Answer:
Part A: Hexacyanoferrate (III)
Part B: DiammintetraaquoCupperate (II)
Part C: Dichlorobis(ethylenediamine) Chromate (II)
Part D: Triaquocarbonylnickel (II) Sulphate
Part E: Potassium Dicarbonatedifluoroplatinate (II)
Explanation:
For naming the complex ions there is a specific rule
Nomenclature of the complex ions are as follow
- write a correct formulae
- Indicate the oxidation number of metal in the complex
- The oxidation number should write in the roman numeral in perenthasis after metal name
- Ligand named before the metal ion
- Ligan can be named in following order
* 1st negative, 2nd neutral, 3rd positive
* If there are more than 2 same charged ligand the write in
alphabetical order.
- Write prefix i.e di, tri, tetra for multiple monodentate ligands
- Anions name end at ido the replace the final name.
- Neutral ligands named as their usual name, but there are some exceptions such as
NH3 named as ammine
H2O names aqua or aquo
CO named ascarbonyl
NO named as nitrosyl
- If the complex is an anion, then name of the central atom will end in -ate, and its Latin name will be used except for mercury
- The name of full complex will end with cation or anion with separate word.
Keeping the rules in mind the complexes named as following.
_________________________
Part A:
[Fe(CN)₆]³⁻
Name of the Complex : Hexacyanoferrate (III)
___________________
Part B:
[Cu(NH₃)₂(H₂O)₄]²⁺
Name of the Complex : DiammintetraaquoCupperate (II)
_______________________
Part C
CrCl₂(en)₂
Name of the Complex : Dichlorobis(ethylenediamine) Chromate (II)
________________________
Part C
[Ni(H₂O)₃(CO)]SO₄
Name of the Complex : Tetraaquocarbonylnickel (II) Sulphate
______________________
Part E
K₄[Pt(CO₃)₂F₂]
Name of the Complex : Potassium Dicarbonatedifluoroplatinate (II)
Answer:
Yes
Explanation:
The process of respiration and burning are similar in the following ways: Both respiration and burning require oxygen. Energy is released during both respiration and burning. The products produced (carbon dioxide and water) are the same for both respiration and burning.
Answer:
27oz
Explanation:
Let the volume quantity of the 10% solution = x
The volume quantity of the 15% alcohol solution is known as = 18oz
Therefore the volume of the 12% product = 18oz + x
Therefore
0.10 × x + 0.15 × 18oz = 0.12 × (18oz+x)
0.15×18oz - 0.12×18oz = 0.12×x - 0.10×x
.
0.03*18oz = 0.02x
Dividing both sides by 0.02
x = 27oz
Verifying the answer, we have
0.1 × 27 + 18 × 0.15 = 0.12 × 45
2.7 + 2.7 = 5.4
Answer:
B
Explanation:
This is because conductivity depends on the mobility of electrons.
the solid particles take up the intermolecular spaces in the liquid.
