Analyze the Name of complex Compound.
T<span>etracarbonylplatinum(iv) chloride
So, there are,
4 Carbonyl groups = 4 CO = (CO)</span>₄
1 Platinum Metal = 1 Pt = Pt
Unknown Chloride atoms = ?
In complexes positive part is always named first, so the sphere containing Pt and carbonyl ligands is written first,
[Pt (CO)₄]
The charge on sphere is +4 because CO ligand is neutral, and Pt has a Oxidation state of four as written in name (IV),
So,
[Pt (CO)₄]⁴⁺
Now, in order to neutralize +4 charge we should add 4 Chloride ions, So,
[Pt (CO)₄] Cl₄
Answer:
Unchanged.
Explanation:
Hello,
In this case, for the experiment you performed, as you previously removed the sodium chloride, its % salt calculation will not be affected by the lost of the sand. It is important to remember that such experiments must be performed carefully as the committed error is a high-impact factor regarding to the experiment's performance. It is important for you to remember that to separate the sodium chloride, it earlier dissolution in water allowed to be dissolved leveraging the insolubility of sand in water, thus, it was decanted easily.
Best regards.
The lone pair of electrons in Theobromine are shown in RED color in attached figure.
Details:
In chemistry there are two types of electrons.
i) Bonding Pair Electrons:
These are those electrons which are being shared and are involved in making bonds. They are also called as sharing electrons. In given structure all the solid bonds either single or double are made up of bonding pair electrons.
ii) Lone Pair / Non-Bonding Pair Electrons:
Those electrons which doesn't take part in bonding and are not shared between atoms. In given structure the lone pair of electrons are found only on nitrogen atoms (single lone pair) and oxygen atoms (two lone pair of electrons) respectively.
Hybridization of Nitrogen Atoms;
N₁ = Sp²
N₂ = Sp³
N₃ = Sp³
N₄ = Sp³
Hybridization of Oxygen Atoms;
O₁ = Sp²
O₂ = Sp²
Answer:
It’s Photosynthesis
Explanation:
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