From,
RAM=element×its relative abudance/total abudance
=((107×13)+(12×109))/25
The answer is=107.96
A reaction occurs between the two gases Chlorine monofluoride (ClF) and Fluorine (F₂) when they are added together and as a result of the reaction a compound named, Chlorine trifluoride (ClF₃) is formed.
The reaction which occurs by addition of Chlorine monofluoride (ClF) and Fluorine (F₂) is as follows -
ClF (g) + F₂ (g) = ClF₃ (l)
When one molecule of Chlorine monofluoride (ClF) reacts with one molecule of Fluorine (F₂) gas, both the gases react together to form one molecule of Chlorine trifluoride (ClF₃) which is a liquid. Therefore, the above reaction is already balanced.
Chlorine trifluoride (ClF₃) is a greenish-yellow liquid which acts as an important fluorinating agent and is also an interhalogen compound (compounds that are formed by mixing two different halogen compounds together). Other than it's liquid state ClF₃ also can exist as a colorless gas. This compound ClF₃ is a very toxic, very corrosive and powerful oxidizer used as an igniter and propellent in rockets.
Learn more about Chlorine monofluoride (ClF) here-
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Answer:
(a) Pair 1: H₂S and HS⁻
Pair 2: NH₃ and NH₄⁺
(b) Pair 1: HSO₄⁻ and SO₄⁻
Pair 2: NH₃ and NH₄⁺
(c) Pair 1: HBr and Br⁻
Pair 2: CH₃O⁻ and CH₃OH
(d) Pair 1: HNO₃ and NO₃⁻
Pair 2: H₃O⁺
Explanation:
When an acid loses its proton (H⁺), a conjugate base is produced.
When a base accepts a proton (H⁺), it forms a conjugate acid.
(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.
NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺
(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.
The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.
(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.
CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.
(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.
H₂O gains a proton to form the conjugate acid H₃O⁺.
Answer:
Mg
Explanation:
Mg + Cu2+ — Mg2+ + Cu
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
<em>Consider the following reactions.
</em>
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized.
The purpose is that it shows you all the elements we know and it arranges them in groups that are also the same kind like Chemistry.
