Answer:
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A way of building knowledge about the world around us through observation and experimentation.
Answer:
The correct answer is - yes, 4.57 g of solute per 100 ml of solution
Explanation:
The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.
Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.
Then solublity will be :
32. g solute / 700 ml solution = y / 100 ml solution
⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.
Thus, the answer is 4.57 g of solute per 100 ml of solution.
Characteristics of a Precipitate:
A precipitate is characterized by the following properties:
Appears as a solid species.
Settled down at the bottom of the reaction pot.
Insoluble in the corresponding solvent.
Answer:
3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.
