1-5 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_H &= 1.0078\ut{u}\\ 1\ut{u} \times1\ut{mol}&= 1\ut{g}\\ N_A&=6.02214076\times10^{23} \ut{EA/mol}\\ M_H&=3.30\ut{kg}\\ \end{cases} $$ $$ \begin{aligned} \Ans&=\cfrac{3.30\ut{kg}}{1.0078\ut{u}}\times\cfrac{1\ut{u}\cdot1\ut{mol}}{1\ut{g}} \times\cfrac{1000\ut{g}}{1\ut{kg}} \times\cfrac{6.02214076\times10^{23}\ut{EA}}{1\ut{mol}}\\ &=\cfrac{3.30\cdot1000\cdot6.02214076\times10^{23}}{1.0078}\ut{EA}\\ &\approx1.97192543243\times10^{27}\ut{EA}\\ &\approx1.97\times10^{27}\ut{EA}\\ \end{aligned} $$