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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Answer:
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Explanation:
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Answer:
21.02moles of KBr
Explanation:
Parameters given:
Number of moles BaBr₂ = 10.51moles
Complete reaction equation:
BaBr₂ + K₂SO₄ → KBr + BaSO₄
Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:
The balanced equation is:
BaBr₂ + K₂SO₄ → 2KBr + BaSO₄
From the equation:
1 mole of BaBr₂ produces 2 moles of KBr
∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr
Single replacement because only one letter is being switched out in the reaction
