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Answer:
81.0°C
Explanation:
<em>Kb benzene = 2.5°C/m</em>
<em />
The addition of a solute to a pure solvent produce an elevation in boiling point regarding to boiling point of pure solvent. This phenomenon follows the equation:
ΔT = Kb×m×i
<em>where ΔT represents the increasing in boiling point, Kb is the elevation boiling point constant of the solvent (2.5°C/m for benzene), m is molality of solution (Moles solute / kg solvent) and i is Van't Hoff factor (1 for a non-electrolyte solute as naphthalene).</em>
<em />
Moles of 11.5g of naphthalene (Molar mass: 128.17g/mol) are:
11.5g × (1mol / 128.17g) =<em> 0.0897 moles of naphthalene in 250.0g = 0.250kg of benzene.</em>
Molality is:
0.0897 moles of naphthalene / 0.250kg of benzene = 0.359m
Replacing in the equation:
ΔT = Kb×m×i
ΔT = 2.5°C/m×0.359m×1
ΔT = 0.90°C
That means the solution prepared has an elevation in boiling point of 0.90°C. As boiling point of pure benzene is 80.10°C, boiling point of the solution is:
80.10°C + 0.90°C =
<h3>81.0°C</h3>
Answer:
Removing energy (cooling) atoms and molecules decreases their motion, resulting in a decrease in temperature
Answer:
1.67 moles of N2 produce 2.5moles of NaN3
Explanation:
Equation of reaction:
2Na + 3N₂ → 2NaN₃
From the equation of reaction, 3 moles of N₂ produces 2 moles of NaN₃
How much is gotten from 2.5moles of 2NaN₃?
2 moles of N₂ = 3 moles of NaN₃
X moles of N₂ = 2.5 moles of NaN₃
X = (2.5 × 2) / 3
X = 1.67moles of N₂
1.67 moles of N₂ produces 2.5moles of NaN₃