Calculate the number of moles and the mass of negative ions present in 15.67 g of magnesium nitrate, Mg(NO3)2.
molar mass of Mg(NO3)2 = 148.3 g/mol
no. of moles = 15.67/148.3 = 0.105 moles
One mole of Mg(NO3)2 ---> Mg 2+ + 2 NO3- conrtains one mole of Mg and 2 mol of NO3-
no.of moles of Anion = 0.105*2 = 0.21 mol
mass of NO3 2- = 62+2 =64
weight -= 0.21* 64 = 13.44 gram
Calculate the number of moles and the mass of negative ions present in 15.67 g of...
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of oxygen divided by the number of moles of magnesium prior to
rounding off AND the theoretical value of Mgo . Show your work
!!
Determine the percent error using the number of moles
of oxygen divided by the number of moles of magnesium prior to
rounding off AND the theoretical value of MgO 2 . Show your work
!!
Date: Mass of crucible: Mass of crucible + magnesium: Mass of...
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