To determine the number of moles of ions present in a known volume of solution, follow this example:
HCl dissociates completely in water into H+ and Cl-, because this is a strong acid, and only strong acids, bases, and ionic compounds have the ability to dissociate completely.
This means one equivalent of HCl will generate one equivalent of H+ and Cl- ions; the same number of moles of HCl will generate the same number of moles for H+ and Cl-
HCl --> H+ + Cl-
Now determine the number of moles in the volume of your solution. Remember that 1M is another way to say 1 mole/L.
(2moles HCl/ 1L) x (1L) = 2 moles HCl
Since the equation states that 1 equivalent of HCl is 1 H+, the final answer is:
(2moles HCl/ 1L) x (1L) x (1 mole H+/1mole HCl) = 2 moles H+
Well, molarity (M) is defined as (moles of solute)/(Liters of solution). So first you should convert 200 mL to liters by dividing 200 by 1000 = 0.200 L. Then plug your info into the formula,
0.25 M = X/(0.200 mL), multiply by 0.200 mL to get X. Your answer should be
0.050 moles of HCl.
1 M = 1 mole/L
if you want 1 L then you should need 1 mole
Molarity = moles of solute/Liters of solution (50 ml = 0.05 Liters )
10 M HCl = moles HCl/0.05 Liters
= 0.5 moles HCl
0.0004 mole HCl
2 moles -OH
.2 moles
0.00833 moles of CO3
How many moles of H+ ions are present in the following aqueous solution? 1,410 mL of 0.32 M nitric acid
3(2.6) = 7.8 moles of ions
one
There are two moles of sodium ions in two moles of sodium chloride.
Molarity is moles/liter, so in order to find the moles of a substance in a given volume, simply multiply molarity with volume (in liters). n=M*V
.21 moles/liter * .0655 L = 0.013755 moles AlCl3 .013755 * 3 moles Cl / 1 mole AlCl3 = 0.041265 moles Cl * avagadro's number = number of chloride ions
5,7 moles (SO4)3-.
2.03mol
I suppose that the answers are: - 0,9 moles aluminium ions - 2,7 moles chloride ions
0.00833 moles of CO3
How many moles of H+ ions are present in the following aqueous solution? 1,410 mL of 0.32 M nitric acid
There are 2.8604 moles of ammonium ions in 6.955.
0.1 Moles
3(2.6) = 7.8 moles of ions
Since the formula shows two sodium atoms in each formula unit of sodium sulfate and this compound normally completely ionizes in water solution, the number of sodium ions will be twice the number of moles of the salt; in this instance, 1.0 moles of sodium ions.
Molarity = moles of solute/Liters of solution 0.300 M Na3PO4 = moles Na3PO4/2.50 Liters = 0.75 moles Na3PO4