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[citation needed] A cheese's fat content is expressed as the percentage of fat in the cheese's dry matter (abbreviated FDM or FiDM), which excludes the cheese's water content. [7] For example, if a cheese is 50% water (and, therefore, 50% dry matter) and has 25% fat, its fat content would be 50% fat in dry matter. [8]
An aqueous solution containing 2 g of glucose and 2 g of fructose per 100 g of solution contains 2/100=2% glucose on a wet basis, but 2/4=50% glucose on a dry basis.If the solution had contained 2 g of glucose and 3 g of fructose, it would still have contained 2% glucose on a wet basis, but only 2/5=40% glucose on a dry basis.
For the dry bulk density, the sample is oven dried and weighed, giving the mass of soil solids, M s. The relationship between these two masses is M t = M s + M l, where M l is the mass of substances lost on oven drying (often, mostly water). The dry and wet bulk densities are calculated as Dry bulk density = mass of soil/ volume as a whole
In chemistry, the mass fraction of a substance within a mixture is the ratio (alternatively denoted ) of the mass of that substance to the total mass of the mixture. [1] Expressed as a formula, the mass fraction is:
= 1000 kg/m 3: ounce (avoirdupois) per cubic foot oz/ft 3: ≡ oz/ft 3: ≈ 1.001 153 961 kg/m 3: ounce (avoirdupois) per cubic inch oz/in 3: ≡ oz/in 3: ≈ 1.729 994 044 × 10 3 kg/m 3: ounce (avoirdupois) per gallon (imperial) oz/gal ≡ oz/gal ≈ 6.236 023 291 kg/m 3: ounce (avoirdupois) per gallon (US fluid) oz/gal ≡ oz/gal ≈ 7.489 ...
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 ...
Change in volume with increasing ethanol fraction. The molar volume of a substance i is defined as its molar mass divided by its density ρ i 0: , = For an ideal mixture containing N components, the molar volume of the mixture is the weighted sum of the molar volumes of its individual components.
Example 1: If a block of solid stone weighs 3 kilograms on dry land and 2 kilogram when immersed in a tub of water, then it has displaced 1 kilogram of water. Since 1 liter of water weighs 1 kilogram (at 4 °C), it follows that the volume of the block is 1 liter and the density (mass/volume) of the stone is 3 kilograms/liter.