Recent content by mathemalia

this whole post just reminds more about Belonging lol

what's the outcome of the separation?

I think is overrated because its hard but because its 50%

Background: Electron of atoms can absorb electromagnetic energy when they wanna become more 'energetic' and release EM when wanna become less energetic. But ever atom, has different electron configurations and shell numbers, so its absorption of EM waves will be very specific to itself, it'll...

Lead is aloner, he's too fat and (heavy metal) and so is chloride, he bloody smells, no one wants them. so you use them to teach for each other. Barium and calcium are good mates. Sulfates differentiates between their friendship however, making both become very angry and pale white (ppt)...

I will be nasty and say 8 marks on a whole titration experiment write up with calculations

Length contraction is a direct consequence of time dilation: Some dude on a platform measures the length of it, l0, then he measures how long it took for some other dude on some fast ass train to cover the length of the platform, the time he measures is not the proper time. Thus, v= l0/tv The...

Very easy my friend. Say you have 50 gram or 50mL (both work) and the mass of phosphates in the solution is 0.05 grams (both have to be expressed as grams or mL ,not one milligrams and the other grams etc), so the ratio becomes: Phosphate : Water 0.05 : 50 So you say okay, what if you i had...

You don't have to work with moles, do you? I mean for every 2 units, there is one water condensed out. The first and last units are,however, not connected by both sides. So if they ask for mass for n units of glucose: m(n cellulose)= M(glucose)*n - M(water)*(n-1) ?

Hey there, are we supposed to know the levels that make water of high quality or bad quality. Like for pH, healthy water is 6-8, and dead water is <1. Are we supposed to know all these levels for hardness, salinity etc thanks

ugghh...I didn't integrate =/ 3 marks there.

I = 10^-12 * e^{0.1L} L =\frac{ln\frac{I}{10^{-12}}}{0.1} If you double I, it becomes 2I: L1 = \frac{ln\frac{2I}{10^{12}}}{0.1} L1 = \frac{ln2 + lnI - ln10^{12}}{0.1} L1 = \frac{ln2}{0.1} +\frac{ ln\frac{I}{10^{-12}}}{0.1}

I = 10^-12 * e^{0.1L} L =\frac{ln\frac{I}{10^{-12}}}{0.1} If you double I, it becomes 2I: L1 = \frac{ln\frac{2I}{10^{12}}}{0.1} L1 = \frac{ln2 + lnI - ln10^{12}}{0.1} L1 = \frac{ln2}{0.1} +\frac{ ln\frac{I}{10^{-12}}}{0.1} L1 = \frac{ln2}{0.1} + L

just sub P=2(r+theta) for both sides, and you should end up like this 0 < theta < 2pi...(i think)

Thanks for that...and my mentioned quote says ln2/0.1 ... thats what i got from what i remember