ile atomów węgla zawartych jest w 10g K₄[Fe(CN)₆]?
mK₄[Fe(CN)₆]=4*39u+56u+6*12u+6*14u=368u
368g K₄[Fe(CN)₆]---------6*6,02*10²³ at. węgla
10g K₄[Fe(CN)₆]----------------x at. węgla
x = 9,88*10²² at. węgla
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mK₄[Fe(CN)₆]=4*39u+56u+6*12u+6*14u=368u
368g K₄[Fe(CN)₆]---------6*6,02*10²³ at. węgla
10g K₄[Fe(CN)₆]----------------x at. węgla
x = 9,88*10²² at. węgla