Molarity = moles of solute/Liters of solutionOr, for our purposes....,Liters of solution (volume) = moles of solute/MolarityVolume (liters) = 0.150 moles HCl/4.00 M HCl= 0.0375 liters = 37.5 milliliters======================
In order to get 10 percent HCl how much liters of water is needed when combined with 0 Celsius degrees 0.7 atmosphere pressure and 160 liters of HCl it will take a lot of thinking. The answer to this question is 1.64L.
First get moles HCl, then get the grams HCl.Molarity = moles of solute/Liters of solution3.8 M HCl = X moles/17.2 Liters= 65.36 moles HCl===============so,65.36 moles HCl (36.458 grams/1 mole HCl)= 2382.89 grams HCl======================and,put this amount, 2382.89 grams, 65.36 moles HCl into the 17.2 liters of solution ( water, I suppose ) 2382.89 grams = 2.38 kilograms. I suppose you could weigh out the HCl in a flask of some sort.
There are approximately 1.04 liters in one kilogram.
"Pounds" means weight. "Liters" means space. The number of pounds in 50 liters depends on what substance is in them. If the liters are empty, then there are no pounds at all in them.
Molarity = moles of solute/Liters of solutionSo, get moles HCl.73 grams HCl (1 mole HCl/36.458)= 2.00 moles HCl---------------------------Molarity = 2.00 moles HCl/2 Liters= 1 M HCl=======
31 g of HCl divided by the molar mass of HCl to find how many mols. HCl --> H(1) + Cl(1) = molar mass of HCl 1.008(1) + 35.45(1)= 36.428 molar mass 31/36.428= 2.97 which is the mols of solute The mols of solute divided by kg of solute = molality 2.97/.500= 1.7 molality
Molarity = moles of solute/Liters of solution ( 50 ml = 0.05 Liters ) 12 M HCl = moles HCl/0.05 Liters = 0.60 moles HCl
Molarity = moles of solute/Liters of solutionOr, for our purposes....,Liters of solution (volume) = moles of solute/MolarityVolume (liters) = 0.150 moles HCl/4.00 M HCl= 0.0375 liters = 37.5 milliliters======================
Molarity = moles of solute/liters of solution Molarity = 0.597 moles HCl/0.169 liters = 3.53 M HCl ------------------
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
1 (kg / liters) per second = 3600 (kg / liters) per hour. So, multiply kg/l per second by 3,600 to get kg/l per hour.
In order to get 10 percent HCl how much liters of water is needed when combined with 0 Celsius degrees 0.7 atmosphere pressure and 160 liters of HCl it will take a lot of thinking. The answer to this question is 1.64L.
First get moles HCl, then get the grams HCl.Molarity = moles of solute/Liters of solution3.8 M HCl = X moles/17.2 Liters= 65.36 moles HCl===============so,65.36 moles HCl (36.458 grams/1 mole HCl)= 2382.89 grams HCl======================and,put this amount, 2382.89 grams, 65.36 moles HCl into the 17.2 liters of solution ( water, I suppose ) 2382.89 grams = 2.38 kilograms. I suppose you could weigh out the HCl in a flask of some sort.
1 Liter = 1 kg --> 3.5 Liters = 3.5 kg
Balanced equation. NaOH + HCl -> NaCl + H2O all one to one. find moles HCl. 11 grams HCl (1 mole HCl/36.458 grams ) = 0.3017 moles HCl Moles HCl same as moles NaOH Molarity = moles of solute/Liters of solution 1.06 M NaOH = 0.3017 moles NaOH/liters of solution = 0.2846 Liters this is equal to..... 285 milliliters of NaOH needed
Molarity = moles of solute/liters of solution, so... 0.400M HCl (X mols HCl/0.250L ) = 0.100 moles HCl