An analytical procedure requires a solution of chloride ions ,How many grams of CaCl2 must be disolved to make2.15L of 0.0520 M C
I assume you meant 0.0520 M Cl^-.
You want mols Cl^- = M x L = approx 0.12 but that's just an estimate.
Then mols CaCl2 = 0.118/2 since 1molecule of CaCl2 supplies 2 Cl^- ions.
g CaCl2 = mols CaCl2 x molar mass CaCl2.
you have exposed electrodes of light bulb in a solution of H2SO4 such that the light bulb is on you add dilute solution and bulb goes dim which of the following could be the solution
Ba(OH)2
NaNO3
K2SO4
Cu(NO3)2
none of the above
Thank you very much
You didn't express your post very well but I think you have H2SO4 and are adding one of the four to it. If the light dims (and in fact it will go essentially out if you add to the exact equivalence point) Ba(OH)2 this will form BaSO4 and that is insoluble. So the answer is a.
Ba(OH)2 + H2SO4 ==> BaSO4 + 2H2O
To determine how many grams of CaCl2 are needed to make 2.15L of a 0.0520 M chloride ion (Cl-) solution, we need to use the equation:
Molarity (M) = moles of solute / volume of solution (L)
The molar mass of CaCl2 is 111.0 g/mol, and it dissociates in water to form two chloride ions (Cl-).
Step 1: Calculate the moles of chloride ions (Cl-) needed.
Since the ratio of chloride ions to CaCl2 is 2:1, we will multiply the volume of the solution (in liters) by the molarity and divide by 2 to get the moles of chloride ions:
moles of Cl- = (0.0520 M) x (2.15 L) / 2
Step 2: Calculate the moles of CaCl2 needed.
Since each mole of CaCl2 produces 2 moles of chloride ions, we can multiply the moles of chloride ions by 2:
moles of CaCl2 = moles of Cl- x 2
Step 3: Calculate the grams of CaCl2 needed.
Multiply the moles of CaCl2 by the molar mass of CaCl2:
grams of CaCl2 = moles of CaCl2 x molar mass of CaCl2
Now let's plug in the values and calculate the answer:
moles of Cl- = (0.0520 M) x (2.15 L) / 2
moles of Cl- = 0.05602
moles of CaCl2 = 0.05602 x 2
moles of CaCl2 = 0.11204
grams of CaCl2 = 0.11204 x 111.0 g/mol
grams of CaCl2 = 12.43344
Therefore, approximately 12.43 grams of CaCl2 must be dissolved to make 2.15L of a 0.0520 M chloride ion solution.