What is the [OH-] in a solution if the [H*] = 1.2 x 10-3 M?

We know that [OH⁻] * [H⁺] = 10⁻¹⁴

plugging the value of [H⁺]

[OH⁻] * 1.2 * 10⁻³ = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ * (10³/1.2)

[OH⁻] = 833.3 * 10⁻¹⁴

[OH⁻] = 8.33 * 10⁻¹²

chemistry sucks a**

explanation:

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally