Homework Section

[ShawmMoe]

I need an answer ASAP for this! Very very very important! Going to hand in my biggest assignment made every tonight! Need to answer this final question;
How can you rewrite Ks, so that you can calculate pH using Ks?
I really need an answer of this, and the one who finds me the answer will get a reward; 90K SA:MP money.
please answer ASAP.
Thanks.

[Brad.]

I need an answer ASAP for this! Very very very important! Going to hand in my biggest assignment made every tonight! Need to answer this final question;
How can you rewrite Ks, so that you can calculate pH using Ks?
I really need an answer of this, and the one who finds me the answer will get a reward; 90K SA:MP money.
please answer ASAP.
Thanks.

This is copied out of one of my old e-text books - hope it helps you:

[code]
The pH of a solution depends on the concentration of H+(aq) ions present.

A shorthand method of denoting concentration of hydrogen ions is [H+(aq)]. The square brackets surrounding the ion symbol mean 'concentration in moles per litre'.

When the concentration of hydrogen ions is expressed using powers of 10, the pH is equal to the negative of the power number.
Example:

[H+(aq)] = 0.01 mol l-1 = 10-2 mol l-1

So in the example above, the pH = 2. This is because the concentration of H+ ions are expressed as a power of 10, where the power is -2.

The exact pH of a solution can be calculated using:

    pH = -log[H+(aq)]

For most solutions the pH lies in the range 0 to 14.
Example:

In a solution of pH 2, what is the concentration of H+(aq)?

pH of solution = 2

[H+(aq)] = 10-2 mol l-1
Example:

A salt solution has a hydrogen ion concentration of 10-9 mol l-1. What is its pH?

[H+(aq)] = 10-9 mol l-1

pH of solution = 9

Calculations involving the ionic product

In pure water the concentration of hydrogen ions is equal to the concentration of hydroxide ions.

The pH of pure water is 7.

This means that the concentrations of both the hydrogen and hydroxide ions are 10-7 mol l-1.

This is written:

[H+(aq)] = [OH-(aq)] = 10-7 mol l-1

Multiplying the concentration of hydrogen ions and the concentration of hydroxide ions gives the ionic product for water:

[H+(aq)]

[Gerrie = Banned (Daco)]

5/0 -> plus infinity?

Yeah.

[Janar]

Word 'dye', how is that written in Present Continuous?
As word 'die' is 'dying' in Present Continuous.
According to the -ing end rules, it should by 'dying' also.
Can someone confirm this? My English teacher didn't know this either.

[Jubin]

Dyeing:
the use of dye to change the color of something permanently
http://wordnetweb.princeton.edu/perl/webwn

[Janar]

Okay, thanks!

[ShawmMoe]

Math question to Jubin
Pictures uploaded..
You know what to do on the first one..
Second: Calculate using the square of a two-size.
Third and last: Calculate using two square sum multiplied with two-size difference..
(Sorry for my english)..

[Jubin]

Dude, those are way too easy, so why won't you at least try to do them yourself, and post a picture of your calculation I can tell you what you did wrong ( if you did something wrong)

[Nighteyes]

Wow, never learned the -21 (x) bla..blaa, this really hard, tho.

[ShawmMoe]

Ok..
Here we go:
The first one: You showed me how to do it when it's -, and I understood it, but dunno when it's +.. And yes, I tried, I always get a different answer than my calcuator...
Uploaded picture of the 2nd and 3rd.
Thanks in advance.

[Jubin]

2nd it should be +25T²

 and 3d is correct.

[ShawmMoe]

May I ask, how do I know when I shall use + or - in such case?

[Jubin]

May I ask, how do I know when I shall use + or - in such case?

well (a+b)²=a²+2ab+b²
and (a-b)²=a²-2ab+b²

so if you see + inside of the brackets it will be +
and if you see - inside of the brackets it will be -

[ShawmMoe]

Oh I see.. Now it makes much more sence.
Bur what about the 1st math question? The one I didnt know how to solve...
Calcuator gives a different answer than mine.. A whole different.

[Jubin]

Well first of all, let's say a=2 and b=3 there fore we get [(3*2)/(2*3+2)]+[(2*2)/(4*3+2*2)]= 1 so whatever we do with later, it all still must equal to 1.

Let's take the 2a/4b+2a, well there we can see that it is 2a/2(2b+a) because 2(2b+a)= 4b+2a. So 2a/2(2b+a) is equaled to a/2b+a

Now we have (3a/2b+a) +(a/2b+a) as the denominator is the same (2b+a) we can write it on the same fraction

so it will be (3a+a)/(2b+a) = 4a/2b+a that should be all, now let's see if the test goes right. 4*2/2*3+2= 8/8= 1
Whoa, test comes out, therefore 4a/2b+a is the right answer.