Answer:
Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.
Carbons starting from the left end:
- sp²
- sp²
- sp²
- sp
- sp
Refer to the sketch attached.
<h3>Explanation</h3>
The hybridization of a carbon atom depends on the number of electron domains that it has.
Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.
Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.
- A carbon atom with four electron domains is sp³ hybridized;
- A carbon atom with three electron domains is sp² hybridized;
- A carbon atom with two electron domains is sp hybridized.
Starting from the left end (H₂C=CH-) of the molecule:
- The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
- The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
- The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
- The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
- The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.
The first answer is -.595454 the second answer is -1.9488
Answer:
See Explanation Below
Explanation:
A) The rate law can only be on the reactant side and you can only determine it after you get the net ionic equation because of spectators cancelling out. So in this case the rate law is k=[CH3Br]^1 [OH-]^1. The powers are there because the rxn is first order.
B) Since the rxn is first order anything you do to it will be the exact same "counter rxn" per say so since you are decreasing the OH- by 5 the rate will decease by 5
C) The rate will increase by 4 since you are doubling both you have to multiply them both.