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.
Salts is the correct awnser
Answer:
Explanation:
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In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:
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Answer:
It effects all of the cellular respiration process
Explanation:
It inhibits the Glycolysis. It replaces the phosphate groups that is needed for making Pyruvate and ATP.
The five strategies could include; 1) research the question of clean water availability on line and in books (2 strategies), 2) interview or ask those affected by lack of availability of clean water and find out how it affects them 3) consult with other groups ie political groups also concerned with this question and find out what they did to ensure availability to clean water and lastly to take actions to fight for the accessibility of clean water for certain groups of people like the First Nations people on reserves in Canada.
