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February 24, 2013

## How Do I calculate the pH of water after adding 500 ml of 1 M acetic acid (Ka = 1.8 x10-5) to 500 ml of H2O?

How do I calculate the change in pH, following (i) the addition of 500 ml of 1M Ba(OH)2 to 500 ml of 2O, and (ii) the addition of 500 ml of 1 M acetic acid (Ka = 1.8 x10-5) to 500 ml of H2O.

Ka= [H+]^2/[CH3COOH] so [H+]=sqrt([0.5]xKa) then pH=-log[H+]

## What has changed in gaming technology to cause such a massive increase in detail?

I was just watching the trailer of Halo Reach X10 and the tech guys kept talking about how they could do way more this time around following Halo 3. But what really changed for these people technology wise for this to be possible?

The main advance that allows them to show more detail is very simple. Everything 3d in games is based on polygons, you can see here how her face is made of polygons http://tinypic.com/view.php?pic=2ah7ixe&s=5

basically the more polygons the more detail. Advancements in software have allowed us to make things with many more polygons without effecting performance or anything else. This results in a much more detailed 3d environment.

-Here is an example of a model with a high number of polygons

and here is one with fewer polygons:

http://i.imgur.com/0gNYi.jpg

see the difference?

## How do I go about solving this problem?

The average density of the planet Uranus is 1.27 x10^3 kg/m3. The ratio of the mass of Neptune to that of Uranus is 1.19. The ratio of the radius of Neptune to that of Uranus is 0.969. Find the average density of Neptune.

Assuming that when you wrote :
• “The ratio of the mass of Neptune to that of Uranus is 1.19.”
you meant:
• “The ratio of the mass of Neptune to that of Uranus is 1.19 : 1″ ….

Then …

Let:
• Du = density of Uranus, Dn = density of Neptune
• Mu = mass of Uranus, Mn = mass of Uranus
etc.

Given that:
• Du = 1.27 x 10³ kg/m³
• Mn : Mu = 1.19 : 1 ? Mn = 1.19Mu
• Rn : Ru = 0.969 : 1 ? Vn : Vu = 0.969³ : 1³ ? Vn = 0.969³(Vu)

Dn = Mn ÷ Vn

Dn = (1.19Mu) ÷ (0.969³(Vu))

Dn = (1.19/0.969³)(Mu/Vu)

Dn = (1.19/0.969³)(Du)

Dn = (1.19/0.969³)(1.27×10³) kg/m³

Dn = 1.66103… X 10³ kg/m³

Dn = 1.66 x 10³ kg/m³ (3 sd)

Hope this helps! Cheers!

.

## What is the distance of the average american school gymnasium?

My gym teacher said i could run 10 laps around my gym in 2:16 . I wanna know how big the average one is so i could convert it to meters, x10 and i get my time for X meters;

Thanks, soule

(also, it’d be appreciated if you had the time to do my calculations. Thanks!)