The balanced equation for the reaction between MnSO4 and Na2CO3 is: MnSO4 + Na2CO3 -> MnCO3 + Na2SO4.
Colourless, or slightly white
I must have the same sheet as you as I have the exact same question. Okay, The RFM of Manganese (Mn) is 54.9 The RFM of MnSO4 4H20 is 223 The percentage of Mn in MnSO4 4H2O is 54.9/223 X 100 and that is 24.6 % 0.1 Moles of MnSO4 4H20 is 223/100 X10 = 22.3 24.6 % (percentage of Mn in MnSO4 4H2O) of 22.3 (0.1 Moles of MnSO4 4H2O) is 5.49g Then you do 5.49g/54.9g (RFM of Mn) which is 0.1 Moles of Mn, which is your answer.
The molar mass for MnSO4 (manganese(II) sulfate) is approximately 151.0 g/mol. This value is calculated by adding the atomic masses of one manganese atom (Mn) and one sulfate ion (SO4).
Mn(s) + H2SO4(aq) ------> MnSO4(aq) + H2(g) [molecular] Mn(s) + 2H+(aq) -----> Mn2+(aq) + H2(g) [net ionic]
The balanced equation for the reaction between MnSO4 and Na2CO3 is: MnSO4 + Na2CO3 -> MnCO3 + Na2SO4.
The conventional name of MnSO4 is manganese(II) sulfate.
For the anhydrous MnSO4 the mass is 167,61 g.
MnSO4 is the only one that directly reacts with O2 at BOD?With alkaline: MnSO4 + 2OH = MnOH2 +SO44Mn(OH)2 + 2 H20 + O2 = 4 Mn(OH)3
No, MnSO4 is not Epsom salt. Epsom salt is actually magnesium sulfate (MgSO4). MnSO4 is manganese(II) sulfate, which is a different compound with manganese and sulfate ions.
MnSO4.
To produce one ton of manganese sulfate (MnSO4), you would need roughly 700 kilograms of manganese (Mn). This is because manganese sulfate contains about 32.5% manganese by weight.
Formula: MnSO4
MnSO4
Yes, MnSO4 (manganese(II) sulfate) is soluble in cold water.
In MnSO4, the oxidation number of Mn (manganese) is +2, the oxidation number of S (sulfur) is +6, and the oxidation number of each O (oxygen) is -2.
colourless is farbloss