3.16 X 10^-11 M is the [OH-] concentration when H3O+ = 1.40 *10^-4 M.
Explanation:
data given:
H30+= 1.40 X 10^-4 M\
Henderson Hasslebalch equation to calculate pH=
pH = -log10(H30+)
putting the values in the equation:
pH = -log 10(1.40 X 10^-4 M)
pH = 3.85
pH + pOH =14
pOH = 14 - 3.85
pOH = 10.15
The OH- concentration from the pOH by the equation:
pOH = -log10[OH-]
10.5= -log10[OH-]
[OH-] = 10^-10.5
[OH-] = 3.16 X 10^-11 is the concentration of OH ions when hydronium ion concentration is 1.40 *10^-4 M.
Hi there!
We can use the following conversions to solve:
Total mass --> amount of mols --> amount of atoms (Avogadro's number)
Begin by calculating the amount of boron trifluoride in 3.61 grams:
3.61 g * (1 mol BF₃ / 67.8 g) ≈ 0.0532 mol BF₃
Use avogadro's number to convert:
0.0532 mol * 6.02× 10²³atoms / 1 mol = 3.203 × 10²² atoms
Rubidium Chromate is the compound
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I Believe this is the right answer:
Get a periodic table of elements. ...
Find your element on the periodic table. ...
Locate the element's atomic number. ...
Determine the number of electrons. ...
Look for the atomic mass of the element. ...
Subtract the atomic number from the atomic mass.
Hoped this helped!
