Answer:
134.8 mmHg is the vapor pressure for solution
Explanation:
We must apply the colligative property of lowering vapor pressure, which formula is: P° - P' = P° . Xm
P° → Vapor pressure of pure solvent
P' → Vapor pressure of solution
Xm → Mole fraction for solute
Let's determine the moles of solute and solvent
17.5 g . 1 mol/180 g = 0.0972 moles
82 g . 1mol / 32 g = 2.56 moles
Total moles → moles of solute + moles of solvent → 2.56 + 0.0972 = 2.6572 moles
Xm → moles of solute / total moles = 0.0972 / 2.6572 = 0.0365
We replace the data in the formula
140 mmHg - P' = 140 mmHg . 0.0365
P' = - (140 mmHg . 0.0365 - 140mmHg)
P' = 134.8 mmHg
The reaction will produce 12.1 g Ag₂S.
<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S
<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)
× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S
Identify<span> each </span>bond<span> as either </span>polar<span> or nonpolar.</span>
Answer: 4) Concentrate it
Explanation:
