0}{\pi}\displaystyle \sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)\cosh ((k+1/2)\pi)}$$
With:
$$k=0,1,2,3,...$$
Sums of infinite series can be a little scary but we can evaluate it in two ways:
1. Numerically:
Just the first term (k=0) gives:
$$T_c=0.2537 u_0$$
Then adding a few terms (k=1, k=2, etc) immediately shows the sum converges on:
$$T_c=0.25 u_0$$
2. Analytically:
Various databases of analytically determined infinite sums exist and this one shows that:
$$\displaystyle \sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)\cosh ((k+1/2)\pi)}=\frac{\pi}{8}$$
So that:
$$T_c=\frac{2u_0}{\pi}\times \frac{\pi}{8}=\frac{u_0}{4}=25 C$$
Plots:
For L=1, first 5 terms:
The squiggly edge is due to having used only 5 terms, using more terms smooths that edge more and more.
[Edited on 23-9-2016 by blogfast25]
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MeshPL
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Well, the problem with the Non-competition question, was that it seemed obvious that the center had the average temerature from all the sides,
allowing anybody with a bit of knowledge to guess (but not be sure) that the center is 25C hot. This is why you've received 2 correct answers but no
correct explanations. (How many people coreectly guessed the answer, is probably a higher number).
I guess, that if one was able to calculate accurately the temperature of any given point of that metal sheet, they really wouldn't be a N00b.
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aga
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Holy schmoly !
I just guessed that if three are at 0 and one is at 100, then the middle bit would be 1/4 of 100 = 25
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blogfast25
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End of current Noob competition: Friday 30 September, midnight UTC.
No more questions will be posted before that date.
Thanks all for playing! 
[Edited on 27-9-2016 by blogfast25]
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aga
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Thanks for taking the time to post & mark the questions !
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blogfast25
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Answer to NC #7:
$$S=kC_R=60$$
Concentration after addition of R:
$$C=\frac{0.25 \times C_R+0.05}{0.25}=C_R+0.2$$
$$S=kC_R+0.2k=120$$
Subtracting first equation from third gives:
$$120-60=60=0.2k\implies k=300$$
$$\implies C_R=\frac{60}{300}=0.2\mathrm{M}$$
Alternatively: since as the signal is twice as high after the addition, the concentration after addition must be twice as high as before the addition:
$$C=2C_R$$
$$2C_R=C_R+0.2\implies C_R=0.2\mathrm{M}$$
<hr>
Answer to NC #8:
At time t the concentration of HCl(aq) is:
$$[HCl]_t=\frac{0.75\times 0.500-0.192}{0.75}=0.244\mathrm{M}$$
$$\implies v_{H_2}(t)=\frac{0.244}{0.500}\times 20=9.76\mathrm{mmol/s}$$
[Edited on 1-10-2016 by blogfast25]
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blogfast25
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And the winner is:
Lefauxcheux10 with 97.5 %
Runner up: aga with 89.9 %
Third place: Charlie A with 81.0 %
<hr>
Congrats to all! Lefaucheux10 to collect prize via U2U.
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CharlieA
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Thanks
blogfast: Thanks for a job well-done. I don't know how many "papers" you had to "grade", but I appreciate all of your efforts. When I taught chemistry
for a few years, the part I liked least was grading papers.
To the other winners, congratulations!
To all who participated, I hope you enjoyed this at least as much as I did.
CharlieA
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aga
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Woohoo !
Thanks for taking the time to make a great quiz.
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