10cm^3 of solution b is added to a beaker.
distilled water is added to the beaker until the final volume in the beaker is 100cm^3
the ph of the solution is measured before and after distilled water is added.
table 4 sows the results(im jus going to write wats in the table):

| volume of solution in beaker: | ph of solution: |
| 10cm^3 | 3 |
| 1000cm^3 | x |
calculate the value of x

can someone me on this question plz, i feel like its easy, but i just forgot how to do it lol

4.0.

Explanation:

∵ pH = - log[H⁺] = 3.

∴ log [H⁺] = - 3.

∴ [H⁺] = 0.001 M.

no. of millimoles before the dilution = no. of millimoles after the dilution

(MV) before dilution = (MV) after dilution

∴ M after dilution = [H⁺] = (MV) before dilution/ V after dilution = (10.0 cm³)(0.001 M)/(100.0 cm³) = 0.0001 M.

∴ pH of 100.0 cm³ solution = - log[H⁺] = - log(0.0001) = 4.0.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

the correct answer is:

a. formation of precipitation