Question

un amestec ce contine 16,8 l de hidrogen si 5,6 litri azot
a.are masa 4.25g
b. e mai greu ca aerul
c.contine 1.2014 * 10 ^22 atomi
d are 3 ori mai putine molecule de N2 fata de H2

Answer 1

n = V/Vm = 16,8/22,4 = 0,75 moli H2
n = V/Vm = 5,6/22,4 = 0,25 moli N2

m H2 = 0,75×2 = 1,5g
m N2 = 0,25×28 = 7g

m amestec = 1,5+7 = 8,5g

a. fals
b. Maer = 28,9g/mol ---> fals
c. 1 mol H2........2×6,022×10^23 atomi
0,75 moli H2......x = 9,033×10^23 atomi

1mol N2..............2×6,023×10^23 atomi
0,25 moli N2......x = 3,011×10^23 atomi

nr atomi total = 12.044×10^23 atomi

c - fals

d. nr molecule H2 = 0,75×6,022×10^23 = 4,5165×10^23
nr molecule N2 = 0,25×6,022×10^23 = 1,5055×10^23

1,055×3 = 4,5165

d - adevarat