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What mass of Cu metal is required to react with 6.8 drops of 16M HNO3? Assume 20 drops = 1 mL. Cu (MM 63.54 g/mol), (16M = 16 mol/L) Cu(s)+4HNO3(aq) a Cu(NO3)2(aq) + 2NO2(g)+ 2H20(0) Hint x drops HNO3 X Cu XLX mol X drops mL mol mol Give answer to 2 significant figures. Do not use scientific notation. After the number, include one space and then the unit, g. Do calculation all at once, round off at the end.
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