W probce siarczanu(VI) sodu znajduje sie okolo 3,01 * 1022kationow sodu. Oblicz mase probki.
mNa2SO4=142u
Na2SO4 ---> 2Na(+) + SO4(2-)
142g Na2SO4 ------ 12,04*10^23 Na(+)
xg Na2SO4 ----------0,301*10^23 Na(+)
x = 3,55g Na2SO4
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mNa2SO4=142u
Na2SO4 ---> 2Na(+) + SO4(2-)
142g Na2SO4 ------ 12,04*10^23 Na(+)
xg Na2SO4 ----------0,301*10^23 Na(+)
x = 3,55g Na2SO4