we know that,
molarity=(given mass/molar mass)*(1000/volume of solution in ml)
all the dats are given,only we have to find out the given mass or the present mass of the KCl.
1.25=(given mass/74.5)*(1000/750)
=> given mass=(1.25*750*74.5)/1000
=69.84g.
About 0.013
Balanced equation and potassium limits and drives the reaction.2K + Cl2 -> 2KCl6.75 grams K (1 mole K/39.10 grams)(2 mole KCl/2 mole K)(74.55 grams /1 mole KCl)= 12.9 grams potassium chloride produced==============================
.250M/.350L = .714 mol .714 mol x 58g/1mol = 41.7g 41.7g NaCl is the answer
The mass is 234,8 g.
There are 23.5 grams of solute contained for every 1,000,000 liters of solution that contains 21.7 ppm of SnC12. Therefore, there is 0.00008365 grams of solute in 3.5 liters of solution.Ê
Tablet Klor Con contains potassium chloride in it. 20 milliequivalent of potassium chloride equals to 1.5 grams of potassium chloride.
About 0.013
50 g of potassium chloride are dissolved in 100 g water at cca. 80 oC.
I Don't knows Sorry
Balanced equation and potassium limits and drives the reaction.2K + Cl2 -> 2KCl6.75 grams K (1 mole K/39.10 grams)(2 mole KCl/2 mole K)(74.55 grams /1 mole KCl)= 12.9 grams potassium chloride produced==============================
4 milliequivalents of sodium chloride solution is a solution having 0,2338 g in 1 L.
438 grams.
.250M/.350L = .714 mol .714 mol x 58g/1mol = 41.7g 41.7g NaCl is the answer
For 0,5 mol the answer is 26,745 g.
Answering "http://wiki.answers.com/Q/How_many_grams_of_calcium_chloride_are_needed_to_produce_10_g_of_potassium_chloride"
You get salt at the bottom of the pan!
This is (mass of solute) divided by (mass of total solution) expressed as a percentage. The solute is what you are dissolving into the solution. Example: you have 90 grams of water, and you add 10 grams of salt (sodium chloride). The water is the solvent, sodium chloride is the solute, and the solution is salt water. 90 grams + 10 grams = 100 grams (mass of total solution). (10 grams) / (100 grams) = 0.1 --> 10% mass mass percent concentration.