4.79 x 10⁻⁷moldm⁻³
Explanation:
Given parameters:
pOH of RbOH = 6.32
Unknown:
Molarity of the base = ?
Solution:
The pH or pOH scale is used for expressing the level of acidity alkalinity of aqueous solutions.
pOH = -log[OH⁻]
we know the pOH to be 6.32
6.32 = -log[OH⁻]
[OH⁻] = inverse log₁₀(6.32)
[OH⁻] = 4.79 x 10⁻⁷moldm⁻³
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The answer is NH3 I hope this helps y’all out....
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Answer:
26.67 mol HCl
Explanation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.
To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:
- 8.89 mol Al(OH)₃ *
= 26.67 mol HCl
Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.
