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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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If you add a solution of NaOH to a solution of H₂CO₃, two reactions occur, one after the other. Write the chemical equations for
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We have a solution of NaOH and H₂CO₃

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3 years ago
How many moles of water are in 1/4 cup of water
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3.4752 moles of water

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5 0
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Hund's rule states that electrons must spread out within a given subshell before they can pair
Temka [501]

Answer:

Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.

Explanation:

If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.

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