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dsp73
3 years ago
9

What are some of the uses of halogens?​

Chemistry
2 answers:
BartSMP [9]3 years ago
8 0

Answer:

Halogens are used in, water and sanitation, plastics, pharmaceutical, pulp and paper, textile, military and oil industries. Bromine, chlorine, fluorine and iodine are chemical intermediates( bleaching agents and disinfectants). Chlorine is used to purify water. Chlorine also is part of salt, sodium chloride, which is one of the most widely used chemical compounds. Fluorine is used in fluorides, which are added to water supplies to prevent tooth decay.

hopes this helps~

djverab [1.8K]3 years ago
7 0

Answer:

What are some uses of halogen elements? Chlorine is used to purify water. Chlorine also is part of salt, sodium chloride, which is one of the most widely used chemical compounds. Fluorine is used in fluorides, which are added to water supplies to prevent tooth decay.

Explanation:

Halogen, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts). They were given the name halogen, from the Greek roots hal- (“salt”) and -gen (“to produce”), because they all produce sodium salts of similar properties, of which sodium chloride—table salt, or halite—is best known.

Because of their great reactivity, the free halogen elements are not found in nature. In combined form, fluorine is the most abundant of the halogens in Earth’s crust. The percentages of the halogens in the igneous rocks of Earth’s crust are 0.06 fluorine, 0.031 chlorine, 0.00016 bromine, and 0.00003 iodine. Astatine and tennessine do not occur in nature, because they consist of only short-lived radioactive isotopes.

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Answer:

The answer to your question is the letter C) 5648 kJ/mol

Explanation:

Data

                C₁₂H₂₂O₁₁  +  12 O₂  ⇒   12 CO₂  +  11 H₂O

H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol

H° O₂ = 0 kJ / mol

H° CO₂ = -393.5 kJ/mol

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Formula

ΔH° = ∑H° products - ∑H° reactants

Substitution

ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)

ΔH° = -4722 - 3143.8 + 2221.8

Result

ΔH° = -5644 kJ/mol

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