If 1mol ------- is ----------- 6,02×10²³
so x ------- is ----------- 1,39×10²⁴
Answer:
The thermodynamic parameter which is of significance in this case is the 'Reduction Potential' for molecular bromine which is ~ +1.1 v vs N.H.E. In other words, it is a strong oxidizing agent. The bromine will oxidize sulfur compounds in which the valence of sulfur is lower than six to sulfate.
There are many possible reactions. Here is one possible example:
Na2 S2O3 + 4Br2 + 5 H2O = 2NaHSO4 + 8 HBr
To reduce a haloalkane, it has to be eliminated first. This is because 2 bromopropanes are a saturated compound that can not be reduced by a saturated compound. Dehydrohalogenation of the haloalkanes into the form of propene and hydrogen bromide can lead to the elimination. The propene is then reduced to propane.
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