PH asam 1. HNO3 Basa. ? 0,2 M -4 2. HF 0,15 M Kaz 6,6 +10 3. Ba(OH)2 0,01 M N2H4 4. N₂ Ha 0,6 м kh= 1,3 x 10​

x = 0 9 x 2 + 9 y = 81 9. 0 2 + 9 y = 81 9 y = 81 y = 9 81 ​ y = 9 ​
y = 0 9 x 2 + 9.0 = 81 9 x 2 = 81 x 2 = 9 81 ​ x 2 = 9 x = ± 3 ​

Answered by Ryan2 | 2024-06-10

The intercepts of the equation 9 x 2 + 9 y = 81 are found by setting x and y to zero. The y-intercept is ( 0 , 9 ) and the x-intercepts are ( 3 , 0 ) and ( − 3 , 0 ) . Thus, the intercept points are ( 0 , 9 ) , ( 3 , 0 ) , and ( − 3 , 0 ) .
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Answered by Ryan2 | 2025-02-18

Jawaban:cari tentukan pH dari masing-masing larutan. 1. HNO3 0,2 M (Asam Kuat) HNO3 adalah asam kuat, jadi akan terionisasi sempurna. [H+] = 0,2 MpH = -log[H+] = -log(0,2) = -log(2 x 10^-1) = - (log 2 + log 10^-1) = - (0,3010 - 1) = 0,699 ≈ 0,7 pH = 0,7   2. HF 0,15 M (Ka = 6,6 x 10^-4) HF adalah asam lemah, jadi kita perlu menggunakan Ka untuk menghitung [H+]. Ka = [H+]^2 / [HF] 6,6 x 10^-4 = [H+]^2 / 0,15 [H+]^2 = (6,6 x 10^-4) x 0,15 = 9,9 x 10^-5 [H+] = akar(9,9 x 10^-5) ≈ 9,95 x 10^-3 M pH = -log[H+] = -log(9,95 x 10^-3) ≈ 2,00 pH ≈ 2,00   3. Ba(OH)2 0,01 M (Basa Kuat) Ba(OH)2 adalah basa kuat, jadi akan terionisasi sempurna. Ba(OH)2 -> Ba2+ + 2OH- [OH-] = 2 x 0,01 M = 0,02 M = 2 x 10^-2 M pOH = -log[OH-] = -log(2 x 10^-2) = - (log 2 + log 10^-2) = - (0,3010 - 2) = 1,699 ≈ 1,7 pH = 14 - pOH = 14 - 1,7 = 12,3 pH = 12,3   4. N2H4 0,6 M (Kb = 1,3 x 10^-6) N2H4 adalah basa lemah, jadi kita perlu menggunakan Kb untuk menghitung [OH-]. Kb = [OH-]^2 / [N2H4] 1,3 x 10^-6 = [OH-]^2 / 0,6 [OH-]^2 = (1,3 x 10^-6) x 0,6 = 0,78 x 10^-6 = 7,8 x 10^-7 [OH-] = akar(7,8 x 10^-7) ≈ 8,83 x 10^-4 M pOH = -log[OH-] = -log(8,83 x 10^-4) ≈ 3,05 pH = 14 - pOH = 14 - 3,05 = 10,95 pH ≈ 10,95

Answered by ara1412 | 2025-08-21