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1 mole of KOH is 56.1074 grams
1.350 g/56.1074 g =0.02406 (2.406 x 10^-2) moles of KOH
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∙ 11y ago
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How many moles are in 234.1 grams of KOH?
Molar mass of KOH = 39.1+16.0+1.0 = 56.1Amount of KOH = mass of sample / molar mass = 32/56.1 = 0.570mol
How many grams of solid KOH is necessary to prepare 220.0 ml of 0.500 M KOH solution?
Molarity = moles of solute/Liters of solution ( 220.0 ml = 0.220 Liters ) 0.500 M KOH = moles KOH/0.220 Liters = 0.110 moles KOH (56.108 grams/1 mole KOH) = 6.17 grams solid KOH needed
What is the molarity of a solution with 0.2 moles of KOH in 100 mL of water?
Molarity = moles of solute/Liters of solution get moles KOH 6.31 grams KOH (1 mole KOH/56.108 grams) = 0.11246 moles KOH 0.250 M KOH = 0.11246 moles KOH/XL 0.11246/0.250 = 0.4498 liters = 450 milliliters
How many moles of KOH are contained in 750. mL of 5.00 M KOH solution?
Molarity = moles of solute/volume of solution 0.500 M KOH = moles KOH/125 ml 62.5 millimoles, or to answer the question precisely, 0.0625 moles KOH
What is the concentration of a HBr solution in 12.0 mL of the solution is neutralized by 15.0 mL of a 0.25 M KOH solution?
Balanced equation. KOH + HBr -> KBr + H2O everything is one to one, so... Molarity = moles of solute/liters of solution ( change ml to liters ) 0.25 M KOH = moles KOH/0.015 liters = 0.00375 moles of KOH this is as many moles that you have of HBr, so... Molarity of HBr = 0.00375 moles/0.012 liters = a concentration of HBr that is 0.31 M
How many moles are in 234.1 grams of KOH?
Molar mass of KOH = 39.1+16.0+1.0 = 56.1Amount of KOH = mass of sample / molar mass = 32/56.1 = 0.570mol
How many grams of solid KOH is necessary to prepare 220.0 ml of 0.500 M KOH solution?
Molarity = moles of solute/Liters of solution ( 220.0 ml = 0.220 Liters ) 0.500 M KOH = moles KOH/0.220 Liters = 0.110 moles KOH (56.108 grams/1 mole KOH) = 6.17 grams solid KOH needed
How many moles of potassium hydroxide are required to prepare 300 mL of 0.250 M solution?
Molarity = moles of solute/liters of solution ( 300 ml = 0.300 liter ) 0.250 molar KOH = moles KOH/0.300 liters = 0.075 moles KOH
What is the molarity of a solution with 0.2 moles of KOH in 100 mL of water?
Molarity = moles of solute/Liters of solution get moles KOH 6.31 grams KOH (1 mole KOH/56.108 grams) = 0.11246 moles KOH 0.250 M KOH = 0.11246 moles KOH/XL 0.11246/0.250 = 0.4498 liters = 450 milliliters
How many moles are present in two hundredths of a gram of KOH?
None. Moles live underground.
How many moles of KOH are contained in 750. mL of 5.00 M KOH solution?
Molarity = moles of solute/volume of solution 0.500 M KOH = moles KOH/125 ml 62.5 millimoles, or to answer the question precisely, 0.0625 moles KOH
How many grams of solid KOH is necessary to prepare 60ml of 2M KOH solution?
Moles KOH = Molarity x Volume = 0.214 moles/liter x 0.0602 liters = 0.0129 moles KOH. Remember, 60.2 mL = 0.062L
How many moles are in 35.28 g of koh?
35.28 (g) / 56.1056 (g/mol) = 0.6288 mol KOH
What is the concentration of a HBr solution in 12.0 mL of the solution is neutralized by 15.0 mL of a 0.25 M KOH solution?
Balanced equation. KOH + HBr -> KBr + H2O everything is one to one, so... Molarity = moles of solute/liters of solution ( change ml to liters ) 0.25 M KOH = moles KOH/0.015 liters = 0.00375 moles of KOH this is as many moles that you have of HBr, so... Molarity of HBr = 0.00375 moles/0.012 liters = a concentration of HBr that is 0.31 M
In the reaction MgCl2 plus 2KOH---Mg(OH)2 plus 2KCl how many moles of KOH react with 1 mole of MgCl2?
MgCl2 + 2KOH ==> Mg(OH)2 + 2KCl1 mole MgCl2 reacts with 2 moles KOH 2 moles KOH x 56.1 g/mole = 112.2 g KOH = 100 g KOH (to 1 significant figure based on 1 mole)
What mass of KOH is needed to make 400mL of 725 M KOH?
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
How many moles of base are needed to neutralize 2 mol of acid for HNO3 plus KOH?
Since both the acid and the base have equivalent weights equal to their formula weights, 2 moles of KOH are needed to neutralize 2 moles of nitric acid.
