# Convert tonne / gallon to ounce / gallon

Learn how to convert 1 tonne / gallon to ounce / gallon step by step.

## Calculation Breakdown

Set up the equation
$$1.0\left(\dfrac{tonne}{gallon}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{ounce}{gallon}\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)$$
$$\text{Left side: 1.0 } \left(\dfrac{tonne}{gallon}\right) = {\color{rgb(89,182,91)} \dfrac{1016.0469088}{3.785411784 \times 10^{-3}}\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(89,182,91)} \dfrac{1016.0469088}{3.785411784 \times 10^{-3}}\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}$$
$$\text{Right side: 1.0 } \left(\dfrac{ounce}{gallon}\right) = {\color{rgb(125,164,120)} \dfrac{28.349523125}{4.54609}\left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(125,164,120)} \dfrac{28.349523125}{4.54609}\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(\dfrac{tonne}{gallon}\right)={\color{rgb(20,165,174)} x}\left(\dfrac{ounce}{gallon}\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{1016.0469088}{3.785411784 \times 10^{-3}}} \times {\color{rgb(89,182,91)} \left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} \dfrac{28.349523125}{4.54609}}} \times {\color{rgb(125,164,120)} \left(\dfrac{{\color{rgb(230,179,255)} kilo}gram}{cubic \text{ } meter}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} \dfrac{1016.0469088}{3.785411784 \times 10^{-3}}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} \dfrac{28.349523125}{4.54609}} \cdot {\color{rgb(125,164,120)} \left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{1016.0469088}{3.785411784 \times 10^{-3}}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} \dfrac{28.349523125}{4.54609}} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{{\color{rgb(230,179,255)} k}g}{m^{3}}\right)}}$$
$$\text{Conversion Equation}$$
$$\dfrac{1016.0469088}{3.785411784 \times 10^{-3}} = {\color{rgb(20,165,174)} x} \times \dfrac{28.349523125}{4.54609}$$
Switch sides
$${\color{rgb(20,165,174)} x} \times \dfrac{28.349523125}{4.54609} = \dfrac{1016.0469088}{3.785411784 \times 10^{-3}}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{4.54609}{28.349523125}\right)$$
$${\color{rgb(20,165,174)} x} \times \dfrac{28.349523125}{4.54609} \times \dfrac{4.54609}{28.349523125} = \dfrac{1016.0469088}{3.785411784 \times 10^{-3}} \times \dfrac{4.54609}{28.349523125}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times \dfrac{{\color{rgb(255,204,153)} \cancel{28.349523125}} \times {\color{rgb(99,194,222)} \cancel{4.54609}}}{{\color{rgb(99,194,222)} \cancel{4.54609}} \times {\color{rgb(255,204,153)} \cancel{28.349523125}}} = \dfrac{1016.0469088 \times 4.54609}{3.785411784 \times 10^{-3} \times 28.349523125}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{1016.0469088 \times 4.54609}{3.785411784 \times 10^{-3} \times 28.349523125}$$
Rewrite equation
$$\dfrac{1.0}{10^{-3}}\text{ can be rewritten to }10^{3}$$
$$\text{Rewrite}$$
$${\color{rgb(20,165,174)} x} = \dfrac{10^{3} \times 1016.0469088 \times 4.54609}{3.785411784 \times 28.349523125}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx43042.04533\approx4.3042 \times 10^{4}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{tonne}{gallon}\right)\approx{\color{rgb(20,165,174)} 4.3042 \times 10^{4}}\left(\dfrac{ounce}{gallon}\right)$$