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svetlana [45]
2 years ago
15

Part A Name the complex ion [Fe(CN)6]^3- . The oxidation number of iron is +3. Part B Name the complex ion [Cu(NH3)2(H2O)4]^2+ .

The oxidation number of copper is +2. Part C Name the complex CrCl2(en)2 . The oxidation number of chromium is +2. Part D Name the salt [Ni(H2O)3(Co)]SO4 . The oxidation number of nickel is +2. Part E Name the salt K4[Pt(CO3)2F2] given that the carbonate ion acts as a monodentate ligand in the complex. The oxidation number of platinum is +2.
Chemistry
1 answer:
zvonat [6]2 years ago
5 0

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)

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2 years ago
Scientists tracking a flock of ducks found that, on average, the ducks flew 740 km south in 12 hours. What were the speed and ve
Verizon [17]

1) 17.1 m/s

2) 17.1 m/s south

Explanation:

1)

The speed of an object is a scalar quantity indicating "how fast" is the object moving, regardless of its direction of motion.

Being a scalar quantity, it has only a magnitude, while it doesn't have a direction.

The speed of an object is given by the formula:

v=\frac{d}{t}

where:

d is the distance covered by the object, which is the total length of the path covered by the object

t is the time taken to cover the distance d

In this problem we have:

d=740 km =740,000 m is the distance covered by the ducks

t=12 h  \cdot 60 \cdot 60 =43,200 s is the time taken

Substituting, we find:

v=\frac{740,000}{43,200}=17.1 m/s

2)

The velocity of an object is a vector quantity indicating how fast the object is moving in a certain direction.

Being a vector, velocity has both a magnitude and a direction.

The magnitude of the velocity is given by:

v=\frac{d}{t}

where

d is the displacement, which is the shortest distance in a straight line between the initial position and the final position of motion

t is the time taken for the displacement d to occur

And the direction is equal to the direction of the displacement.

In this problem:

d=740 km = 740,000 m south is the displacement

t=12 h  \cdot 60 \cdot 60 =43,200 s is the time taken

Substituting, we find:

v=\frac{740,000}{43,200}=17.1 m/s south

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2 years ago
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Vlad [161]

Answer:

I and IV

Explanation:

Increasing the number of particles at one side of the reaction (H2 in this case) results in the shifting of the equilibrium to the side with lesser number of particles, so in this case the equilibrium will shift to the left (towards the reactants)

A decrease in temperature will always function to favor the exothermic reaction, and since the backwards reaction is exothermic, the equilibrium will shift to the left (towards the reactants).

Option II and V will shift the equilibrium to the products, and adding a catalyst has no effect on the equilibrium.

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The work done by effort is called​
Travka [436]

Answer:

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