Mainly, you need to find out how many liters there are in a mole. This varies from one substance to another.
2.5 g per mL = 2,500 g per Litre.
You begin with 1.63 g of MgCl2•xH2O. You end up with 0.762 g MgCl2. That means you lost 1.63 - 0.762 = 0.87 g of H2O. 0.762 g MgCl2 / 95.21 g (molar mass) = .00800 mol MgCl2 0.87 g H2O / 18.02 g (molar mass) = .048 mol H2O x = mol solute / mol solvent x = .048 mol H2O / .00800 mol MgCl2 x = 6 MgCl2•6H2O
Formula for table salt = NaCl 1.00 g = 1.00 / 58.45 mol = 1.71 x 10-2 mol Mole Na required = 1.71 x 10-2 mol Mass of Na = 1.71 x 10-2 mol x 23.0 g /mol = 0.393 g = 393 mg
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.
98.08 g/mol
Density of HCl = 1.186 g/mL Molar Mass HCl = 36.46g/mol We want a concentration in mol/L so we will first convert density into g/L (1.186 g/mL)(1000mL/1L) = 1186 g/L We must now know what density 37% of that is (0.37)(1186 g/L) = 438.82 g/L Now divide this density by the molar mass to cance out the g and give you mol/L (concentration) C = (438.82g/L)/(36.46g/mol) C = 12.04 M Since HCl is monoprotic, 1M = 1N. Therefore, 37% HCl is ~12N
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
Multiply the molarity (M, which is in mol/L) with the volume (in L) to get the number of moles needed. Then multiply the result with the molar mass. If you look at the units they will cancel to give an answer in grams. (mol/L)*(L)=mol, (mol)*(g/mol)=g So for the numerical answer you get (0.0552 mol/L)*(0.750 L)*(119.00 g/mol)= 4.93 g KBr
mol/L
C [mol/L] = w% * rho [g/mL] *1000 [mL/L] / (100% * M [g/mol] ) = = w% * rho [g/mL] *10 [mL/L%] / ( M [g/mol] ) = = 15.00 % * 1.04 [g/mL] *10 [mL/L%] / ( 158.034 [g/mol] ) = = 0.987 [mol/L] the molarity of the solution of potassium [in mol per liter] equals the content of KMnO4 by mass in percents multiply by density in g/mL multiply by 10 [mL/(L*%)] divide by molar mass of KMnO4 (M = 158.034 g/mol) This gives 0.987 mol/L. Thus the molarity equals approximately 1 M.
Examples are: mol/L, g/L, g/1oo mL.
First we need to convert grams(g) to moles(mol) and mL to L, since M(molarity)=mol/L.Ag=107.868N=14O=16 x3=48AgNO3 =169.868 g/mol42.5g x 1 mol/169.868g = .25mol AgNO3100mL x 10-3 L/ 1mL = .1LM=mol/L.25mol/.1L = 2.5M AgNO3
Molarity
convert to moles: 0.45 g x 1 mol/36.46 g/mol = 0.0124 mol M = mol/L M = 0.0124/3.5 M = 0.0035 [H+] = 0.0035 pH = -log(0.0035) pH = 2.46
[4.00 (g) / 40.0 (g/mol)] / 2.00 (L) = [0.100 (mol)] / 2.00 (L) = 0.0500 mol/L
This measure is the concentration of NaCl expressed in mol/L, g/L, g/100 g (percentage).
Perform the following conversions. Always convert to tenth place. . 2.35 mol Li to g Li the answer is 16.31135g or 16.3g 7.73 g Co to mol Co the answer is .131g or 0.131g 5.19 g Kr to mol Kr the answer is .062g or 0.062g 0.3630 mol As to g As the answer is 27.20g or 27.2g