E- is a negative electron
Realize that pH + pOH = 14
so, 9 + pOH = 14 -> pOH = 5
pOH = -log[OH-]
5 = -log[OH-]
plug it into a calculator and you get 1.0 x 10^-5
alternatively, use [OH-] = 10^-pOH to get the same answer
[OH-] = 1.0 x 10^-5
Explanation:
oh shi sorry I wish I could help but I'm stupid
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
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