Answer:
The solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Explanation:
Step 1: Given data
The volume of water sample = 46.0 mL.
The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.
Step 2: Calculate the solubility of X in water
46.00 mL of water sample contains 0.87 g of the mineral compound X.
To calulate how many grams of the mineral compound 1.0 mL of water sample contains:
0.87 g/46.0 mL = 0.0189 g.
This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Answer: D) 140g
Explanation: no. of moles of NaCl = molarity X volume in litres = 2 X 1.2 = 2.4, and molar mass or mass of 1 mole of NaCl = 58.44 g, so 2.4 moles NaCl = 140.256 g
Answer: pOH = 4.68
Explanation:
pOH = 14 - pH
pH = - Log [H+]
= - Log [4.8 x 10^-10]
= -(-9.32)
pH =+9.32
Therefore, pOH= 14 - 9.32
= 4.68
