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A mixture of KCl and KNO3 is 44.20 percent potassium by mass the percentage of kcl in the mixture is?
40%
How do you prepare 3.5M KCl solution?
3.5M means 3.5 moles of KCl. 1 mole is the combined molecular weight of the compound per litre. Molecular weight of K (potassium) = 39.10g Molecular weight of Cl (chlorine) = 35.45g So molecular weight of KCl = (39.10 + 35.45) = 74.55g That means that 1 mole of KCL = 74.55 grams per litre If 1 mole of KCL contains 74.55g then 3.5M of KCL will contain 74.55g x 3.5 and so 3.5M of KCL = 260.925g/L
When 200 mL of water are added to 100mL of 12 percent KCL solution the final concentration of KCL is?
4.0%
How many ppm is 2 percent KCl?
20000 mg/l
What if a potassium chlorate sample is contaminated with KCL would the experimental percent oxygen be higher or lower than the theoretical percent oxygen?
the experimental % oxygen would be lower because there would be more KCL in the simple than oxygen
A mixture of KCl and KNO3 is 44.20 percent potassium by mass the percentage of kcl in the mixture is?
40%
How do you prepare 3.5M KCl solution?
3.5M means 3.5 moles of KCl. 1 mole is the combined molecular weight of the compound per litre. Molecular weight of K (potassium) = 39.10g Molecular weight of Cl (chlorine) = 35.45g So molecular weight of KCl = (39.10 + 35.45) = 74.55g That means that 1 mole of KCL = 74.55 grams per litre If 1 mole of KCL contains 74.55g then 3.5M of KCL will contain 74.55g x 3.5 and so 3.5M of KCL = 260.925g/L
When 200 mL of water are added to 100mL of 12 percent KCL solution the final concentration of KCL is?
4.0%
What is the Empirical formula of the compound that is 52.7 percent K 47.3 percent Cl?
KCl
What is the equivalent weight of Ag?
equivalent weight of silver nitrate = 169.87 so 0.1 N Ag NO3 = 16.987 gm /litre of AgNO3 now equivalent weight of KCl = 74.55 so 0.1 N KCl = 7.455 gm/litre so 0.1 N AgNO3 = 0.1N KCl = 7.456 gm of KCl [ not mg ]
How many ppm is 2 percent KCl?
20000 mg/l
How much 75.66MKCL is required toprepare 1L of solution?
I did not know that you could get a concentration of 75.66 M KCl, but; Molarity = moles of solute/Liters of solution 75.66 M KCl = moles KCl/1 liter = 75.66 moles of KCl 75.66 moles KCl (74.55 grams/1 mole KCl) = 5640 grams KCl that is about 13 pounds of KCl in 1 liter of solution. This is why I think there is something really wrong with this problem!
Powder of nacl and kcl is dssolved in 80 ml of water and trested with 277 ml of 0.264 molar AgNO3 what is the percent by weight of the NaCl?
Sodium chloride contain chlorine and sodium.
How many g of KCl do you need to make 2.5 L of a 10 weight volume solution?
For a 10 % solution you need 250 g KCl.
What if a potassium chlorate sample is contaminated with KCL would the experimental percent oxygen be higher or lower than the theoretical percent oxygen?
the experimental % oxygen would be lower because there would be more KCL in the simple than oxygen
Help with stoichiometry show steps How many grams of KClO3 must be decomposed to yield 30 grams of oxygen?
2 KClO3 -> KCL + 3O2 Molar weight of O2 = 32 grams/mole (so close it doesn't matter) 30 grams/32grams/mole = 0.9375 moles Molar weight of KCL = 39+35.5 = 74.5 grams/mole (Want more accuracy? Do it yourself?) now if we have 3 moles of O2 then we have 2 moles of KCl. If we have one mole of O2 then we have 2/3 moles of KCL What ever moles we have of O2 we must multiply it by 2/3 to get the moles of KCl So we have 0.9375moles of O2 x 2/3 = 0.625 moles of KCl So 0.625 moles of KCl x 74.5 grams/mole KCl = 46.5625 grams KCl
How much kcl require for prepaire 3M kcl?
Molar mass of KCl = 74.55g/mol.ie, if you dissolve 74.55g KCl in 1litre (1000 ml) of water, it will be 1M KCl solution.If you want to make 3M KCl solution,Dissolve 3 ×74.55 = 223.65g KCl in 1litre (1000 ml) of water.If you want to make different molar solutions of KCl, just calculate as per below given equation.Weight of KCl to be weighed =Molarity of the solution needed × Molecular weight of KCl (ie, 74.55) × Volume of solution needed in ml / 1000.To prepare 3M KCl in 1 litre, it can be calculated as follows,3 mol × 74.55 g/mol × 1000 ml / 1000 ml = 223.65gByPraveen P Thalichalam, Kasaragod (Dist), Kerala.
