The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell.
The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.
<span>Iron Oxide</span><span> Fe2O3 = 5 atoms.</span>
Answer:
A reaction at equilibrium implies that the concentration/amount of reactants & products present in that reaction don't change but will remain constant. It could also mean that the forward reactions and reverse reactions will occur at the same rates.
Explanation:
A reaction at equilibrium implies that the concentration/amount of reactants & products present in that reaction don't change but will remain constant. It could also mean that the forward reactions and reverse reactions will occur at the same rates.
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
<u>Answer:</u> The rate law for the reaction is
<u>Explanation:</u>
In a mechanism of the reaction, the slow step in the mechanism always determines the rate of the reaction.
For the given chemical reaction:
The intermediate reaction of the mechanism follows:
<u>Step 1:</u>
<u>Step 2:</u>
As, step 1 is the slow step. It is the rate determining step
Rate law for the reaction follows:
Hence, the rate law for the reaction is