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.
In one mole of hydrated manganese II sulphate, there is one mole of manganese atoms. Therefore, in 0.1 moles of hydrated manganese II sulphate, there will be 0.1 moles of manganese.
Mn(s) + H2SO4(aq) ------> MnSO4(aq) + H2(g) [molecular] Mn(s) + 2H+(aq) -----> Mn2+(aq) + H2(g) [net ionic]
MnSO4+Na2CO3
Various type of salts are formed with H2SO4.. They are:K2SO4 , Na2SO4 , MnSO4 , BaSO4
The balanced chemical equation would be K4FeC6N6 + KMnO4 + H2SO4 = KHSO4 + Fe2SO43 + MnSO4 + HNO3 + CO2 + H2O.
MnSO4
Manganese sulfate has an ionic bond.
AnswerMnSO4 is Manganese Sulfate... I think that this is what you were after?
MnSO4.
Formula: MnSO4
In all sulphate compounds 'S' is in 6+ oxidation state.
Mn(s) + H2SO4(aq) ------> MnSO4(aq) + H2(g) [molecular] Mn(s) + 2H+(aq) -----> Mn2+(aq) + H2(g) [net ionic]
Manganese(II) Fluoride = MnF2Manganese(III) Fluoride = MnF3Manganese(IV) Fluoride = MnF4
Manganous sulfate solution for 1L Dissolve one of the following in 1 L of distilled water: 480 g MnSO4 • 4H2O -or- 400 g MnSO4 • 2H2O -or- 364 g MnSO4 • H2O
MnCrO4 is manganese(II) chromate
The chemical equation is:Al2(SO4)3 + 3 Mn = 2 Al + 3 MnSO4
BaNO3 + MnSO4 --> MnNO3 + BaSO4 (Manganes(II) nitrate and barium sulfate) (double replacement reaction)