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What is the percent yield of water if we start with 5 g C, Hg and produce 6.13 g H,O? The end point in a titration of a 58 mL
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Answer #1

According to HOMEWORKLIB RULES we have to answer only first one problem

HCl + NaOH = NaCl + H2O

Number of moles NaOH = Molarity * volume in L

= 0.83 * 25/1000

= 0.02075 moles NaOH

At equivalence mole of NaOH is equal to mole of HCl

0.02075 moles NaOH *1/1

= 0.02075 moles HCl

Molarity = number of moles / volume in L

= 0.02075 moles HCl/ 58 ML*1 L/1000 ml

= 0.35775862 M

= 0.36 M

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