Ad
related to: mg g to 100g sugar conversion equation
Search results
Results From The WOW.Com Content Network
Glycemic load of a 100 g serving of food can be calculated as its carbohydrate content measured in grams (g), multiplied by the food's GI, and divided by 100. For example, watermelon has a GI of 72. A 100 g serving of watermelon has 5 g of available carbohydrates (it contains a lot of water), making the calculation (5 × 72)/100=3.6, so the GL ...
Unit conversion formula from mmol/L to mg/dL [5] m g / d L = m m o l / L × m o l e c u l a r w e i g h t ÷ 10 {\displaystyle mg/dL=mmol/L\times molecular\ weight\div 10} Since the molecular mass of glucose C 6 H 12 O 6 is 180.156 g/mol, the factor between the two units is about 18, so 1 mmol/L of glucose is equivalent to 18 mg/dL.
The conversion between the two anomers can be observed in a polarimeter since pure α-d-glucose has a specific rotation angle of +112.2° mL/(dm·g), pure β-d-glucose of +17.5° mL/(dm·g). [64] When equilibrium has been reached after a certain time due to mutarotation, the angle of rotation is +52.7° mL/(dm·g). [ 64 ]
For a 2 hour GTT with 75 g intake, a glucose level below 7.8 mmol/L (140 mg/dL) is normal, whereas higher levels indicate hyperglycemia. Blood plasma glucose between 7.8 mmol/L (140 mg/dL) and 11.1 mmol/L (200 mg/dL) indicate " impaired glucose tolerance ", and levels at or above 11.1 mmol/L at 2 hours confirm a diagnosis of diabetes.
In chemistry, acid value (AV, acid number, neutralization number or acidity) is a number used to quantify the acidity of a given chemical substance.It is the quantity of base (usually potassium hydroxide (KOH)), expressed as milligrams of KOH required to neutralize the acidic constituents in 1 gram of a sample.
The osmol gap is typically calculated with the following formula (all values in mmol/L): = = ([+] + [] + []) In non-SI laboratory units: Calculated osmolality = 2 x [Na mmol/L] + [glucose mg/dL] / 18 + [BUN mg/dL] / 2.8 + [ethanol/3.7] [3] (note: the values 18 and 2.8 convert mg/dL into mmol/L; the molecular weight of ethanol is 46, but empiric data shows that it does not act as an ideal ...
An earlier definition, used especially for chemical elements, holds that an equivalent is the amount of a substance that will react with 1 g (0.035 oz) of hydrogen, 8 g (0.28 oz) of oxygen, or 35.5 g (1.25 oz) of chlorine—or that will displace any of the three.
A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). This is incorrect because the unit "%" can only be used for dimensionless quantities. Instead, the concentration should simply be given in units of g/mL.