Question

va rog e urgent dau coroana ​

Answer 1

a.

1 mol .............. 2 * 6.023×10²³ atomi

x moli ............ 12.046×10²³ atomi         x = 1 mol S₂

$n_{S_2}=1\ mol\\M_{S_2}=2\ {\cdot}\ 32=64\ g/mol\\\\1\ mol .......64\ g\implies m_{S_2}=64\ g$

b.

1 mol .......... NA molecule (NA = 6.023×10²³ particule - ioni, atomi, molecule)

$6.023{\cdot}10^{26}\ molec.\ CO_2\implies 6023{\cdot}10^{23}\ molec.\\\\1\ mol.........6.023{\cdot}10^{23}\\x................6023{\cdot}10^{23}\\\\x=1000\ moli = 1\ kmol\\\\n_{CO_2}=1\ kmol\\M_{CO_2}=12+(2\ {\cdot}\ 16)=44\ g/mol\\\\n = \frac{m}{M}\implies m=n\ {\cdot}\ M=1\ {\cdot}\ 44=44\ kg\ CO_2$

c.

$n_{HNO_3}=2.5\ moli\\M_{HNO_3}=1+14+(3\ {\cdot}\ 16)=63\ g/mol\\\\n=\frac{m}{M}\implies m=n\ {\cdot}\ M=2.5 \not moli\ {\cdot}\ 63\ \frac{g}{\not mol}=157.5\ g\ HNO_3$

d.

$n_{H_2}=0.5\ kmoli\\M_{H_2}=2{\cdot}1=2\ g/mol\\\\n=\frac{m}{M}\implies m=n\ {\cdot}\ M=0.5\not kmoli}\ {\cdot}\ 2\ \frac{g}{\not mol}=1\ kg\ H_2$