Answer:
The solution becomes diluted.
Explanation:
When you add water to a solution, the number of moles of the solvent stays the same while the volume increases. Therefore, the molarity decreases.
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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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Elements of Group 1 and group 2 in the periodic
table contain elements so reactive that they are never found in the free state
<u>Explanation</u>:
The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.
These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.
To determine the heat dissipated when a substance freezes, we multiply the heat of fusion of the substance to the mass of the substance that freezes. We calculate as follows:
Heat = -3.16 (64/32.06) = - 6.32 kJ
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