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In the standard system the conversion is that 1 gallon = 231 cubic inches and 1 inch = 2.54 cm, which makes a gallon = 3785.411784 millilitres exactly. For nutritional labeling on food packages in the US, the teaspoon is defined as exactly 5 ml, [22] giving 1 gallon = 3840 ml exactly. This chart uses the former.
The specific weight, also known as the unit weight (symbol γ, the Greek letter gamma), is a volume-specific quantity defined as the weight W divided by the volume V of a material: = / Equivalently, it may also be formulated as the product of density, ρ, and gravity acceleration, g: = Its unit of measurement in the International System of Units (SI) is newton per cubic metre (N/m 3), with ...
Sometimes specific volume is expressed in terms of the number of cubic centimeters occupied by one gram of a substance. In this case, the unit is the centimeter cubed per gram (cm 3 /g or cm 3 ·g −1). To convert m 3 /kg to cm 3 /g, multiply by 1000; conversely, multiply by 0.001. Specific volume is inversely proportional to density.
Volume fraction, which is widely used in chemistry (commonly denoted as v/v), is defined as the volume of a particular component divided by the sum of all components in the mixture when they are measured separately. For example, to make 100 mL of 50% alc/vol ethanol solution, water would be added to 50 mL of ethanol to make up exactly 100 mL.
The factor–label method can convert only unit quantities for which the units are in a linear relationship intersecting at 0 (ratio scale in Stevens's typology). Most conversions fit this paradigm. An example for which it cannot be used is the conversion between the Celsius scale and the Kelvin scale (or the Fahrenheit scale). Between degrees ...
United States standard drinks of beer, malt liquor, wine, and spirits compared. Each contains about 14 grams or 17.7 ml of ethanol. A standard drink or (in the UK) unit of alcohol is a measure of alcohol consumption representing a fixed amount of pure alcohol.
The quantity "1 ppm" can be used for a mass fraction if a water-borne pollutant is present at one-millionth of a gram per gram of sample solution. When working with aqueous solutions, it is common to assume that the density of water is 1.00 g/mL. Therefore, it is common to equate 1 kilogram of water with 1 L of water.
V eq is the volume of titrant (ml) consumed by the crude oil sample and 1 ml of spiking solution at the equivalent point, b eq is the volume of titrant (ml) consumed by 1 ml of spiking solution at the equivalent point, 56.1 g/mol is the molecular weight of KOH, W oil is the mass of the sample in grams. The normality (N) of titrant is calculated as: