Answer:
Repeated SN2 reactions occur leading to the formation of a racemic mixture
Explanation:
S-2-iodooctane is a chiral alkyl halide with an asymmetric carbon atom. The presence of an asymmetric carbon atom implies that it can rotate plane polarized light and thus lead to optical isomerism. The two configurations of the compound are R/S according to the Cahn-Prelong-Ingold system.
However, when S-2-iodooctane is treated with sodium iodide in acetone, repeated SN2 reactions occur since the iodide ion is both a good nucleophile and a good leaving group. Hence a racemic modification is formed in the system with time hence we end up with (±)- Iodooctane.
Explanation:
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6.21 x 10^3 = (Move decimal point 3 spaces to the right)
6210
6210 (0.1050)
652.05
Answer:
4KNO3 ==> 2K2O + 2N2 + 5O2
Explanation:
It's a decomposition, but not a simple one.
KNO3 ==> K2O + N2 + O2 I don't usually do this, but I think the easiest way to proceed is to balancing the K and N together. That will require a 2 in front of KNO3
4KNO3 ==> 2K2O + 2N2 + 5O2
Now you have (3*4) = 12 oxygens. Two are on the K2O. So the other 10 must be on the O2
That should do it.
B and e
first we need to balance the NH3 hence first we do E and multiplying the coefficient by 2. that will leave us with N2+H2–>2NH3.
N2 and H2 is balanced and now all that is left to do is the balance H2 by 3 as there is 6H on RHS hence we need 6H on LHS
