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SELECTED ARTICLES. 
discontinued when the chloride of oxide is reduced to half its 
previous volume, the residue being then a mixture of neutral 
metallic chloride, with a small quantity of chlorate, the forma- 
tion of which appears to me accidental. 
When on the other hand, chloride of lime or the neutral 
chlorides of potassa and soda prepared from chloride of lime 
by double decomposition are distilled, the product is water, 
containing barely appreciable traces of hypochlorous acid or 
chlorine. The following formulae express this difference in 
the results: 
Ch 2 + Ch 2 K0 = Ch'K -f Ch 2 ; 
Thus the excess of chlorine by its affinity for oxygen, may 
favour the decomposition of the chloride of oxide, which 
always has a tendency to transform itself into metallic chlo- 
ride and hypochlorous acid. Without the free chlorine, we 
could not account for the formation of hypochlorous acid; for 
we have, 
2Ch 2 KO = Ch 2 + Ch 2 K + KO, 
which does not express the reaction, because the residue of 
the distillation is not alkaline, as it should be, were the for- 
mula correct. 
It is scarcely possible to explain the phenomena of the dis- 
tillation of the chlorides of potassa and soda with excess of 
chlorine, by Balard's hypothesis; for according to it, we must 
suppose that the free chlorine first decomposes the hypochlorite 
and then reacts on the oxide thus set free, so as to form with 
it a metallic chloride, and hypochlorous acid, a complicated 
and scarcely admissible reaction. 
We know too that when liquid hypochlorous acid is poured 
on a metallic chloride of the first section, the chloride of so- 
dium for instance, it is decomposed with effervescence due to 
the escape of chlorine and there remains a bleaching chloride 
of soda, similar to that prepared in the ordinary way. This 
reaction is readily explained by the following formula: 
Ch 2 + Ch 2 Na= Ch 2 + Ch 2 NaO. 
