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
Ain't is the correct answer
Answer:
if the object sank then that object has a greater density then water. if the object floated then its density is lower then water.
Explanation:
lets say object 1 has a density of 24/cm3. the density is greater then water (1.0000g/cm3) so it would sink. now lets say object 2 has a density of 0.79383g/cm3 since it's less then the density of water (1.0000g/cm3) it would float.
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