The oxidizing agent is the one that is reduced in the reaction. In this reaction, the charge of Cu falls from +2 to zero charge (neutral atom in the right side). Hence, CuO is the oxidizing agent. The reducing agent, the one being oxidized is carbon from zero charge to +4. The answer is CuO.
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
Answer & Explanation:
In physics, a contact force is a force that acts at the point of contact between two objects, in contrast to body forces. Contact forces are described by Newton's laws of motion, as with all other forces in dynamics. Contact force is the force in which an object comes in contact with another object. Contact forces are also direct forces. Contact forces are ubiquitous and are responsible for most visible interactions between macroscopic collections of matter. Pushing a car up a hill or kicking a ball or pushing a desk across a room are some of the everyday examples where contact forces are at work. In the first case the force is continuously applied by the person on the car, while in the second case the force is delivered in a short impulse.
1 Hydrogen 1s1
2 Helium 1s2
3 Lithium 2s1
Answer
is: The molar solubility of calcium phosphate is 108s⁵ = Ksp.
<span>
Balanced chemical reaction: Ca</span>₃(PO₄)₂(s) → 3Ca²⁺(aq) + 2PO₄³⁻(aq).<span>
[Ca²</span>⁺] =
3s(Ca₃(PO₄)₂) =
3s.<span>
[PO</span>₄³⁻] = 2s.<span>
Ksp = [Ca²</span>⁺]³ · [PO₄³⁻]².<span>
Ksp = (3s)³ · (2s)².
Ksp = 108s</span>⁵.
s = ⁵√(Ksp ÷ 108).
