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3 years ago

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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Answer:

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Explanation:

We know the equation:

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Mass = No. of moles × mass of element

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3 years ago

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skad [1K]

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2 years ago

4. The picture shows a full moon.

Anuta_ua [19.1K]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

139%

GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

$2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}$

$\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}$

$\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}$

$\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}$

Based on no. of iron reacted,

$\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)$

n = m/M

$\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}$

% composition of $FeCl_3$

= $(9.75 / 10.39)^{*} 100$

= 94%

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3 years ago