Answer:
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Explanation:
Reduction half reaction
2H_2O + CrO_4^2- + 3e -> CrO_2^- + 4OH^-
Oxidation half reaction
2OH^- + SO_3^2- -> SO_4^2- + H_2O + 2e
Balanced overall equation
H_2O + 2CrO_4^2- + 3SO_3^2- -> 3SO_3^2- + 2CrO_2^- + 2OH^-
Answer:
2 sig figs.
Explanation:
Sig Fig Rules:
Any non-zero digit is a significant figure.
Any zeros between 2 non-zero digits are significant figures.
Trailing zeros after the decimal are significant figures.
Fluorine (F), Chlorine (Cl) and Iodine (I) are all found in the same group on the Periodic Table because they have similar physical properties. Since they are all Halogens, they have 7 valence electrons in their outer shell. In order to get a total of 8, they naturally combine with elements of the same isotope (itself), so D comes close to being correct, but it's not the best answer choice.
Because coefficients numbers make the elements to be stable.
