Oblicz, ile gramów tlenu powstanie podczas rozkładu 60 g tlenku rtęci(II).
434u 32u
2HgO --> 2Hg + O₂
60g x
z proporcji wyliczam x
x = 4,42g
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
2HgO -> 2Hg + O2
434g ------- 32g
60g --------- xg
x= 32*60/434= 4,42g O2
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434u 32u
2HgO --> 2Hg + O₂
60g x
z proporcji wyliczam x
x = 4,42g
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
2HgO -> 2Hg + O2
434g ------- 32g
60g --------- xg
x= 32*60/434= 4,42g O2