Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
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Answer:
Mass = 5.56 g
Explanation:
Given data:
Mass of Cl₂ = 4.45 g
Mass of NaCl produced = ?
Solution:
Chemical equation:
2Cl₂ + 4NaOH → 3NaCl + NaClO₂ + 2H₂O
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 4.45 g/ 71 g/mol
Number of moles = 0.063 mol
Now we will compare the moles of Cl₂ with NaCl.
Cl₂ : NaCl
2 : 3
0.063 : 3/2×0.063 =0.095 mol
Mass of NaCl:
Mass = number of moles × molar mass
Mass = 0.095 mol × 58.5 g/mol
Mass = 5.56 g
