Answer:
7.07
Explanation:
HA = weak acid = 0.053
A+ = conjugate base = 0.045
Ka = 7.2x10^-8
Ka = [H+][A-]/HA
7 2x10^-8 = [H+][0.045]/0.053
[H+] = 7.2x10^-8 x 0.053/0.045
= 8.48x10^-8
PH = -log[H+]
= -log[8.48x10^-8]
PH = -[login.48 + log10^-8]
PH = -0.928 - (-8)log10
= 7.07
For every 1 molecule of Magnesium hydroxide or Mg(OH)2 there will be 2 molecules of HCl neutralized.
If molar mass of magnesium hydroxide is 58.3197g/mol, the amount of mol in 5.50 g magnesium hydroxide should be: 5.50g/ (<span>58.3197g/mol)= 0.0943mol.
Then, the amount of HCl molecule neutralized would be: 2* </span>0.0943mol= 0.18861 mol
If molar mass of HCl is 36.46094 g/mol, the mass of the molecule would be: 0.18861 mol* 36.46094g/mol = 6.88grams
Low because it is not diverse. It is just seed corn.
Answer
solubility product = 3.18x 10^-7
Explanation:
We were given the pressure in torr then we need to convert to atm for consistency, ten we have
21torr/760= 0.0276315789 atm
21 Torr = .0276315789 atm
P = i M S T
M = P / iRT
Where p is osmotic pressure
T= temperature= 25C+ 273= 298K
for XY vanthoff factor i = 2
S = 0.0821 L-atm / mol K
M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)
M = 0.000564698046 mol/liters
solubility= 0.000564698046 mol/liters
Ksp = [X+][Y-]
Ksp = X^2
Ksp = [Sr^+2] * [SO4^-2]
Ksp = X^2
Ksp = (0.000564698046)^2
Ksp = 3.18883883 × 10-7
Ksp = 3.18x 10^-7
solubility product = 3.18x 10^-7
Therefore, the solubility product of this salt at 25 ∘C∘C is 3.18x 10^-7
