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^-
When the block of iron is placed in water the volume of water that is displaced is 27.0 cm³
<u><em> calculation</em></u>
The volume water that is displaced is equal to volume of block of the iron
volume of block of iron = length x width x height
length= 3 cm
width = 3 cm
height = 3 cm
volume is therefore = 3 cm x 3 cm x 3 cm = 27 cm³ therefore the volume displaced = 27 cm³ since the volume of water displaced is equal to volume of block.
