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evablogger [386]
3 years ago
7

Explain the difference between fixed and variable oxidation

Chemistry
1 answer:
dem82 [27]3 years ago
7 0

Fixed vs Variable Oxidation is given below.

Explanation:

1.In its compounds, hydrogen has an oxidation number of +1, except. hydrides where the. oxidation number of hydrogen is -1. In their compounds, the metals with fixed oxidation states have the oxidation number that. corresponds with the fixed oxidation number.

A variable oxidation state is a value that determines the charge of the atom depending on certain conditions.

2. Ox­i­da­tion state of el­e­ments is con­sid­ered to be of the most im­por­tant in the study of chem­istry. For some el­e­ments, this fig­ure is con­stant known as fixed oxidation , while for oth­ers it is vari­able is called variable oxidation state.

3. MgCl2 :  magnesium is in Group IIA and all elements in Group IIA have fixed oxidation numbers of +2

FeCl2 :  iron has a variable oxidation number of either +2 or +3 and is not fixed

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actually its D) movement of tectonic plates

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What do all four of these types of molecules have in common?
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D. They all contain carbon as an important part of their structure.
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What do chemists study
ANTONII [103]

Answer:

Chemists investigate the properties of matter at the level of atoms and molecules. They measure proportions and reaction rates in order to understand unfamiliar substances and how they behave, or to create new compounds for use in a variety of practical applications.

Explanation:

3 0
3 years ago
Read 2 more answers
How many grams N2F4 can be produced from 225 g F,?​
zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

          2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

          2F₂     +    N₂   ----------->  N₂F₄

        2 mole                              1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

                 2F₂          +       N₂   ----------->  N₂F₄

        2 mole (38 g/mol)                        1 mole (104 g/mol)

                 76 g                                           104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

                  76 g of F₂ ≅ 104 g of N₂F₄

                   225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

                  X of N₂F₄ = 104 g x 225 g / 76 g

                  X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

7 0
3 years ago
Be sure to answer all parts. Sulfur dioxide is released in the combustion of coal. Scrubbers use lime slurries of calcium hydrox
scoray [572]

<u>Answer:</u> The balanced chemical equation is written below and \Delta S^o for the reaction is -160.6 J/K

<u>Explanation:</u>

When calcium hydroxide reacts with sulfur dioxide, it leads to the formation of calcium sulfate and water molecule.

The chemical equation for the reaction of calcium hydroxide and sulfur dioxide follows:

Ca(OH)_2(s)+SO_2(g)\rightarrow CaSO_3(s)+H_2O(l)

To calculate the entropy change of the reaction, we use the equation:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}]

For the given reaction:

\Delta S^o_{rxn}=[(1\times \Delta S^o_{CaSO_3(s)})+(1\times \Delta S^o_{H_2O(l)})]-[(1\times \Delta S^o_{Ca(OH)_2(s)})+(1\times \Delta S^o_{SO_2(g)})]

Taking the standard entropy change values:

\Delta S^o_{CaSO_3(s)}=101.4Jmol^{-1}K^{-1}\\\Delta S^o_{H_2O(l)}=69.9Jmol^{-1}K^{-1}\\\Delta S^o_{Ca(OH)_2(s)}=83.4Jmol^{-1}K^{-1}\\\Delta S^o_{SO_2(g)}=248.5Jmol^{-1}K^{-1}

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(1\times (101.4))+(1\times (69.9))]-[(1\times (83.4))+(1\times (248.5))]\\\\\Delta S^o_{rxn}=-160.6J/K

Hence, the balanced chemical equation is written above and \Delta S^o for the reaction is -160.6 J/K

3 0
3 years ago
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