솔루션피아

$$ \begin{cases} 1\ut{u}&=1\ut{g/mol}\\ 1\ut{mol}&=6.022~140~76\times10^{23}\\ m_H&=1.0\ut{u}\\ m_O&=16\ut{u} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} m_{(H_2O)}&=1.0\ut{u}\times2+16\ut{u}\\ &=18\ut{u}\\ &=18\ut{u}\times\frac{1\ut{g}}{1\ut{u}\cdot1\ut{mol}}\times\frac{1\ut{kg}}{1000\ut{g}}\\ &=\frac{18}{6.022~140~76\times10^{23}\cdot1000}\ut{kg}\\ &\approx2.988970320912924\times10^{-26}\ut{k..

$$ \begin{cases} 1\ut{slug}&=14.59\ut{kg}\\ 1\ut{U.S.~bushel}&=35.238\ut{L}\\ \end{cases} $$ $$ \begin{cases} A_p\ut{\$}&=\text{Cost of pig }\ut{3.108~slug}\\ A_c\ut{\$}&=\text{Cost of corn }\ut{U.S.~bushel}\\ \cfrac{A_p}{A_c}&= 6.50\\ \end{cases} $$ $$ \begin{aligned} 3.108\ut{slug}&=3.108\ut{slug}\times\frac{14.59\ut{kg}}{1\ut{slug}}\\ &=\frac{1133643}{25000}\ut{kg}\\ \end{aligned} $$ $$ \begi..

$$\ab{a}$$ $$ \begin{aligned} \text{water in En}&=1\ut{MorningCup}+1\ut{TeaCup}+6\ut{TableSpoon}+1\ut{Dessertspoon}\\ &=2\ut{TeaCup}+1\ut{TeaCup}+6\ut{TableSpoon}+1\ut{Dessertspoon}\\ &=3\ut{TeaCup}+6\ut{TableSpoon}+1\ut{Dessertspoon}\\ &=24\ut{TableSpoon}+6\ut{TableSpoon}+1\ut{Dessertspoon}\\ &=30\ut{TableSpoon}+1\ut{Dessertspoon}\\ &=60\ut{DessertSpoon}+1\ut{Dessertspoon}\\ &=61\ut{DessertSpoo..

$$ \rho = 1\ut{g/cm^3} $$ $$ \ab{a} $$ $$ \begin{aligned} m &= \rho V\\ &= 1\ut{g/cm^3} \cdot 1\ut{m^3}\\ &= \cfrac{1\ut{g}}{1\ut{cm^3}} \cdot \cfrac{1\ut{kg}}{1000\ut{g}}\cdot\(\cfrac{100\ut{cm}}{1\ut{m}}\)^3\cdot 1\ut{m^3} \\ &=\cfrac{100^3}{1000}\ut{kg}\\ &=1000\ut{kg} \end{aligned} $$ $$ \ab{b} $$ $$ \begin{aligned} \Ans&=\cfrac{4800\ut{m^3}}{15.7\ut{h}}\cdot\cfrac{1000\ut{kg}}{1\ut{m^3}}\cd..

$$ \begin{cases} 1\ut{acre} &= 4047\ut{m^2}\\ 1\ut{ft} &= 12\ut{in}\\ 1\ut{in} &= 2.54\ut{cm}\\ S &= 17\ut{km^2}\\ h &= 2.0\ut{in}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} V &= S \cdot h \\ &= (17\ut{km^2}) \cdot (2.0\ut{in})\\ &=\bra{17\ut{km^2} \cdot \(\cfrac{1000\ut{m}}{1\ut{km}}\)^2 \cdot \cfrac{1\ut{acre}}{4047\ut{m^2}}}\cdot\(2.0\ut{in}\cdot\cfrac{1\ut{ft}}{12\ut{in}}\)\\ &=\cfrac{17..

$$ \begin{cases} x_A &= 12\ut{m}\\ y1_{A} &= 6.0\ut{m}\\ y2_{A} &= 3.0\ut{m}\\ z_A &= 20\ut{m}\\ L_A : L_B &= 12:1\\ L_A : L_C &= 144:1\\ \end{cases} $$ $$ \begin{aligned} V_A &= S_{xy} \times L_z\\ &= (S1_{xy} + S2_{xy})\times L_z\\ &=\(6.0\times12+\cfrac{1}{2}\times12\times3.0\)\times20\\ &=1800\ut{m^3} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} V_A : V_B &= (L_A)^3 : (L_B)^3 \\ &= 12^3:1^..

$$ \begin{cases} \rho_A &= 10\ut{EA/U.S.~pint}\\ \rho_B &= 30\ut{EA/U.S.~pint}\\ V &= 1.0\ut{m}\times20\ut{cm}\times20\ut{cm}\\ 1\ut{U.S.~pint}&=0.4732\ut{L} \end{cases} $$ $$ \begin{aligned} V&=100\cdot20\cdot20\ut{cm^3}\\ &=40000\ut{cm^3} \end{aligned} $$ $$ \begin{aligned} \Ans &= N_B - N_A \\ &=\rho_B \cdot V - \rho_A \cdot V\\ &= V\(\rho_B - \rho_A\)\\ &=40000\ut{cm^3}\cdot20\ut{EA/U.S.~pin..

$$ \begin{cases} 1\ut{AU}&=1.50\times10^8\ut{km}\\ c&=3.0\times10^8\ut{m/s} \end{cases} $$ $$ \begin{aligned} c&=\cfrac{3.0\times10^8\ut{m}}{1\ut{sec}}\times\cfrac{1\ut{km}}{1000\ut{m}}\times\cfrac{1\ut{AU}}{1.50\times10^8\ut{km}}\times\cfrac{3600\ut{sec}}{1\ut{h}}\\ &=\cfrac{3.0\cdot3600}{1000\cdot1.50}\ut{AU/h}\\ &=\cfrac{36}{5}\ut{AU/h}\\ &=7.2\ut{AU/h} \end{aligned} $$

$$ \begin{cases} 1\ut{mol}&=6.02\times10^{23}\ut{EA}\\ \end{cases} $$ $$ \begin{aligned} N^3 &= 0.2\ut{mol}\\ N &= \sqrt[3]{0.2\times6.02\times10^{23}}\\ &\approx 4.93789854178\times10^7\\ &\approx 4.9\times10^7\ut{EA}\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=4.9\times10^7\ut{cm}\\ &=4.9\times10^7\ut{cm}\times \cfrac{1\ut{m}}{100\ut{cm}}\times\cfrac{1\ut{km}}{1000\ut{m}}\\ &=4.9\times10^2\ut{..

$$ \begin{cases} 1\ut{diplacement~ton}&=7\ut{barrel~bluk}\\ 1\ut{freight~ton}&=8\ut{barrel~bluk}\\ 1\ut{register~ton}&=20\ut{barrel~bluk}\\ 1\ut{barrel~bluk}&=0.1415\ut{m^3}\\ 1\ut{m^3}&=28.378\ut{U.S~bushel} \end{cases} $$ $$A = 73\ut{diplacement~ton}$$ $$ \ab{a}$$ $$ \begin{aligned} \Ans &= 73\ut{freight~ton}-73\ut{diplacement~ton}\\ &=73\cdot\(1\ut{freight~ton}-1\ut{diplacement~ton}\)\\ &=73\..