Recent content by jacee

I want to be a fashion designer, and my aunt is telling me that if I want to continue in this line of work I have to used something I printed off the internet. She says "that's what real designers use." Is this true? Or do they dray freehand?

I know it's certainly a matter of opinion, but I am trading in my Env2 soon for an upgrade. I'm just not sure what I want yet, and was wondering if anyone had any opinion about what phones they like or do not like. I text constantly and use the internet frequently, those are the main uses for my...

You did the first two right, but the second two are solutions of a strong base, not a strong acid. There are two common ways to find the pH, I think this one is easier: C) Since you know the concentration of strong base, you can find pOH, which is -log[OH-] (works just like pH) pOH=...

You did the first two right, but the second two are solutions of a strong base, not a strong acid. There are two common ways to find the pH, I think this one is easier: C) Since you know the concentration of strong base, you can find pOH, which is -log[OH-] (works just like pH) pOH=...

First figure out the limiting reactant by calculating how many moles of product you get from each starting material. For the 55.0g of K2PtCl4, convert that to moles by dividing by the molar mass (415.1 g/mol) to get .132 moles K2PtCl4. You get 1 mole of product for ever mole of K2PtCl4, so...

Not quite. There are a couple of ways to do it, but to get the mass in grams, this is a good one. You have 12 molecules, you need to see how many moles that it (it is much much less than 1 mole), so you DIVIDE by avogradro's number: 12 molecules * (1 mole/6.022x10^23 molecules)= 1.99x10^-23...