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How many grams of potassium permanganate are in 2.20 moles?
To calculate the grams of potassium permanganate in 2.20 moles, you would need to know the molar mass of potassium permanganate. The molar mass of potassium permanganate (KMnO4) is about 158.034 g/mol. So, 2.20 moles of KMnO4 is equal to 2.20 moles x 158.034 g/mol = 347.67 grams of potassium permanganate.
He formula for potassium dichromate is KËÃCrËÃOËÈ How many grams of oxygen will be in 2 moles of potassium dichromate?
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
How many moles in potassium bicarbonate?
You did not describe the amount of potassium bicarbonate amount in grams in your question. But if you are about 1 gram of potassium bicarbonate it will be 0.0099 moles in one gram of potassium bicarbonate. 0.0199 moles in 2 grams of potassium bicarbonate.
Calculate the molarity of a solution of potassium permanganate and density is 1.04 gmL which contans 15.00 percent by mass?
C [mol/L] = w% * rho [g/mL] *1000 [mL/L] / (100% * M [g/mol] ) = = w% * rho [g/mL] *10 [mL/L%] / ( M [g/mol] ) = = 15.00 % * 1.04 [g/mL] *10 [mL/L%] / ( 158.034 [g/mol] ) = = 0.987 [mol/L] the molarity of the solution of potassium [in mol per liter] equals the content of KMnO4 by mass in percents multiply by density in g/mL multiply by 10 [mL/(L*%)] divide by molar mass of KMnO4 (M = 158.034 g/mol) This gives 0.987 mol/L. Thus the molarity equals approximately 1 M.
A certain amount of hydrogen peroxide was dissolved in 100 ml of water and then titrated with 1.68m kmno4 how much h2o2 was dissolved if the titration required 22.3 ml of the kmno4 solution?
To calculate the amount of hydrogen peroxide (H2O2) dissolved, you can use the equation: moles of KMnO4 = moles of H2O2. Firstly, calculate the moles of KMnO4 (given concentration and volume). Then, use the balanced chemical equation to determine the moles of H2O2, as they have a 1:1 stoichiometric ratio. Finally, convert moles of H2O2 to grams using the molar mass to find the amount dissolved in the solution.
How many grams of potassium permanganate are in 2.20 moles?
To calculate the grams of potassium permanganate in 2.20 moles, you would need to know the molar mass of potassium permanganate. The molar mass of potassium permanganate (KMnO4) is about 158.034 g/mol. So, 2.20 moles of KMnO4 is equal to 2.20 moles x 158.034 g/mol = 347.67 grams of potassium permanganate.
If you have 17.34 moles of potassium permanganate how many kilograms of the substance do you have?
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.
Can you dissolve .35 moles of Potassium Permanganate (KMnO4) into 500 mL of water?
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.
How many atom of oxygen are in 17.7g of potassium promangenate?
Assuming that the questioner meant to write "permanganate" instead of "promangenate", the answer may be found as follows: The formula of potassium permanganate is KMnO4, showing four atoms of oxygen per formula unit. The formula unit mass of KMnO4 is 158.03; therefore, 17.7 g of potassium permanganate contains [4(17.7)/158.03](Avogadro's Number) or 2.70 X 1023 atoms of oxygen, to the justified number of significant digits.
How many moles of KMnO4 are present in 75.0 mL of a 0.0950 M solution?
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 ------------------------------------
How many moles are in 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
What is the word equation for potassium permanganate when heated?
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.
What is the number of oxygen atoms in 90 g of potassium permanganate?
In order to find the actual number of Oxygen atoms in that substance, you must first find the molar mass of that substance by adding the atomic masses of all the atoms which make up that compound. Then, divide the mass of the substance you have by that molar mass. After that, multiply that amount by how many Oxygen atoms are in a molecule of the substance to find how many moles of oxygen you have. Then finally, multiply the number of moles of Oxygen by Avogadro's number (6.022*10^23).
What is the molarity of 0.1 normal KMnO4?
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.
How many grams are in 34.7 moles of potassium?
34,7 moles of potassium 1 356,7 g.
How many moles of potassium iodide are there in 50 grams?
For this you need the atomic (molecular) mass of KCl. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. KCl= 74.6 grams50.0 grams KCl / (74.6 grams) = .670 moles KCl
He formula for potassium dichromate is KËÃCrËÃOËÈ How many grams of oxygen will be in 2 moles of potassium dichromate?
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
