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C1 x V1 = C2 x V2

.266 M x V1 = .075 M x 150 mL

V1 = (.075 M x 150 mL)/.266 M = 42.3 mL

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Q: How many milliliters of a 0.266 m lino3 solution are required to make 150.0 ml of 0.075 m lino3 solution?
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What mass of each product results if 750 mL of 6.00 M H3PO4 reacts to the equation?

What I would first is to find the amount of mols of H3PO4. Since you know that 1 M means there is 1 mol of solution in every 1 liter, you can figure out the moles of the solution. The solution is 6M, so that means that to find the moles of solute in the solution you multiply .750 (750 milliliters converted to liters) by 6. I think you did something similar, but .0075 L makes 7.5 milliliters. This will give you 4.5 mols of H3PO4.Now you can use stoichiometry to find the grams of Ca3(PO4)2. The balanced equation states that it takes 2 mols of H3PO4 to get 1 mol of Ca3(PO4)2:4.5 mol H3PO4 * 1 mol Ca3(PO4)2 / 2 mol H3PO4 = 2.25 mol Ca3(PO4)2Find the molar mass of Ca3(PO4)2 to convert the mols to grams:3*40 + 2*31 + 8*16 = 310 grams / mol2.25 mol Ca3(PO4)2 * 310 grams Ca3(PO4)2 / 1 mol Ca3(PO4)2 = 697.5 grams Ca3(PO4)2You can use the same process to find the mass of water.4.5 mol H3PO4 * 6 mol H2O / 2 mol H3PO4 = 13.5 mol H2O13.5 mol H2O * 18 g H2O / 1 mol H2O = 243 g H2OHope that helped! NB this is not my work just saw it on another website


What is the pH of a 0.100 molar HCl solution?

The answer depends on several unspecified variables, most importantly the final molarity of the solution, which depends on the final volume. You can calculate the value yourself using the formula: pH = -log[H+] where [H+] is the final concentration of H+ ions in solution. For HCl, [H+] is equal to molarity. So, for example, if you add 50.0 ml of 1.0M HCl to 950 ml of deionized water, your final concentation is: (50.0 ml/1000 ml) * (1.0M) = 0.05M Therefore: pH = -log[0.05] = 1.3