Catod(-)
K⁺
2H20 +2e⁻ ---> H2 +2OH⁻
We can also say, that
K⁺ +OH⁻ +H2 = KOH +H2
At the cathode KOH and H2 are formed.
Hello!
The basic equations to solve this is
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
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Find pHpH = -log(1 * 10^-1)
pH = 1
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Find pOH1 + pOH = 14
pOH = 13
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Find OH-[OH-] = 10^(-pOH)
[OH-] = 1 * 10^-13mo/L
The answer is
![[OH-] = 1 * 10^{-13} mol/L](https://tex.z-dn.net/?f=%5BOH-%5D%20%3D%201%20%2A%2010%5E%7B-13%7D%20mol%2FL)
Hope this helps!
To find the molecular formula from the empirical formula, you need to find a multiple (x) that will give you the molar mass of the compound which in the question is 54 g/mol.
If C₂H₃ is the empirical formula
molar mass of empirical formula = (12 × 2) + (1 × 3) g/mol
= 27 g/mol
let x = multiple
let molecular formula = C₂ₓ H₃ₓ
multiple = molecular mass ÷ empirical mass
= 54 g/mol ÷ 27 g/mol
= 2
If molecular formula = C₂ₓ H₃ₓ
then molecular formula = C₂₍₂₎H₃₍₂₎
= <span>C₄H</span>₆