What product is formed at the positive electrode (anode) when 0.001 mol dm–3 H2SO4 (aq) is
1 answer:
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Answer: 1. moles
2. 90 mg
Explanation:
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus moles of ozone is removed by =
moles of sodium iodide.
Thus moles of sodium iodide are needed to remove
moles of
2.
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by = moles of sodium iodide.
Mass of sodium iodide= (1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of .