Answer: To describe an object’s motion, you must have a distance, km, mm, m, a speed, and time to show how long, seconds, minuets, hours, days.
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;
![K=Ae^{\frac{-E}{RT}}](https://tex.z-dn.net/?f=K%3DAe%5E%7B%5Cfrac%7B-E%7D%7BRT%7D%7D)
If
&
![K_2=Ae^{\frac{-E_2}{RT}} -------equation 2](https://tex.z-dn.net/?f=K_2%3DAe%5E%7B%5Cfrac%7B-E_2%7D%7BRT%7D%7D%20-------equation%202)
![\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7BK_1%7D%7BK_2%7D%20%3D%20e%5E%7B%5Cfrac%7B-E_1-E_2%7D%7BRT%7D)
![E_1= E_2-RT*In(\frac{K_1}{K_2})](https://tex.z-dn.net/?f=E_1%3D%20E_2-RT%2AIn%28%5Cfrac%7BK_1%7D%7BK_2%7D%29)
![E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})](https://tex.z-dn.net/?f=E_1%3D%2050%2C000-8.314%2A310.15%2AIn%28%5Cfrac%7B3.50%2A10%5E3%7D%7B1%7D%29)
![J/mol](https://tex.z-dn.net/?f=J%2Fmol)
E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol
11. ionic charge +1, helium.
12. ionic charge 2-, neon.
13. ionic charge 3+, neon.
18
In chemistry, a group is a column of elements in the periodic table of the chemical elements. There are 18 numbered groups in the periodic table; the f-block columns are not numbered.
Answer:
the second one I think...