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murzikaleks [220]
3 years ago
6

Eddie and val observed the picture of an athlete running in a race. eddie stated the picture shows potential energy being transf

ormed to kinetic energy. val stated that it shows chemical energy being transformed to mechanical energy. which best explains who is correct?
a) only eddie is correct because the picture does not show a transformation that starts with chemical energy.
b) only eddie is correct because the picture does not show a transformation that ends with mechanical energy.
c) both eddie and val are correct because chemical energy is a type of potential energy and mechanical energy includes kinetic energy.
d) both eddie and val are correct because chemical energy is a type of kinetic energy and mechanical energy includes potential energy.
Chemistry
2 answers:
Lelu [443]3 years ago
5 0

Answer:

C. Both Eddie and Val are correct because chemical energy is a type of potential energy and mechanical energy includes kinetic energy.

lidiya [134]3 years ago
3 0

C. Both Eddie and Val are correct because chemical energy is a type of potential energy and mechanical energy includes kinetic energy.

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Now, for an element to displace hydrogen, it means the particular element is more electropositive than hydrogen on the activity series. All the elements in the options are in a greater position relative to hydrogen on the activity series except silver. This means it cannot displace hydrogen from a dilute mineral acid

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4 0
2 years ago
From these two reactions at 298 K, V2O3(s) + 3CO(g) → 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K V2O5(s) + 2CO(g) → V2O3(s)
gregori [183]

Answer:

ΔG° = -133,1 kJ

Explanation:

For the reactions:

<em>(1) </em>V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

<em>(2) </em>V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K

By Hess's law it is possible to obtain the ΔH° and ΔS° of:

2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Substracting -(1)-(2), that means:

ΔH° = -369,8 kJ - (-234,2 kJ) = <em>-135,6 kJ</em>

ΔS° = - 8,3 J/K - 0,2 J/K =<em> -8,5 J/K</em>

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

<em>ΔG° = -133,1 kJ</em>

I hope it helps!

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