Do my Penguin for me

[Cube]

Edit: Heh... Cube's answers are probably better, I was still writing when he posted.

 

I just used the define: function in google.

[Supergrunch]

I just used the define: function in google.

Oh, I used the define: function in my head...

[KKOB]

Does anyone know the mole equations for A Level chem. i have n=m/mr and n=c*v but can't think of any others :S

 

i'm trying to work out how much NiCl2 6H2O i need to make a 1L 80microM solution of the NiCl.

 

grrrrrrrr

[Supergrunch]

Those are the only equations you need.

 

Are you saying you need 1 dm^3 of a 0.08 mol dm^(-3) solution of NiCl, and you need to know what mass of NiCl.6H20 crystals to add?

 

If this is the case, you need 0.08 moles of NiCl (as you are making 1 dm^3).

 

So we need to work out the mass of 1 mole of NiCl crystals, which is Mr(Ni) + Mr(Cl) + 6Mr(H20). (using values from a periodic table this will give you the value in grams)

 

Now, we want 0.08 of a mole, so simply multiply the calculated value by 0.08 to give the grams of crystals required.

[KKOB]

i need 1000ml of 0.000080 (80x10-6) solution.

 

the MR of all that is 231.7, and when i do the math i end up with 0.019g which doesn't look right to me . . .

[Supergrunch]

i need 1000ml of 0.000080 (80x10-6) solution.

 

the MR of all that is 231.7, and when i do the math i end up with 0.019g which doesn't look right to me . . .

I think that's right, 0.00008 mol dm^(-3) is an insanely dilute solution.

[KKOB]

I think that's right, 0.00008 mol dm^(-3) is an insanely dilute solution.

 

Ah fair enough then

 

i'll put down my workings and submit then thanks thanks thanks :D:D

[kellett03]

just a bit of a maths problem, i have been stumped on his question for some time now, thanks in advanced

 

edit: thanks heaps ginger chris

 

do you differentiate first or do u sub in the 2?

[Ginger_Chris]

Always differentiate first. If you sub in the x first you'll get dy/dx=0.

 

y = ax^3 - bx + 3

dy/dy = 3ax^2 - b

d2y/dx2= 6ax

 

subbing in values:

5 = 12a -b

12 = 12a

 

a=1

b=7

[kellett03]

another maths problem,

 

the point(1,5) is a point of inflection on the curve y = ax^3 + bx^2 + x + 2

 

find the values of 'a' and 'b'

[Ginger_Chris]

Very similar to the previous question.

Differentiating it once will give you the gradient.

Differentiating it again will give you the 'bend'. The point of inflection is where the bend changes, ie d2x/d2y=0. use this to find x, then sub into the original to find y.

[Supergrunch]

Very similar to the previous question.

Differentiating it once will give you the gradient.

Differentiating it again will give you the 'bend'. The point of inflection is where the bend changes, ie d2x/d2y=0. use this to find x, then sub into the original to find y.

By the way, do happen to know of a sitution where the second derivative is 0 at a point that isn't an inflection? Because apparently such a thing exists, as if you find the second derivative to be 0 then you can't guarantee an inflection.

 

(Kellet, this isn't relevant to your question so don't let it confuse you)

[Ginger_Chris]

By the way, do happen to know of a sitution where the second derivative is 0 at a point that isn't an inflection? Because apparently such a thing exists, as if you find the second derivative to be 0 then you can't guarantee an inflection.

 

(Kellet, this isn't relevant to your question so don't let it confuse you)

 

Hmm, I'd guess that could occur when there are two identical roots to d2y/dx2=0, that way you have two (or four, or six..) inflections at a point, resulting in no inflection. Kinda of like y=x^3 (x=0), where you get a minimum and maximum at the same point, so you don't get either, just an inflection.

An odd number of inflections at a point would give an inflextion, just like y=x^4 gives a minimum at x=0, even though there are 2 minimums and a maximum technically at the same point.

That help?

[kellett03]

sorry to be such a annoying person but another maths question

 

a shop keeper found that if he orders lots of x computers, his storage and handling costs for a year will be $C, where

 

C = 15x + 24000/x + 6000

 

how many should she order at one time to minimise his costs

[Ginger_Chris]

To minimise costs you have to find a minimum. To do that, find what value of x satisfies dC/dx=0. To check if its a minimum either do a rough sketch of the graph, or find what sign d2C/dx2 is at the given value of x.

 

Doing the rough sketch of it, the costs don't really change between 20 and 80, but the actual minimum is about 40. (in questions like this, if you ever find the answer to be a fraction or decimal, because you can only have a whole number of computers, round it either up or down depending on the question with (with reasoning)).

 

On a compuetely useless note, the question also doesn't make any sense, why would you be minimising costs? The shop keeper should really be maximising profits. Unless thats costs per computer, in which is should be maximised, but that would make

 

Profit =((yearly taking per year per computer) - C)*x

which you should be maximising. sigh, another non-nonsensical maths question using an inappropriate model to describe something just because the numbers give you a nice answer.

[Paj!]

*stands with his mouth open*

 

I can't do Chemistry for shit, and I'm in the "special" Maths class.

 

Good to know theres people here to rely on if i ever need help!

[Atomic Boo]

Ive been doing these questions for the past hour or so, different variations but here's one im 'stuck' on.

 

Factorise

5x squared + 27x + 10

 

fairly easy i know. (how do you right powers on a keyboard?)

[Problematique]

5x^2 + 27x + 10

 

Basically I do trial and error for these in my head...

 

(5x + 2)(x + 5)

 

You keep the same constants in the factorised form (regardless of the coefficient of x^2) and then just fiddle about. It's not the most efficient way but it works.

[The fish]

5x^2 + 27x + 10

 

Basically I do trial and error for these in my head...

 

(5x + 2)(x + 5)

 

You keep the same constants in the factorised form (regardless of the coefficient of x^2) and then just fiddle about. It's not the most efficient way but it works.

 

Damn it, you beat me too it, I just wrote out a whole thing.

 

I disagree: I suggest dividing by the coefficient of x^2: for a start, it gives you smaller numbers to juggle...

Also, in an exam, it's best to put things in their simplified form, in my experience.

 

(x + 0.4)(x + 5) = 0

[Atomic Boo]

Hehe, I actually had the right answer written down but its just i calculated it wrong!

Thanks anyway guys.

[Supergrunch]

I wouldn't recommend making the x coefficient 1 unless all the coefficients will remain as integers, as factorising fractions can be more difficult.

[Problematique]

I wouldn't recommend making the x coefficient 1 unless all the coefficients will remain as integers, as factorising fractions can be more difficult.

 

Word. It's easy once you've practised it a few times, after a while you can kind of spot it straight away. It's uncanny.

[conzer16]

Anyone speak French here?

 

How do I say I was in French?

 

ie; passé composé of etre?

[ReZourceman]

Anyone speak French here?

 

How do I say I was in French?

 

ie; passé composé of etre?

 

Its "Je" something. Lol.

[Letty]

Anyone speak French here?

 

How do I say I was in French?

 

ie; passé composé of etre?

 

Already posted in meaningless, but here too if you miss it!

 

J'etait.