How many moles are there in 500cm3 of a 0.5mol/dm³ solution of sulfuric acid?

0.25moles

Explanation:

There are 1000 for 1

Therefore in 1000 of 0.5 mol/ solution has = 0.5 moles

Therefore 500 contains = 0.5/1000 x 500 = 0.25moles

to be fair, you have to have a very high iq to understand rick and morty. the humour is extremely subtle, and without a solid grasp of theoretical physics most of the jokes will go over a typical viewer's head. there's also rick's nihilistic outlook, which is deftly woven into his characterisation- his personal philosophy draws heavily from narodnaya volya literature, for instance. the fans understand this stuff; they have the intellectual capacity to truly appreciate the depths of these jokes, to realise that they're not just funny- they say something deep about life. as a consequence people who dislike rick & morty truly are idiots- of course they wouldn't appreciate, for instance, the humour in rick's existential catchphrase "wubba lubba dub dub," which itself is a cryptic reference to turgenev's russian epic fathers and sons. i'm smirking right now just imagining one of those addlepated simpletons scratching their heads in confusion as dan harmon's genius wit unfolds itself on their television screens. what fools.. how i pity them.

and yes, by the way, i do have a rick & morty tattoo. and no, you cannot see it. it's for the ladies' eyes only- and even then they have to demonstrate that they're within 5 iq points of my own (preferably lower) beforehand. nothin personnel kid

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.