Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
Answer:
c) 2.5 mL
Explanation:
Solution
Doctors order = 0.125g
and
The liquid suspension concentration = 250 mg/5ml
= 0.250g/5ml
Or 0.05g/ml
Amount of ml of suspension required = 0.125g/(0.05g/ml) = 2.5ml
Condensation occurs through the release of energy
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