<span>Answer:
</span><span>
</span><span>
</span><span>Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)</span><span />
<span>Explanation:
</span>
<span>1) Combine the cation Li⁺ (aq) with the anion Cl- (aq) to form LiCl(s).
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<span>LiCl is a solid soluble substance, a typical ionic compound. So, it will reamain as separate ions in the product side: Li⁺ + CL⁻</span>
<span>2) Combine the anion OH⁻ with the cation H⁺ to form H₂O(l).
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<span>Since, the ionization of H₂O is low, it will remain as liquid in the product side: H₂O(l)</span>
<span>3) Finally, you can wirte the total ionic equation:
</span>
Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)
<span>an oxide of iron, magnesium, aluminum, and chromium</span>
use variable
1K₂MnF₆ + aSbF₅⇒ bKSbF₆ + cMnF₃ + dF₂
K, left=2,right=b⇒b=2
Mn, left=1, right=c⇒c=1
Sb, left=a, right=b⇒a=b=2
F, left=6.1+5a, right=6b+3c+2d
equation:
6+5(2) = 6(2)+3(1)+2d
16=15+2d
1=2d
d=0.5
So the reaction equation becomes:
1K₂MnF₆ + 2SbF₅⇒ 2KSbF₆ + 1MnF₃ + 0.5F₂ x2
2K₂MnF₆ + 4SbF₅⇒ 4KSbF₆ + 2MnF₃ + F₂
