This value is 116,88 g.
19
2 moles or 117 gram of NaCl is precepitated
One atom of chlorine per molecule of NaCl.
117 grams of NaCl
(1.84 L)(0.200 mol/L) = 0.368 moles NaCl
1.31*.3=.393 mols NaCl. This also equals .393(22.99+35.45)=22.97 g NaCl.
The formula unit of sodium chloride (NaCl) contain 60,33 % chlorine.
The recommended concentration is 20 +/- 5 mg iodine/kg NaCl.
0.95% * 500 g = 4.75 g NaCl
You need 58,44 mg of ultrapure NaCl; dissolve in demineralized water, at 20 0C, in a thermostat, using a class A volumetric flask of 1 L.
58 g NaCl = 58 (g) / 58.44 (g/mol NaCl) = 0.9925 (mol NaCl) = 0.9925 (mol Cl-) = 0.9925 * 35.45 (g/mol Cl-) = 35.2 g Cl-
how much is keratine produced?