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Answer:
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
Explanation:
Let's consider the molecular equation between aqueous copper(II) chloride and aqueous sodium phosphate.
3 CuCl₂(aq) + 2 Na₃PO₄(aq) ⇒ 6 NaCl(aq) + Cu₃(PO₄)₂(s)
The complete ionic equation includes all the ions and insoluble species.
3 Cu²⁺(aq) + 6 Cl⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ 6 Na⁺(aq) + 6 Cl⁻(aq) + Cu₃(PO₄)₂(s)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and insoluble species.
3 Cu²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Cu₃(PO₄)₂(s)
<span>The following acids do NOT ionize completely in solution because they are not strong acids are as follows:
a HBr
b HI
c H2SO3
d H3N
e HNO2
f HF
</span>
Answer:
The total energy change, ΔE, in kilojoules = -61.93 kJ
Explanation:
Relationship between ΔH, ΔE and work done is given by first law of thermodynamics.
ΔE = ΔH - PΔV
Where,
ΔH = Change in enthalpy
ΔE = Change in internal energy
PΔV = Work done
Given that,
ΔH = -75.0 kJ = -75000 J
P = 43.0 atm
ΔV = Final volume - initial volume
= (2.00 - 5.00) = -3.00 L
PΔV = 43 × (-3.00) = -129 L atm
1 L atm = 101.325 J
-129 L atm = 129 × 101.325 = -13071 J
So ,
ΔE = ΔH - PΔV
= (-75000 J) - ( -13071 J)
= -75000 J + 13071 J
= -61929 J
Total energy change, ΔE = -61.929 kJ
