What weight of potassium periodate (FW = 230.0) is required to convert 3.00 mg of Mn2+ to permanganate, assuming a 100%

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answerhappygod
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What weight of potassium periodate (FW = 230.0) is required to convert 3.00 mg of Mn2+ to permanganate, assuming a 100%

Post by answerhappygod »

What weight of potassium periodate (FW = 230.0) is required to
convert 3.00 mg of Mn2+ to permanganate, assuming a 100% excess is
desired?
(Answer: 62.8 mg)
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