Answer:
The balanced equations for those dissociations are:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻ (aq)
H₂SO₄ (aq) → 2H⁺(aq) + SO₄⁻²(aq)
Explanation:
As a strong base, the barium hidroxide gives OH⁻ to the solution
As a strong acid, the sulfuric acid gives H⁺ to the solution
Ba(OH)₂, is a strong base so the dissociation is complete.
H₂SO₄ is considerd a strong acid, but only the first deprotonation is strong.
The second proton that is released, has a weak dissociation.
H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺ (aq) + SO₄⁻² (aq) Ka
The solution for the question above is:
C = 0.270
<span>V = 0.0275L </span>
<span>n = ? </span>
<span>Use the molar formula which is: C = n/V </span>
<span>Re-arrange it to: n = CV </span>
<span>n = (0.270)*(0.0275) </span>
<span>n = 0.007425 mols </span>
<span>(more precise) n = 7.425 x 10^-3 mols
</span>
7.425 x 10^-3 mols is the answer.
CONDUCTION!!! :) Conduction is direct surface to surface contact.
It is only a physical change as you are only changing the state of the substance (from liquid to gas)
Answer:
moles PbI₂ produced = = 0.9 mole (1 sig. fig.)
Explanation:
Given Pb(NO₃)₂ + 2KI −−> PbI₂ + 2KNO₃
300g ?moles
moles KI = 300g KI/166g/mol = 1.81 mole KI
Rxn ratio for KI and PbI₂ => 2:1
∴ moles PbI₂ produced = 1/2(1.81 mole) = 0.9036144 mole (calculator answer)
= 0.9 mole (1 sig. fig.)
