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
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>
<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>
<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>
<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>
<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
I have attached a paper with the answer. hope you understand. let me know if you have any question
Answer:
Oxygen
Explanation:
I'm not completely sure about the explosion part but I know oxygen fuels fire.
