I think the answer is it represent the shape of orbitals that electrons can occupy within an atom.
Answer:
In the first combination neutralization takes place to give a salt. So, solution 'a' is neutral in nature.
In the solution 'c', both salts are resulted by the combination of weak base and strong acid. The combination of these salts suppresses the acidity.
In last combination basic nature is observed due to the presence of CN⁻ ions. Thus, the solution 'd' is basic in nature.
Out of the five given solutions, 0.0100 M in HF and 0.0100 M in KBr is most acidic. Therefore, solution 'b' is most acidic in nature.
Explanation:
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.
Answer:
1. an educated guess
2. data
3. what changes in experiment
4. what stays the same in both groups
5. the group where nothing changes, normal
6. group with independent variable, what's being tested
The product formed when HCl and CH2Cl2 react is CH4
H
H-C- H methane structure
H
HCl react with CH2Cl2 to form methane (CH4) and chlorine gas(Cl2)
that is,
2HCl(g) + CH2Cl2(l) = CH4 (g) +2Cl2 (g)
