Anhydrous magnesium sulfate, $MgSO_4$ (aka in the lab as $"magsulf"$) is commonly used as a drying agent in phase extractions. It has next to no solubility in organic phases such as ether and methylene chloride. Why should this be important?
Possibly, the best means would be by complexometry. So, perhaps you would take thesalt, $ZnSO_4*7H_2O$, or one of its several hydrates, and perform a titration with $"EDTA"$................. And $"EDTA"$ reacts...
And so we gots.... $Ba^(2+) + SO_4^(2-) rarr BaSO_4(s)darr$ ...as the net ionic equation. The magnesium and nitrate ions are along for the ride.....and stay in solution. The complete metathesis...
$54.64$ $"g"$ anhydrate With $"100.0 g"$ of the heptahydrate, use the molar mass to determine how many mols you have. Mols are interconvertible, as one mol...
are balanced with respect to mass and charge. For every reactant particle, there is a corresponding product particle. And for every reactant charge, there must be a corresponding product...
In $"MgSO"_4$, the number of oxygen atoms is represented by the subscript directly next to the $"O"$. In $"7H"_2"O"$, the coefficient $7$ can be multiplied throughout each element in the...
First, work out the mass of each element that was present in the original compound. Carbon is always present as $CO_2$ in the ratio (12.011 g / 44.0098 g), and...
If we assume that the (23) of the gas has been compared to the density of $H_2$,then the molar mass of the gas becomes $46"g/mol"$ The organic compound burnt...
And $"mass of solute "=" moles"xx"molar mass"$ And so we calculate the product: $"Volume"xx"Concentration"xx"Molar mass of magnesium sulfate"$ $=125xx10^-3cancel(L)xx0.200*cancel(mol*L^-1)xx120.37*g*cancel(mol^-1)$, an answer with units of mass as required. $=??*g$
You specify a $0.5*mol*L^-1$ solution of $MgSO_4(aq)$. Since $"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$, $"Moles of solute"="Concentration"xx"Volume of solution"$. If we have a $1000*mL$ volume, then....... $"Moles of...
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