Pilne !!!!!!!!!!!!!!!!!!!!!!!!!!!! Chemia 1)oBLICZ mase 3,01 * 10²³ atomow potasu 2)oblicz mase 12.04 * 10²³ czasteczek wody. z gory dziekuje . daje naj Pozdrawiam.:)

grzesiek2485

Witam 1/ 1 mol K .....39g K.......................6,02*10(23) at. K ................x..........................3,01*10(23) at. K ------------------------------------------------------------------ x = 39g *3,01*10(23) //6,02*10(23) = 19,5g K 2/ 1 mol H2O....18g H2O....................6,02*10(23) cz. H2O ..................y..........................12,04*10(23) cz. H2O ------------------------------------------------------------------------- y = 18g *12,04*10(23) //6,02*10(23) = 36g H2O ...............................pozdrawiam

elwa124

1 mol ----6.02*10²³atomów x -------- 3.01*10²³ atomów x=0.5 mola 1 mol----39g 0.5 mola----x x=19.5g analogicznie.... 1 mol ----6.02*10²³ x----------12.04*10²³ x= 2 mole 1 mol H2O----18g 2 mole ----- x x=36g

czesio04

1) mK=39u 39g K-----------6,02*10^23 atomów K xg K-------------3,01*10^23 atomów K x = 39*3,01*10^23/6,02*10^23 x=19,5g K 2) mH2O=18u 18g H2O-------6,02*10^23 cząst. H2O xg H2O---------12,04*10^23 cząst. H2O x=18*12,04*10^23/6,02*^23 x=36g H2O