We can get Kp value from this formula:
Kp = Kc(RT)^Δn
when Kc= 0.135
and R = 0.08206 L.atm/mol.K
and Δn = total number of moles of products - total number of moles of reactants
= 1+3-2 = 2
T= tempreature in kelvin = 1615 °C+273 =1888 K
so by substitution:
∴ Kp = 0.135 * (0.08206 * 1888)^2
= 3240.4
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Answer:
Yes, it is possible.
Explanation:
A diprotic acid is an acid that can release two protons. That's why it is called diprotic.
Monoprotic → Release one proton, for example Formic acid HCOOH
Triprotic → Releases three protons, for example H₃PO₄
Polyprotic → Release many protons, for example EDTA
it is a weak acid.
In the first equilibrum, it release proton, and the second is released in the second equilibrium. So the first equilibrium will have a Ka1
H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
The HA⁻ will work as an amphoterous because, it can be a base or an acid, according to this:
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
HA⁻ + H₂O ⇄ OH⁻ + H₂A Kb₂
Answer:
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Explanation:
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