The solubility of potassium permanganate in water at 2o 0C is 64 g/l (or 32 g/0,5 L).
The molar mass of KMnO4 is 158,034 g and 0,35 moles KMnO4 is equal to 55,312 g.
So it is not possible to dissolve o,35 moles KMnO4 in 0,5 L.
Yes.
Okay, a mole of potassium perchlorate contains 6.02x1023 formula units of potassium perchlorate, but you're asking about individual atoms. So, let's look at the formula: KClO3. That's 1 potassium, 1 chlorine, and 3 oxygens, for a total of 5 atoms per formula unit. Now, multiple 5 by Avogadro's number above, to get 30.1x1023, which simplifies to 3.01x1024 atoms.
1
0.1 N KCl is the same as 0.1 M KCl. This requires one to dissolve 0.1 moles per each liter of solution. The molar mass of KCl is 74.6 g/mol. So 0.1 moles = 7.46 gDissolve 7.46 g KCl in enough water to make 1 liter (1000 ml)Dissolve 3.73 g KCl in enough water to make 0.5 liter (500 ml)Dissolve 0.746 g KCl in enough water to make 0.1 liter (100 ml)etc., etc.
Approx 3.29 moles.
How many molecules are in 94 grams of sodium fluoride
This mass is 347,67 for KMnO4.
The compound potassium permanganate has chemical formula KMnO4 Molecular mass of KMnO4 = 39.1 + 54.9 + 4(16.0) = 158.0 Mass of KMnO4 = amount of KMnO4 x Molecular mass of KMnO4 = 2.55 x 158.0 = 403g
Potassium permanganate = KMnO4 Molar mass of KMnO4 K = 1 * 39.10 g = 39.10 g Mn = 1 * 54.94 g = 54.94 g O = 4 * 16.00 g = 64.00 g Total = 158.04 g/mol 17.34 mol KMnO4 * (158.04 g/1 mol KMnO4) = 2740.41 g KMnO4 Convert the grams to kilograms. 1 kg = 1000 g 2740.41 g * (1 kg/1000 g) = 2.74041 kg Therefore, 17.34 moles of potassium permanganate is equal to about 2.74 kilograms.
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Algebraically manipulate, moles of solute = Liters of solution * Molarity Moles KMnO4 = (0.075 Liters)(0.0950 M) = 7.13 X 10 -3 moles KMnO4 ------------------------------------
the charge in KMno4 is zero . it has no unpaired electrons but it is still dark in colored due to charge transfer in KMnO4. from Ayushi Sharma
The formula mass of KMnO4 is 158.0Amount of KMnO4 = mass of sample / molar mass = 100/158.0 = 0.633 molThere are 0.633 moles in 100g of potassium permanganate.
here is the rection... 2KMnO4 + 3H2SO4 = K2SO4 + 2MnSO4 + 3H2O + 2.5O2 (alkaline) and in other conditions... 6 KMnO4 + 9 H2SO4 → 6 MnSO4 + 3 K2SO4 + 9 H2O + 5 O3 The H2SO4/KMnO4 reaction can also produce the oily Mn2O7, which is unstable and can decompose explosively.
1 mol = 118.94 1 mol / 118.94 = 1.70 / x G = 202.10g
The Stoichiometry of molar concentration follows this simple formula: Mol = Volume (dm³) × Molar Concentration (mol/dm³) Hence: Molar Concentration (mol/dm³) = Number of moles (mol)/ Volume (dm³)
there are two moles produced in potassium nitrate.
12 g of potassium is equivalent to 0,307 moles.
Since molecules of potassium contain only single potassium atoms, molecules of iodine contain two atoms, and moles of potassium iodide contain one atom of each element, 2.5 moles of iodine are needed to react completely with 5 moles of potassium.