Lorrie Leisure
Friday, August 12, 2011
Chemistry please help?
What is the m of potium iodide (166.0028 g/mol) is precipitated from 4.84 g of lead(II) iodide(461.009 g/mol) ? __Pb(NO3)2(aq) + __KI(s) __PbI2(s) + __KNO3(aq)
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