In the following reaction, which species is reduced? Au(s) + 3NO3-(aq) + 6H+(aq) → Au3+(aq) + NO(g) + 3H2O (l) H+ N+5 O2- H2O Au

NO3-

Explanation:

Given the reaction equation;

Au(s) + 3NO3-(aq) + 6H+(aq)→Au3+(aq) + 3NO2(g) + 3H2O (l).

We can consider the oxidation states of species on the left and right hand sides of the reaction equation;

Au is in zero oxidation state on the left hand side and an oxidation state of +3 on the righthand side.

NO3- is in oxidation state of +5 on the righthand side and NO2 is in + 4 oxidation state.

H+ is in + 1 oxidation state on both the left and right hand sides of the reaction equation.

Since reduction has to do with a decrease in oxidation number, it follows that NO3- was reduced in the reaction.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.