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2h2o2 â> 2h2o + o2 H= -196kj Given reaction and question below, Is there a faster way to solve this other than calculating the bond energy broken minus bond energy formed
Ope Assume that the bond enthalpies of the oxygen- hydrogen bonds in H2O are not significantly different from those in H202. Based on the valu of ÎHo of the reaction, which of the following could be the bond enthalpies (in kJ/mol) for the bonds broken and formed in the reaction? o-O in O2 500 500 300 300 O-O in H202 O-H (A) 300 (B) 150 (C) 500 (D) 250 500 500 150 150
2h2o2 â> 2h2o + o2 H= -196kj
Given reaction and question below, Is there a faster way to solve this other than calculating the bond energy broken minus bond energy formed
Ope Assume that the bond enthalpies of the oxygen- hydrogen bonds in H2O are not significantly different from those in H202. Based on the valu of ÎHo of the reaction, which of the following could be the bond enthalpies (in kJ/mol) for the bonds broken and formed in the reaction? o-O in O2 500 500 300 300 O-O in H202 O-H (A) 300 (B) 150 (C) 500 (D) 250 500 500 150 150
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Hubert KochLv2
12 Nov 2019
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