Asked by Arlene
How many milliters of 0.500 M magnesium nitrate can be produced from the reaction of 35.5 grams of magnesium bromide.
Answers
Answered by
DrBob222
moles MgBr2 = grams/molar mass = ??
1 mole MgBr2 = 1 mole Mg(NO3)2
M Mg(NO3)2 = moles/L soln
Solve for L and convert to mL.
1 mole MgBr2 = 1 mole Mg(NO3)2
M Mg(NO3)2 = moles/L soln
Solve for L and convert to mL.
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