Explanation:
The reactions which are not truly of first order but become reactions of first order under certain conditions are called pseudo first order reactions.
For example, hydrolysis of ester
The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.
<h3>What is the dissociation enthalpy?</h3>
Given that;
H-H bond energy = 435 kJ mol^-1
H-Cl bond energy = 431 kJ mol^-1
ΔHfO of HCL(g) = -92kJ mol^-1
Bond dissociation enthalpy of the Cl-Cl bond = x
-92 = 435 + 431 + x
x = -92 - (435 + 431)
x = -958 kJ mol^-1
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Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
Answer:
A. The partial pressure for CH4 = 0.0925atm
B. The partial pressure for C2H6 = 0.925atm
C. The partial pressure for C3H8 = 0.346atm
D. The partial pressure for C4H10 = 0.115atm
Explanation:
Total pressure = 1.48atm
Total mole = 0.4+4+1.5+0.5=6.4
A. Mole fraction of CH4 = 0.4/6.4 = 0.0625
The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm
B. Mole fraction of C2H6 = 4/6.4 = 0.625
The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm
C. Mole fraction of C3H8 = 1.5/6.4 = 0.234
The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm
D. Mole fraction of C4H10 = 0.5/6.4 = 0.078
The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm
