Answer:
8.0 moles
Explanation:
Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.
Using the formula: 
Concentration of acid = ?
Volume of acid = 10 mL
Concentration of base = 1.0 M
Volume of base = 40 mL
mole of acid = 1
mole of base = 1
Substitute into the equation:
Concentration of acid = 40/10 = 4.0 M
To determine the number of moles of acid present in 2.0 liters of the unknown solution:
Number of moles = Molarity x volume
molarity = 4.0 M
Volume = 2.0 Liters
Hence,
Number of moles = 4.0 x 2.0 = 8 moles
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
M=D*V
D=620 g/cm³
V=75 cm³
m= 620 g/cm³ * 75 cm³=46500 g
m=46500g
Your answer would be a change in odor! Hope this helps! ;D
