The nuclear composition and electron configuration of an atom of Scandium 45 and the atomic number is 21 the most common isotype of this element is that nuclear consists of 21 protons and 24 neutrons Scandium is a transition metal group 3 period and the d-block of a periodic table
1.regulating ph
2.heating
3.adding some oppositely charged colloids
Rutherford's experiment was the gold foil experiment.
The gold foil experiment was him shooting alpha particles (you could think of this as a Helium atom without its electrons) into a gold foil. The whole experiment was surrounded with something called Zinc Sulfide which sparked when the alpha particles hit it.
Most of the alpha particles went through, approximately 1 in 8000 alpha particles deflected at a large angle (almost right back to where it was shot).
This constant ratio caused him to conclude that:-the atom was mostly empty space (since most alpha particles went through)-there was something very positive in the atom (the proton)-the proton was very dense (since it made something going light speed deflect back at a large angle)-The proton was also very small (since only 1 in 8000 hit it)
Prior to the discovery of the proton, John Dalton's periodic table was used. Having "elements" such as soda and potash. Now that we have discovered the proton and found out that each atom's number of protons is unique, we used that to classify each element's identity.
Only the amount of gas is held constant.
Answer:
T = 215.33 °C
Explanation:
The activation energy is given by the Arrhenius equation:
<u>Where:</u>
k: is the rate constant
A: is the frequency factor
Ea: is the activation energy
R: is the gas constant = 8.314 J/(K*mol)
T: is the temperature
We have for the uncatalyzed reaction:
Ea₁ = 70 kJ/mol
And for the catalyzed reaction:
Ea₂ = 42 kJ/mol
T₂ = 20 °C = 293 K
The frequency factor A is constant and the initial concentrations are the same.
Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:
(1)
By solving equation (1) for T₁ we have:
Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.
I hope it helps you!
