[h3o+] [oh−] ph acidic or basic 3.5×10−33.8×10−7 1.8×10−97.15 +] [oh−] ph acidic or basic 3.5×10−33.8×10−7 1.8×10−97.15

See explanation below

Explanation:

I'm assuming this is a table you need to complete, so, you are not putting this in order, but I already found it in another place. let's do a little summary of the expressions we need to use in order to complete the chart.

To calculate pH we need the following expression:

pH = -log[H3O+] (1)

From this expression we can solve for [H3O+] in case we need it:

[H3O+] = 10^(-pH)   (2)

When we have [OH-] we calculate the pOH and from there, the pH:

pOH = -log[OH-]  (3)

and [OH-]:

[OH-] = 10^(-pOH) (4)

Finally to get the pH from pOH:

14 = pH + pOH

pH = 14 - pOH (5)

With these 5 expressions we can complete the chart. In picture 1, you have the actual chart.

To know if it's acidic or basic, that depends on the value of pH.

If pH <7 it's acidic

If pH >7 it's basic

If pH = 7 it's neutral

First case:

[H3O+] = 3.5x10^-3

pH = -log(3.5x10^-3) = 2.46  (acidic)

pOH = 14 - 2.46 = 11.54

[OH-] = 10^(-11.54) = 2.88x10^-12 M

Second case:

pOH = -log(3.8x10^-7) = 6.42

pH = 14 - 6.42 = 7.58 (it's basic)

[H3O+] = 10^(-7.58) = 2.63x10^-8 M

Third case:

pH = -log(1.8x10^-9) = 8.74 (it's basic)

pOH = 14 - 8.74 = 5.26

[OH-] = 10^(-5.26) = 5.5x10^-6 M

Fourth case:

[H3O+] = 10^(-7.15) =7.08x10^-8 M   (Basic)

pOH = 14 - 7.15 = 6.85

[OH-] = 10^(-6.85) = 1.41x10^-7 M

Hope this can help you

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.