The question is simply asking to convert from units of mass to units of moles. To do this, a conversion factor is needed. For this case, we use the molar mass of glucose which is equal to 180.18 g/mol. We do as follows:
134 g C6H12O6 ( 1 mol / 180.18 g ) = 0.744 mol <span>C6H12O6</span>
Answer:
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" 2 Li + Br2<span> ----> 2 LiBr " .
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Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
I think it is b.heat transfers to his body from the water by conduction
