Answer:
The answer to your question is 1 M
Explanation:
Data
Molarity = ?
mass of CaCl₂ = 222.2 g
Volume = 2 l
Process
1.- Calculate the molar mass of CaCl₂
CaCl₂ = 40 + (35.5 x 2) = 40 + 71 = 111 g
2.- Calculate the moles of CaCl₂
111g of CaCl₂ ---------------- 1 mol
222.2 f of CaCl₂ ---------------- x
x = (222.2 x 1) / 111
x = 222.2 / 111
x = 2 moles
3.- Calculate the Molarity
Molarity = moles / Volume
-Substitution
Molarity = 2/2
-Result
Molarity = 1
2 <span>KOH +1 H3AsO4 →1 K2HAsO4 + 2 H2O</span>
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
