Numărul de moli din:
a). V = 4.48 dm³ N₂
[tex]n=\frac{V}{V_M}=\frac{V}{22.4\ dm^3/mol}=\frac{4.48\ {\not}dm^3}{22.4\ {\not}dm^3/mol}=0.2\ mol\ N_2[/tex]
b). V = 4.48 L NH₃
[tex]n=\frac{V}{V_M}=\frac{V}{22.4\ L/mol}=\frac{4.48\ {\not}L}{22.4\ {\not}L/mol}=0.2\ mol\ NH_3[/tex]
c). 9.033·10²³ molecule N₂
* NA - numărul lui Avogadro (NA = 6.022·10²³ particule (ioni, atomi, molecule)
1 mol .... 6.022·10²³ molecule
x mol .... 9.033·10²³ molecule ⇒ x = 1.5 mol N₂
d). 9.033·10²⁶ molecule NH₃
1 mol ..... 6.022·10²³ molecule
y mol ..... 9.033·10²⁶ ⇒ x = 1.5 · 10³ mol NH₃
Masa a 67.2 dm³ de:
a). N₂
[tex]n=\frac{V}{V_M}=\frac{67.2\ {\not}dm^3}{22.4\ {\not}dm^3/mol}=3\ mol\\n=\frac{m}{\mu}\implies m=n\ \cdot\ \mu=3\ {\not}mol \cdot\ 28\ g/{\not}mol = 84\ g\ N_2[/tex]
SAU
[tex]67.2\ dm^3=0.0672\ m^3\\\rho_{N_2}=1.25\ kg/m^3\\\rho=\frac{m}{V}\implies m=\rho\ \cdot\ V=1.25\ kg/{\not}m^3\ \cdot 0.0672\ {\not}m^3 =0.084\ kg = 84\ g\ N_2[/tex]
b). NH₃
[tex]n=\frac{V}{V_M}=\frac{67.2\ {\not}dm^3}{22.4\ {\not}dm^3/mol}=3\ mol\\n=\frac{m}{\mu}\implies m=n\ \cdot\ \mu=3\ {\not}mol \cdot\ 17\ g/{\not}mol = 51\ g\ NH_3[/tex]
SAU
[tex]67.2\ dm^3=0.0672\ m^3\\\rho_{NH_3}=0.72\ kg/m^3\\\rho=\frac{m}{V}\implies m=\rho\ \cdot\ V=0.72\ kg/{\not}m^3\ \cdot 0.0672\ {\not}m^3 =0.048\ kg \simeq 48\ g\ NH_3[/tex]