1.Oblicz masę 5 atomów fosforu? 2. Ile moli stanowi 12,6 g kwasu azotowego 5? 3.Oblicz masę 14-stu cząsteczek siarczanu VI potasu?

Witaj 1/ 6,02*10^23 at. P................31g 5 at.P..............................x --------------------------------------- x = 5*31g/6,02*10^23 = 25,75*10^(-23)g = 2,58*10^(-22)g 2/ 63g HNO3.........................1mol 12,6g...............................y ------------------------------------------ y = 12,6g*1mol/63g = 0,2 mola HNO3 3/ 6,02*10^23 cz.K2SO4......................174g 14 cz..........................................z --------------------------------------------------- z = 174g*14/6,02*10^23 = 4,05*10^(-21)g .........................pozdrawiam

1. 6,02*10^23 at. fosforu.........31g fosforu 5 at. fosforu.....................xg fosforu x=5*31/6,02*10^23 x=25,75*10^-23g 2. mHNO3=1*1u+1*14u+3*16u=63u 63g HNO3...........1mol 12,6g HNO3.........xmoli x=12,6/63 x=0,2mola HNO3 3. mK2SO4=2*39u+1*32u+4*16u=174u 6,02*10^23 cząst. K2SO4......174g 14 cząst. K2SO4...............xg x=14*174/6,02*10^23 x=40,5*10^-22g