Answer:
Explanation:
As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.
Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>
So, you must just convert the given temperature, 32°F, to kelvin scale.
You can do that in two stages.
- First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80
- Second, convert 0°C to kelvin:
T (K) = T(°C) + 273.15 K= 273.15 K
Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K
Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.
M₁ x V₁ = M₂ x V₂
2.5 x V₁ = 0.50 x 100.0
2.5 x V₁= 50
V₁ = 50 / 2.5
V₁= 20 mL
hope this helps!
She is experimenting with a placebo effect by giving one mouse the sugar. The dependable variable would be the size of the tumor. It depends on the effectiveness of Drug X and the amount given.
I think I found what your looking for
Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:
Rate factor in the presence of catalyst:
Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:
![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
