lithium carbonate valency
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3
You're right, lithium carbonate does "end" in CO3, but a more accurate way of putting it would be that it contains the carbonate anion, CO2−3.
This helps because it provides an important clue about the number of lithium cations that are needed in order to have 1 formula unit of lithium carbonate.
As you know, ionic compounds are neutral, which implies that the overall negative charge coming from the anion(s) must be balanced by the overall positive charge coming from the cation(s).
In this case, you know that the anion carries a 2−charge, so you can say that 1 formula unit of lithium carbonate will contain as many lithium cations as needed to get an overall 2+ charge.
Now, lithium is located in group 1 of the Periodic Table, which means that it forms 1+ cations by giving up its sole valence electron.
You can thus say that in order to have an overall 2+charge, you need to have 2 lithium cations, Li+.
Therefore, 1 formula unit of lithium carbonate will contain 2 lithium cations and 1 carbonate anion.
2×[Li]++[CO3]2−→lithium carbonateLi2CO3
This helps because it provides an important clue about the number of lithium cations that are needed in order to have 1 formula unit of lithium carbonate.
As you know, ionic compounds are neutral, which implies that the overall negative charge coming from the anion(s) must be balanced by the overall positive charge coming from the cation(s).
In this case, you know that the anion carries a 2−charge, so you can say that 1 formula unit of lithium carbonate will contain as many lithium cations as needed to get an overall 2+ charge.
Now, lithium is located in group 1 of the Periodic Table, which means that it forms 1+ cations by giving up its sole valence electron.
You can thus say that in order to have an overall 2+charge, you need to have 2 lithium cations, Li+.
Therefore, 1 formula unit of lithium carbonate will contain 2 lithium cations and 1 carbonate anion.
2×[Li]++[CO3]2−→lithium carbonateLi2CO3
Answered by
1
6x + 7x + 8x = 420
21x = 420
x = 20
therefore, sides of triangle r
6x = 6*20
= 120m
7x = 7*20
= 140m
8x = 8*20
= 160m
s = 420/2
= 210
area = √s (s-a) (s-b) (s-c)
= √210 (210 - 120)(210 - 140)(210 - 160)
= √210(90)(70)(50)
= √66150000
= 8133 (approx.) m2
Hope its correct.......
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