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Experimental data have shown that the rate law for the reaction 2 hgcl2(aq) + c2o4 2 -(aq) → 2 cl-(aq) + 2 co2 (g) + hg2cl2 (s) is: rate = k[hgcl2][c2o4 2 -]2 how will the rate of reaction change if the concentration of c2o4 2 - is tripled and the concentration of hgcl2 is doubled?

Respuesta :

superman1987
When the rate = K [ HgCl2] [C2O4 2-]^2
if the concentration of [ C2O4 2- ] is tripled so we will substitute by [3 (C2O4 2-)] instead of [ C2O4 2- ] 
and if the concentration of [HgCl2] is doubled so we will substitute by [2( HgCl2)] instead of [HgCl2] 
So the new rate will be = K [2(Hgcl2)] [3(C2O4 2-)]^2
BTSJungkook4life

Answer:

The rate will be doubled by a Factor of 18

Explanation:

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