Question

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Answer 1

3.

$m_{CO_2}=8.8\ kg\\M_{CO_2}=44\ g/mol\\\\n=\frac{m}{M}=\frac{8.8\ \not kg}{44\ \frac{\not g}{mol}}=0.2\ kmoli=200\ moli\ CO_2\\\\1\ mol....................6.023{\cdot}10^{23}\ molec.\\200\ kmoli.....................x\\\\x=200\ {\cdot}\ 6.023{\cdot}10^{23}=1204.6{\cdot}10^{23}=1.2046{\cdot}10^{26}\ molec.\ CO_2$

$m_{HNO_3}=0.63\ g\\M_{HNO_3}=63\ g/mol\\\\n=\frac{m}{M}=\frac{0.63\ \not g}{63\ \frac{\not g}{mol}}=0.01\ moli\ CO_2\\\\1\ mol....................6.023{\cdot}10^{23}\ molec.\\0.01\ moli......................\ x\\\\x=0.01\ {\cdot}\ 6.023{\cdot}10^{23}=0.06023{\cdot}10^{23}=6.023{\cdot}10^{20}\ molec.\ HNO_3$

4.

$m_{NH_3}=34\ g\\M_{NH_3}=17\ g/mol\\\\n=\frac{m}{M}=\frac{34}{17}=2\ moli\ NH_3\\\\1\ mol\ NH_3................N_A\ molec.\\2\ moli............................\ x\\\\x=2\ {\cdot}\ N_A=2\ {\cdot}\ 6.023{\cdot}10^{23}=12.046{\cdot}10^{23}\ molec.\ NH_3\\\\17\ g\ NH_3...........1N_A\ atom\ N............3N_A\ atomi\ H\\34\ g\ NH_3 .............\ a.........................\ b\\\\a=2\ {\cdot}\ 6.023{\cdot}10^{23}=12.046{\cdot}10^{23}\\b=6\ {\cdot}\ 6.023{\cdot}10^{23}=36.138{\cdot}10^{23}$