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Bas_tet [7]
3 years ago
12

Ross (he was on break) pours water at 6.0 c into a container with a pressure of 23. torr. he then boils the water to a temperatu

re of 34
c. the volume changes to 18. ml and pressure to 78. torr. what was the initial volume of the water?
Chemistry
1 answer:
kodGreya [7K]3 years ago
8 0
For this question, since the parameters volume, pressure and temperature changes, we can use the combined gas law equation.
It states that PV/T = constant 
Therefore,
\frac{P1V1}{T1} =   \frac{P2V2}{T2}
Where parameters for the first instance are given on the left side and parameters for the second instance are given on the right side of the equation
Temperature in K - T1 = 6 °C + 273 = 279 K
                               T2 = 34 °C + 273 = 307 K
substituting the values in the equation
\frac{23.0 torr*V}{279K} =  \frac{78 torr*18 mL }{307 K}
V = 55.5 mL
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Answer:

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Boiling point order: Methanol (highest) > Methanal > Methane (weakest)

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6 0
3 years ago
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6 0
2 years ago
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The equilibrium constant k for the synthesis of ammonia is 6.8x105 at 298 k. what will k be for the reaction at 375 k?
spayn [35]
Answer is: K <span>be for the reaction at 375 K is 326.
</span>Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁<span><span> = 298 K
</span>T</span>₂<span><span> = 375 K
</span><span>Δ<span>H = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K</span></span></span>·mol.<span>
K</span>₁ = 6,8·10⁵.<span>
K</span>₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.
6 0
3 years ago
Read 2 more answers
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