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STatiana [176]
1 year ago
5

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

Chemistry
1 answer:
kumpel [21]1 year 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-

brainly.com/question/17129650

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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL
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<u>Answer:</u> The volume of concentrated solution required is 9.95 mL

<u>Explanation:</u>

To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

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Putting values in above equation, we get:

0.70=-\log[H^+]

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Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

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M_1\text{ and }V_1 are the molarity and volume of the concentrated nitric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

We are given:

M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

7.0\times V_1=0.199\times 350.0\\\\V_1=\frac{0.199\times 350}{7.0}=9.95mL

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