Question Number 227465 by Spillover last updated on 01/Feb/26
$${How}\:{many}\:{hydrogen}\:{atoms}\:{are} \\ $$$${there}\:{in}\:\mathrm{2}.\mathrm{57}×\mathrm{10}^{−\mathrm{6}} {g}\:\:{of}\:{hydrogen} \\ $$
Answered by Kassista last updated on 02/Feb/26
$$ \\ $$$${M}\left({H}_{\mathrm{2}} \right)=\mathrm{2}\frac{{g}}{{mol}},\:\mathrm{1}\:{mol}\:=\:\mathrm{6}.\mathrm{022}\centerdot\mathrm{10}^{\mathrm{23}} \:{particles} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{2}\:{g}\:−\:\mathrm{6}.\mathrm{022}\centerdot\mathrm{10}^{\mathrm{23}} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}.\mathrm{57}\centerdot\mathrm{10}^{−\mathrm{6}} \:−\:{x}\: \\ $$$$\Leftrightarrow\:{x}\:=\:\frac{\mathrm{6}.\mathrm{022}\centerdot\mathrm{10}^{\mathrm{23}} \centerdot\mathrm{2}.\mathrm{57}\centerdot\mathrm{10}^{−\mathrm{6}} }{\mathrm{2}}\: \\ $$$$=\:\mathrm{7}.\mathrm{73827}\centerdot\mathrm{10}^{\mathrm{17}} \:\overset{×\mathrm{2}} {\Rightarrow}\:\mathrm{1}.\mathrm{547654}\centerdot\mathrm{10}^{\mathrm{18}} \:{atoms}\:{of}\:{H} \\ $$
Answered by Spillover last updated on 02/Feb/26
