Answer:
The answer is 2, 12 g of LiCl in 1 L of solution
Explanation:
We calculate the weight of 1 mol of LiCl:
Weight 1 mol LiCl= Weight Li + Weight Cl =6, 94 g + 35, 45 g= 42,39 g/mol
In a 0.05M solution it means that we have 0.05 moles of compound (in this case, LiCl) in 1 liter of solution:
1 mol---------42, 39 g LiCl
0,05mol-----x=(0,05molx 42, 39 g LiCl)/ 1 mol =2, 1195 g LiCl
Animals that rely on coral for protection and cover, such as grouper, snapper, oysters and clams, would also be negatively impacted. And because this marine life is a vital staple in many peoples' diets, the death of the coral reefs would exacerbate the problem of feeding these groups [source: Skoloff].
Answer:
Lead is added to warm dilute nitric acid. When the carbonate has reacted with the warm acid, more carbonate is added until the carbonate is in excess.
Explanation:
Answer:
4.87g
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Mass of solution = 0.35kg
Molality = 0.238 m
Mass of NaCl =..?
Step 2:
Determination of the number of mole of NaCl in the solution.
Molality of a solution is simply defined as the mole of solute per unit kg of the solvent. It is given as:
Molality = mol of solute /mass of solvent (kg)
With the above formula, we calculate the mole of NaCl present in the solution as follow:
Molality = mol of solute /mass of solvent (kg)
0.238 = mol of NaCl /0.35
Cross multiply
mol of NaCl = 0.238 x 0.35
mol of NaCl = 0.0833 mol
Step 3:
Determination of the mass of NaCl in 0.0833 mol of NaCl.
This is illustrated below:
Number of mole NaCl = 0.0833 mol
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass of NaCl =..?
Mass = number of mole x molar Mass
Mass of NaCl = 0.0833 x 58.5
Mass of NaCl = 4.87g
Therefore, 4.87g of NaCl is contained in the solution.
