How to correctly solve this problem : 4.05Kg+567.95g+100.1g correct and best way

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+ .

zvonat [6]

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)

5 0

2 years ago

If you have 10cm of snow with a volume of 40mL and a density of 0.5 g/mL how many inches of rain is this?

Lelu [443]

Density is the mass of compound divided by its volume can be shown as follows:

$d = 0.5 g /mL = m /V = m /40 = 0.5$

$m = 20 g$

40 mL of snow having 20 g of mass calculated from density.

Now, 10 cm of snow = 3.93 inches = 20 g

As, 10 inches of rain will produce 11 inches of ice as the volume of ice is bigger than rain water.

10 inches rain = 11 inches snow

3.93 inches of snow produced by

$= (3.93 * 10) /11 = 3.57 inches rain.$

Thus, 3.57 incehs of rain produces by 10 cm snow.

4 0

2 years ago

A. Consider four different samples: aqueous LiI, molten LiI, aqueous AgI, and molten AgI. Current run through each sample produc

Rudiy27

Answer:

Explanation:

At the cathode

In case of molten AgI

Silver will be collected

In case of molten LiI

lithium will be collected

in case of aqueous LiI,

hydrogen gas will be collected as reduction potential of H⁺ is more than Li⁺

in case of aqueous AgI,

Silver will be obtained at cathode because reduction potential of silver is more than H⁺

At the Anode

In case of molten NaBr

Bromine will be collected

In case of molten NaF

Fluorine will be collected

in case of aqueous NaBr ,

Bromine will be collected as reduction potential of Br⁻ is less than O⁻²

in case of aqueous NaF ,

oxygen will be obtained because reduction potential of F⁻ is more than O⁻² .

5 0

2 years ago

**PLATO QUESTION, PLEASE ANSWER CORRECTLY, THANK YOU**

Nata [24]

The answers are the following that can be answered using the chart:

A. Which type of hurricane is expected to strike more frequently this century?
Category 4 + 5C. In general, will hurricanes likely become stronger or weaker this century?
Stronger, because there are more stronger ones thant he weaker ones that can result to hurricane.

E. Which types of hurricanes are expected to drop in frequency by more than 25% this century?
Categories 1, 2, 3

4 0

2 years ago

The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Supe

Likurg_2 [28]

Answer:

The correct option is the first option

Explanation:

The equation described in the question is shown below

²³⁹₉₄Pu +¹₀n ⇒ ¹⁰⁰₄₀Zr + blank + 2¹₀n

This equation is a nuclear fission because it involves the splitting of a heavy nucleus, Plutonium (Pu), into smaller nuclei, Zirconium (Zr) and an unknown nuclei.

The law of conservation of matter states that matter can neither be created nor destroyed hence in other to get the missing atom, we must know the total number of subscripts (mass number) and superscripts (atomic number) on both sides.

The total mass number on the reactant side is 239 + 1 = 240

The total atomic number on the reactant side is 94 + 0 = 94

While, The total mass number on the product side is 100 + 2(1) = 102

The total atomic number on the product side is 40 + 2(0) = 40

To determine the missing atom, you subtract the the total atomic number of the product from that on the reactant side; 94 - 40 = 54

The atom with the atomic number 54 is Xenon,

To determine the mass number of the xenon isotope involved, you subtract the the total mass number of the product from that on the reactant side; 240 - 102 = 138.

Hence, the particle that completes the equation is ¹³⁸₅₄Xe, the first option

5 0

2 years ago