Question 1.20 - Chemistry: Molecular Nature Of Matter And Change

Question

Write the conversion factor(s) for

  (a) cm/min to in/s         (b) m³  to in³

  (c) m/s²to km/h²         (d) gal/h to l/min

Step-by-Step Solution

Answer

Answer

Conversions are as follows:     

              $$\begin{gathered} a.) 1 \raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$min$}\right. =0.0423 \raisebox{1ex}{$in$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ b.) 1 m^3 = 61\times {10}^4 i{n}^3 \\ c.) 1 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. = 12960 \raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$h^2$}\right. \\ d.) 1 \raisebox{1ex}{$gal$}\!\left/ \!\raisebox{-1ex}{$h$}\right. =0.0630901963\raisebox{1ex}{$ l$}\!\left/ \!\raisebox{-1ex}{$min$}\right. \end{gathered}$$

Conversion in (a)

$$\begin{gathered} a.) \\ 1 \raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$min$}\right. = \frac{2.54}{60} \raisebox{1ex}{$in$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ Therefore, \\ 1 \raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$min$}\right. = 0.0423 \raisebox{1ex}{$in$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Conversion in (b)

$$\begin{gathered} b.) \\ 1 m^3 ={\left(39.37\right)}^3 i{n}^3 \\ 1 m^3 =61023378 i{n}^3 \\ \mathrm{Therefore}, \\ 1 m^3 = 61\times {10}^4 i{n}^3 \end{gathered}$$

Conversion in (c)

$$\begin{gathered} c.) \\ 1 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. =\left[\frac{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.}\right)}{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${3600}^2$}\right.}\right)}\right] \raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$h^2$}\right. \\ \mathrm{Therefore}, \\ 1 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{s}}^2 $}\right.= 12960\raisebox{1ex}{$ \mathrm{km}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{h}}^2$}\right. \end{gathered}$$

Conversion in (d)

$$\begin{gathered} d.) \\ 1 \raisebox{1ex}{$gal$}\!\left/ \!\raisebox{-1ex}{$h$}\right. =\frac{3.78541178}{60} \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$min$}\right. \\ \mathrm{Therefore}, \\ 1 \raisebox{1ex}{$\mathrm{gal}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{h}$}\right. =0.0630901963 \raisebox{1ex}{$\mathrm{l}$}\!\left/ \!\raisebox{-1ex}{$\min $}\right. \end{gathered}$$