00:02
Alright, so we're going to have to react copper nitric acid.
00:05
So, a reaction here you have copper and nitric 5 acid to give us copper nitrate, and perhaps nitrogen 2 oxide, and also water, okay? so for us to balance this equation, therefore, we're going to put 3, 4 copper.
00:30
Okay? and 3, 4 copper there.
00:34
And then next we're going to count a number of oxygen.
00:36
So we're going to have to put two nitrogen because you have okay so that makes uh hydrogens if we put four there we're going to have eight hydrogens and then we're going to have uh 21 24 oxygens which is correct given these other uh side okay so that is it and then be removed to copper nitrate reacting uh sodium hydroxide to give us copper hydroxide and sedum nitrate okay so perhaps for us to balance this again you have to put two for sodium and two for sodium move this outside and the equation is balanced okay then the last bit is copper hydroxide dissociating or copper hydroxide giving us perhaps copper one oxide and water this is self -pidroxate balancing equation here, okay? because you have two hydrogens, these are the side, two hydrogens, we have two oxygens, two oxygens, so this is balanced, okay? now the other bit is, so that was the first question, the other bit was about, it's 22, because we're given that weight of copper is 0 .51, okay? 0 .51 gram and then relative formula mass is 63 .59, gram per mole so therefore the number of moles from here is going to be 50 .51 divided by 63 .59 and we have 0 .008 more okay now from here now and at a thousand mils of nitric acid solution as six moles okay so therefore 6 .5 ml we'll have you're calculating the number of moles...