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STatiana [176]
2 years ago
5

Complete and balance the following equations. If no reaction occurs, write NR:(d) ClF(g) + F₂(g) →

Chemistry
1 answer:
kumpel [21]2 years ago
8 0

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 is: the hydronium ion concentratio is 1.71×10⁻⁷ mol/dm³ and pH<6.76.

The Kw (the ionization constant of water) at 40°C is 2.94×10⁻¹⁴ mol²/dm⁶ or 2.94×10⁻¹⁴ M².

Kw = [H₃O⁺] · [OH⁻].

[H₃O⁺] = [OH⁻] = x.

Kw = x².

x = √Kw.

x = √2.94×10⁻¹⁴ M².

x = [H₃O⁺] = 1.71×10⁻⁷ M; concentration of hydronium ion.

pH = -log[H₃O⁺].

pH = -log(1.71×10⁻⁷ M).

pH = 6.76.

pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.

5 0
3 years ago
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anastassius [24]

Answer:

The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,

Explanation:

To answer this problem we need to keep in mind the <u>definition of pH</u>:

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So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:

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-5.2 = log [H⁺]

10^{-5.2} = [H⁺]

6.31 x 10⁻⁶ M = [H⁺]

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-5.6 = log [H⁺]

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First, calculate for the amount of heat used up for increasing the temperature of ice.

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Then, solve for the heat needed to convert the phase of water.
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Then, solve for the heat needed to increase again the temperature of water.
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