Answer:
B = CHCl2 + Cl2 --> CHCl3 + Cl
Explanation:
Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.
Free radical chlorination is divided into 3 steps which are:
The initiation step
The propagation step
The termination step
So in reference to the question, propagation step involves two steps.
The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.
The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.
You would find in the attachment the 2 step mechanism.
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hope this helps!
A) electrolyte the others either describe a process, the opposite, or a whole different thing
Answer:
3.75 moles
Explanation:
The chemical equation is 2CH₃OH + 3O₂ -> 2CO₂ + 4H₂O
2 moles of CH₃OH are burned by 3 moles of O₂
For 2.5 moles of CH₃OH are burned by x moles of O₂
Let's solve for x :
2*x=2.5*3 => 2*x=7.5 => x=3.75 moles of O₂ are needed to burn 2.5 moles of CH₃OH
Answer:
No, IR should not soely be used to identify molecules
Explanation:
IR is a method that identifies the functional groups in a molecule by deducing the frequency of stretching and vibration of bonds. Each peculiar type of bond has a frequency for the vibration of each bond represented on the IR spectrum.
However, one method is never enough to identify a compound. A combination of methods must always be used to clear up ambiguities arising from overlapping IR frequencies. Also, interpretation of the nuanced peaks of the fingerprint region in IR spectra is quite challenging and only gives a fair idea of the functional groups present in the compound.
Therefore other methods such as NMR, UV-VISIBLE etc should also be involved in the identification of compounds.