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What is the molarity of a solution prepared by dissolving 2.41 g of potassium iodide KI in 100 mL of water?
Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================
A sample of 0.3220 g of an ionic compound containing the bromide ion Br is dissolved in water and treated with an excess of AgNO3 If the mass of the AgBr precipitate that forms is 0.6964 g?
A sample of 0.3220 g of an ionic compound containing the bromide ion Br is dissolved in water and treated with an excess of AgNO3 If the mass of the AgBr precipitate that forms is 0.6964 g then bromide=0.3220g and AgBr precipitate=0.6964.
What happens when potassium iodide react with 2 moles sulphur in presence of moonlight?
When potassium iodide reacts with sulphur in the presence of moonlight, the reaction results in the formation of potassium sulphide and iodine. Moonlight can act as a catalyst for this reaction, helping to facilitate the conversion of the reactants into the products.
What is the balanced equation when iodine reacts with sodium hydroxide?
6NaOH + 3I2 = 5NaI + NaIO3 + 3H2O Six moles of sodium hydroxide and three moles of diatomic iodine yield five moles of sodium iodide, one mole of sodium iodate, and three moles of water. Cheers!
How many moles in 135 g of iodine?
not sure
How many grams are in 0.02 moles of beryllium iodide moles of beryllium iodide?
To find the number of grams in 0.02 moles of beryllium iodide (BeI2), you would first calculate the molar mass of BeI2, which is 262.83 g/mol. Then, you would multiply the molar mass by the number of moles: 0.02 moles * 262.83 g/mol = 5.26 grams of beryllium iodide.
How many moles of the iodide ions are present in 45.6 ml of a 4.00 x 10-3 M solution of Cobalt (II) iodide?
To find the number of moles of iodide ions in the solution, first calculate the total moles of Cobalt (II) iodide (CoI₂) in 45.6 ml of a 4.00 x 10⁻³ M solution. Moles of CoI₂ = concentration (M) × volume (L) = 4.00 x 10⁻³ mol/L × 0.0456 L = 1.824 x 10⁻⁴ moles. Since each CoI₂ produces two iodide ions (I⁻), the total moles of iodide ions will be 1.824 x 10⁻⁴ moles × 2 = 3.648 x 10⁻⁴ moles of iodide ions.
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
How many grams are in 0.02 moles of beryllium iodide?
0.02 moles of beryllium diiodide = 5,256 grams
How many moles of potassium iodide are needed to make 750 ml of a 1.8 M solution?
To find the number of moles of potassium iodide needed, multiply the volume of the solution (750 ml) by the molarity (1.8 moles/L). First, convert the volume to liters (750 ml = 0.75 L), then multiply 0.75 L by 1.8 moles/L to get 1.35 moles of potassium iodide.
How many moles of iodine are needed to react with 5 moles of potassium to create potassium iodide?
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.
How are moles converted into number of particles?
The moles are converted into a number of particles by multiplying 6.02 by 10(with the power of 23)
HOW TO prepare 0.1 n iodide solution in 500 ml water?
0.1 N iodide would be 0.1 moles of the iodide salt (e.g. KI) per liter of solution. For 500 ml, you would need 0.05 moles of the iodide salt. You need to state the salt (KI, NaI, LiI, etc.) in order to determine the actual mass required.
What mass of the precipitate could be produced by adding 100.0ml of 0.887 m agno3 to a na3po4 solution assume that the sodium phosphate reactant is present in excess?
To find the mass of the precipitate that forms when 100.0mL of 0.887M AgNO3 is added to a Na3PO4 solution, you need to determine the limiting reactant. Since Na3PO4 is in excess, AgNO3 is the limiting reactant. Calculate the moles of AgNO3 using its molarity and volume, then use the mole ratio between AgNO3 and the precipitate to find the moles of the precipitate. Finally, convert the moles of the precipitate to mass using its molar mass.
What is the balance equation on Calcium Nitrate with Potassium Iodide?
When calcium nitrate (Ca(NO₃)₂) reacts with potassium iodide (KI), the balanced chemical equation is: [ \text{Ca(NO}_3\text{)}_2 + 2 \text{KI} \rightarrow \text{CaI}_2 + 2 \text{KNO}_3 ] In this reaction, one mole of calcium nitrate reacts with two moles of potassium iodide to produce one mole of calcium iodide and two moles of potassium nitrate.
How do you find the amount of precipitate in a saturated solution?
first you need to know the number of liters and moles and the equation. you do someting then multiply the liters times the moles. first you need to know the number of liters and moles and the equation. you do someting then multiply the liters times the moles.
What is 2.20 moles of Sn converted into mass grams?
The number 2.20 moles of Sn equals 261.14 grams. This is a taught in biology.
