# 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 $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx7777.7777778\approx7.7778 \times 10^{3}$$
$$\text{Conversion Equation}$$
$$1.0\left(assay \text{ } ton\right)\approx{\color{rgb(20,165,174)} 7.7778 \times 10^{3}}\left(mo(毛)\right)$$