the formula is no. moles is mass / molecular mass. As the number of moles is 1, the mass required will be exactly the same as the molecular mass, which is 58.32g
how much sodium hydroxide in grams must be added to seawater to precipitate 86.9mg of magnesium present?
avagadro number
2.0 M
31.6grams
1
2.04 g
how much sodium hydroxide in grams must be added to seawater to precipitate 86.9mg of magnesium present?
Magnesium chloride: 81,95 g are obtained.
A lot
This is a homogeneous solution of ammonium hydroxide in water.
Molarity = moles of solute/Liters of solution ( 918 ml = 0.918 liters )rearranged algebraically,moles of solute = Liters of solution * Molaritymoles of NaOH = (0.918 l)(0.4922 M)= 0.45184 moles NaOH=======================so,0.45184 moles NaOH (39.998 grams/1 mole NaOH)= 18.1 grams sodium hydroxide needed============================
98g
24 / 40 = x / 4.00 x = 2.4 g Mg
262 - 266
8.62
avagadro number
262 - 266