Question

calculati volumul (c.n) ocupat de :
5 mol de CO2
15mmol de O2
10 kmol de N2
3.55 g de CL2
88kg de CO2
30,11*10²³ molecule de H2
18,066*$10^{26}$ molecule de NH3

E URGENT!!

Answer 1

n = V/Vm => V = n*Vm = 5*22,4 = 112L CO2

n = V/Vm => V = n*Vm = 15*22,4 = 340,4mL = 0,3405L O2

n = V/Vm => V =n*Vm = 10*22,4 = 224m³ = 224000L N2

n = m/M = 3,55/111 = 0,0319 moli CaCl2

n = V/Vm => V = 0,716 L CaCl2

n = m/M = 88/44 = 2 kmoli CO2 = 2000 moli CO2

n = V/Vm => V = 2*22,4*10^3 = 44,8*10^3 L CO2

1 mol H2.....................6,022*10^23 molecule
x moli H2,................. 30,11*10^23

x = 5 moli H2

n = V/Vm => V = n*Vm = 5*22,4 = 112 L H2

1 mol NH3................6,022*10^23 molecule
x moli NH3................18066*10^23 molecule

x = 3000 moli NH3

V = 3000*22,4 = 67,2*10^2 L NH3