Convert assay ton to mo(毛)
Learn how to convert
1
assay ton to
mo(毛)
step by step.
Calculation Breakdown
Set up the equation
\(1.0\left(assay \text{ } ton\right)={\color{rgb(20,165,174)} x}\left(mo(毛)\right)\)
Define the base values of the selected units in relation to the SI unit \(\left({\color{rgb(230,179,255)} kilo}gram\right)\)
\(\text{Left side: 1.0 } \left(assay \text{ } ton\right) = {\color{rgb(89,182,91)} \dfrac{7.0}{2.4 \times 10^{2}}\left({\color{rgb(230,179,255)} kilo}gram\right)} = {\color{rgb(89,182,91)} \dfrac{7.0}{2.4 \times 10^{2}}\left({\color{rgb(230,179,255)} k}g\right)}\)
\(\text{Right side: 1.0 } \left(mo(毛)\right) = {\color{rgb(125,164,120)} \dfrac{3.0}{8.0 \times 10^{5}}\left({\color{rgb(230,179,255)} kilo}gram\right)} = {\color{rgb(125,164,120)} \dfrac{3.0}{8.0 \times 10^{5}}\left({\color{rgb(230,179,255)} k}g\right)}\)
Insert known values into the conversion equation to determine \({\color{rgb(20,165,174)} x}\)
\(1.0\left(assay \text{ } ton\right)={\color{rgb(20,165,174)} x}\left(mo(毛)\right)\)
\(\text{Insert known values } =>\)
\(1.0 \times {\color{rgb(89,182,91)} \dfrac{7.0}{2.4 \times 10^{2}}} \times {\color{rgb(89,182,91)} \left({\color{rgb(230,179,255)} kilo}gram\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{3.0}{8.0 \times 10^{5}}}} \times {\color{rgb(125,164,120)} \left({\color{rgb(230,179,255)} kilo}gram\right)}\)
\(\text{Or}\)
\(1.0 \cdot {\color{rgb(89,182,91)} \dfrac{7.0}{2.4 \times 10^{2}}} \cdot {\color{rgb(89,182,91)} \left({\color{rgb(230,179,255)} k}g\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{3.0}{8.0 \times 10^{5}}} \cdot {\color{rgb(125,164,120)} \left({\color{rgb(230,179,255)} k}g\right)}\)
\(\text{Cancel SI units}\)
\(1.0 \times {\color{rgb(89,182,91)} \dfrac{7.0}{2.4 \times 10^{2}}} \cdot {\color{rgb(89,182,91)} \cancel{\left({\color{rgb(230,179,255)} k}g\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{3.0}{8.0 \times 10^{5}}} \times {\color{rgb(125,164,120)} \cancel{\left({\color{rgb(230,179,255)} k}g\right)}}\)
\(\text{Conversion Equation}\)
\(\dfrac{7.0}{2.4 \times 10^{2}} = {\color{rgb(20,165,174)} x} \times \dfrac{3.0}{8.0 \times 10^{5}}\)
Cancel factors on both sides
\(\text{Cancel factors}\)
\(\dfrac{7.0}{2.4 \times {\color{rgb(255,204,153)} \cancel{10^{2}}}} = {\color{rgb(20,165,174)} x} \times \dfrac{3.0}{8.0 \times {\color{rgb(255,204,153)} \cancelto{10^{3}}{10^{5}}}}\)
\(\text{Simplify}\)
\(\dfrac{7.0}{2.4} = {\color{rgb(20,165,174)} x} \times \dfrac{3.0}{8.0 \times 10^{3}}\)
Switch sides
\({\color{rgb(20,165,174)} x} \times \dfrac{3.0}{8.0 \times 10^{3}} = \dfrac{7.0}{2.4}\)
Isolate \({\color{rgb(20,165,174)} x}\)
Multiply both sides by \(\left(\dfrac{8.0 \times 10^{3}}{3.0}\right)\)
\({\color{rgb(20,165,174)} x} \times \dfrac{3.0}{8.0 \times 10^{3}} \times \dfrac{8.0 \times 10^{3}}{3.0} = \dfrac{7.0}{2.4} \times \dfrac{8.0 \times 10^{3}}{3.0}\)
\(\text{Cancel}\)
\({\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{3.0}} \times {\color{rgb(99,194,222)} \cancel{8.0}} \times {\color{rgb(166,218,227)} \cancel{10^{3}}}}{{\color{rgb(99,194,222)} \cancel{8.0}} \times {\color{rgb(166,218,227)} \cancel{10^{3}}} \times {\color{rgb(255,204,153)} \cancel{3.0}}} = \dfrac{7.0 \times 8.0 \times 10^{3}}{2.4 \times 3.0}\)
\(\text{Simplify}\)
\({\color{rgb(20,165,174)} x} = \dfrac{7.0 \times 8.0 \times 10^{3}}{2.4 \times 3.0}\)
Solve \({\color{rgb(20,165,174)} x}\)
\({\color{rgb(20,165,174)} x}\approx7777.7777778\approx7.7778 \times 10^{3}\)
\(\text{Conversion Equation}\)
\(1.0\left(assay \text{ } ton\right)\approx{\color{rgb(20,165,174)} 7.7778 \times 10^{3}}\left(mo(毛)\right)\)