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iren2701 [21]
3 years ago
11

Vanadium (IV) carbonate

Chemistry
1 answer:
Anna71 [15]3 years ago
5 0

Answer:

Explanation:

1) Vanadium (IV) → V⁺⁴

Carbonate → CO₃⁻²

So , Vanadium (IV) Carbonate = V₂(CO₃)₄  or V(CO₃)₂

2) Tin (II) = Sn⁺²

Nitrite = NO₂⁻

So, Tin (II) Nitrate = Sn(NO₂)₂

3) Cobalt (III) = Co⁺³

Oxide = O⁻²

So , Cobalt (III) Oxide = Co₂O₃

4) Titanium (II) = Tn⁺²

Acetate = CH₃COO⁻

So , Titanium (II) Acetate = Tn(CH₃COO)₂ or Tn(C₂H₃O₂)₂

5) Vanadium (V) = V⁺⁵

Sulfide = S⁻²

So , Vanadium (V) Sulfide = V₂S₅

6) Chromium (III) = Cr⁺³

Hydroxide = OH⁻

So , Chromium (III) Hydroxide = Cr(OH)₃

7) Lithium = Li⁺

Iodide = I⁻

So , Lithium Iodide = LiI

8) Lead (II) = Pb⁺²

Nitride = N⁻³

So , Lead (II) Nitride = Pb₃N₂

9) Silver = Ag⁺

Bromide = Br⁻

So , Silver Bromide = AgBr

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