Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
By using the formula, mass = density x volume, we
calculate mass in grams
20.0 mL CH₃COOH x (1.05 g / mL) = 21.0
g CH₃COOH
To find the moles, molar mass of CH₃COOH = 60.05g/mol<span>
21.0 g </span>CH₃COOH x (1 mole CH₃COOH / 60.05 g CH₃COOH)
= 0.350 moles CH₃COOH
To find molarity,<span>
[</span>CH₃COOH] = moles CH₃COOH / L of solution = 0.350 /
1.40 = 0.250 M CH₃COOH<span>
When </span>CH₃COOH is dissolved in water, it produces
small and equal amounts of H₃O⁺+ and C₂H₃O₂⁻.
<span>
Molarity , </span>CH₃COOH<span> + H</span>₂O <==> H₃O⁺ + C₂H₃O₂⁻
<span>
<span>Initial 0.250 0 0 </span>
Change -x x x
Equilibrium 0.250-x x x
K</span>ₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] = (x)(x) /
(0.250-x) = 1.8 x 10⁻⁵
<span>Since K</span>ₐ is relatively small, we can neglect the -x
term after 0.250 to simplify
<span>x</span>² / 0.250 = 1.8 x 10⁻⁵
x² = 4.5 x 10⁻⁶
<span>
x = 2.1 x 10</span>⁻³<span> = [H</span>₃O⁺]
pH = -log [H₃O⁺] = -log (2.1 x 10⁻³) = 2.68
the system is in constant temperature
so it follows Boyle's law
i.e.P1V1=P2P2
where P1=initial pressure=788 tour
P2=final pressure= ?
V1=initial volume=235 ml
V2=final volume=0.115 ml
Answer:
(c) Gibb's phase rule
Explanation:
Gibb's phase rule :
describe the relationship between the number of degrees fredoom in a closed system, the number of phases in equilibrium and the number of the system components.
the above is no related to the estimation of the vapor pressure.