1m is made by doing 180.2g per 1litre ,
so i assume you make 2m by doing 360.4g per 1litre.
(the rest of the litre being made up by water
You stare at it
One solution is 16 nickels and 2 dimes.
Molar Mass of Carbon + Molar Mass of Silicon = Molar Mass of SiC. 12.0107 + 28.0855 = 40.0962 g / mol.
Solute dissolves in solvent to form solution.
Yes, It is a solution (a+)
148g
100 mL of solution 1 M mixed with 900 mL distilled water
HCl has a molar mass of 36.461 grams per mole. This means that 72.922 grams of HCl are needed per liter of water to make a solution that has a concentration of 2M.
Take twice its molar mass in grams and then dissolve in water up to 1.000 Litre
The term molar it refers a form to know the concentration of a solution, and it is equivalent to a molar unit in a litre of solvent 1 Molar (1M) = 1 mole (molecular weight from the structure you are interested in) / 1000 mL or 1 L. Milimolar is the thousandth part from a solution 1M
i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2
The formula mass of KCl is 75.5 so when 94.375g of it are dissolved in i litre water the solution prepared is 1.25 molar
6.023 X 1023 particles make up a 1M solution.
You need to make this value in a litre. The sum is 1.5/750*100. This gives you a 2 molar solution.
combine 100 mL 6 M HCl with 500 mL H2O
Only a compound has a molar mass not a solution.
The formula of lead (II) nitrate is Pb(NO3)2. This shows that in any solution of lead nitrate only, the molar concentration of nitrate ions will be twice as much as the molar concentration of lead (II) nitrate. Molar concentration is defined as number of moles per liter of solution, and 800 mL is the same as 0.800 liters. Therefore the molar concentration of nitrate ions in the specified solution will be 2(0.027823/0.800) or 6.96 X 10-2 . Only three significant digits are justified because that is the number of digits in 800.