In a man undergoing surgery, it was necessary to aspirate the contents of the uppergastrointestinal tract. After surgery, the fo
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<u>Answer:</u> The vapor pressure of solution is 459.17 mmHg
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For testosterone:</u>
Given mass of testosterone = 7.752 g
Molar mass of testosterone = 288.4 g/mol
Putting values in equation 1, we get:
- <u>For diethyl ether:</u>
Given mass of diethyl ether = 208.0 g
Molar mass of diethyl ether = 74.12 g/mol
Putting values in equation 1, we get:
Mole fraction of a substance is calculated by using the equation:
The formula for relative lowering of vapor pressure will be:
where,
= vapor pressure of solvent (diethyl ether) = 463.57 mmHg
= vapor pressure of the solution = ?
i = Van't Hoff factor = 1 (for non electrolytes)
= mole fraction of solute (testosterone) = 0.0095
Putting values in above equation, we get:
Hence, the vapor pressure of solution is 459.17 mmHg
Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;
If &
E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol