Answer:
No effect
Explanation:
H2(g) + Cl2(g) <=> 2HCl(g)
Decreasing the volume of the re simply means increasing the pressure. As we can see from the above equation, the total number of moles of the reactants and products are the same. There decreasing the volume ( i.e increaing the pressure) will have no effect on the system and hence the equilibrium position is not affected.
Answer:
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Explanation:
The unbalanced equation is
Al(s) + Ni²⁺(aq) ⟶ Ni(s) + Al³⁺(aq)
(i) Half-reactions
Al(s) ⟶ Al³⁺(aq) + 3e⁻
Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)
(ii) Balance charges
2 × [Al(s) ⟶ Al³⁺(aq) + 3e⁻]
3 × [Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)]
gives
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s)
(iii) Add equations
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
<u>3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s) </u>
2Al(s) +3Ni²⁺(aq) + <em>6e</em>⁻ ⟶ 2Al³⁺(aq) + 3Ni(s) + <em>6e⁻
</em>
Simplify (cancel electrons)
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Answer:
the load is between the fulcrum and the handle
Explanation:
Sedimentary rocks form layers.
I believe the answer you are looking for is TRUE!
Forgive me if I am incorrect, hope I helped you out!
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