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Volume to mass conversions for some common cooking ingredients; ingredient density g/mL [note 5] metric cup 250 mL imperial cup ≈284 mL U.S. customary cup ≈237 mL [note 6] g oz g oz g oz water [note 7] 1 [note 8] 249–250 8.8 283–284 10 236–237 8.3 [note 9] granulated sugar 0.8 [20] 200 7.0 230 8.0 190 6.7 wheat flour 0.5–0.6 [20 ...
The system can be traced back to the measuring systems of the Hindus [18]: B-9 and the ancient Egyptians, who subdivided the hekat (about 4.8 litres) into parts of 1 ⁄ 2, 1 ⁄ 4, 1 ⁄ 8, 1 ⁄ 16, 1 ⁄ 32, and 1 ⁄ 64 (1 ro, or mouthful, or about 14.5 ml), [19] and the hin similarly down to 1 ⁄ 32 (1 ro) using hieratic notation, [20] as ...
In US customary units, most units of volume exist both in a dry and a liquid version, with the same name, but different values: the dry hogshead, dry barrel, dry gallon, dry quart, dry pint, etc. The bushel and the peck are only used for dry goods. Imperial units of volume are the same for both dry and liquid goods. They have a different value ...
For example, 20 apples divide into five groups of four apples, meaning that "twenty divided by five is equal to four". This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a ...
The potassium content is 0.72% on a dry matter basis, i.e. 0.72% of the dry matter is potassium. The apple, therefore, contains 158 mg potassium (0.72/100 X 22 g). Dried apple contains the same concentration of potassium on a dry matter basis (0.72%), but is only 32% water (68% dry matter).
The Winchester quart is an archaic measure, [10] equal to 1 ⁄ 16 of a Winchester bushel: 134.4013 cubic inches or 1.937879 imperial quarts or 0.484470 imperial gallons or 2.202443 litres. The 2.5 L bottles in which laboratory chemicals are supplied are sometimes referred to as Winchester quart bottles , although these contain 14% more than a ...
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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.