20.2 g of CuCl2 = .1502 mol CuCl2 M=mol/L M=.1502 mol/L
The Molarity is equal to the number of moles divided by the liters. M=mol/L M=0.120 mol/5.0 L M=0.024 so the molarity is 0.024 M
0.751 mol/0.951 L = 0.790 mol/liter = 0.790 M
25g HCl 1 mol 36.46g HCl =.686 mol M=.686 mol/1.5 L=.457M pH= -log(.457) pH= .34
3 Litre ( = 6.00 mol / 2.00 mol/L )
4.25 grams. .050 M = .050 mol/1 L 5.0 L x .050 mol/L (cancel out L to get mol as a unit)= .25 mol Atomic mass of Ammonia (NH3)= 17 g/mol .25 mol x 17 g/mol (cancel out mol to get g as a unit)= 4.25 g
20.2 g of CuCl2 = .1502 mol CuCl2 M=mol/L M=.1502 mol/L
2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of water.
The Molarity is equal to the number of moles divided by the liters. M=mol/L M=0.120 mol/5.0 L M=0.024 so the molarity is 0.024 M
Molarity (M) = moles of solute (mol) / liter of solution (L)M = mol / LYou have 250 mL of Solution, which is250 mL x ( 1 L / 1000 mL ) = ( 250 / 1000 ) L = .25 LSolute is just what's dissolvedSolvent is just what it's being dissolved inSolution is the solute and the solvent.M = mol / LM = 0.65 mol / 0.25 Liters = 2.6 mol/LThe two numbers that you are given, 0.65 moles and 250 mL both have two significant figures, and the answer is two significant figures (2.6 mol/L)Therefore the answer is 2.6 mol/L.
0.751 mol/0.951 L = 0.790 mol/liter = 0.790 M
0.5 Moles If you have a 0.25 M solution, you have 0.25 mol/dm3, or 0.25 moles in 1 L (0.25 mol/L) If you have 2 L of solution, you have 2 L x 0.25 mol/L = 0.5 mol The L's cancel out, and you're left with moles.
1.52 mol/L * 1.00 L = 1.52 mol
95g NaOH * 1 mol NaOH/ = 2.375 mol NaOH 40g NaOH Molarity (M) = mol / L M = (2.375) / (0.450) M = 5.28
molarity = moles/litre [solution] = 12 mol/6 L [solution] = 2 mol/L = 2 M
0,4 mol NaCl is 23,376 g.2,85 mol NaCl is 166,554 g to 1L.140 mL solution NaCl 2,85 M contain 0,4 mol NaCl.
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