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3 years ago

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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3 years ago

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nevsk [136]

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3 years ago

How many particles are in 12.47 grams of NaCl?

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3 years ago

In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours

Vanyuwa [196]

(a) Power of bulb is 100 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{100}{1000}kW=0.1 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.1 kW\times 2190 h=219 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

$Cost=\$ (219\times 0.15)=\$ 32.85$

Therefore, annual cost of incandescent light bulb is $\$ 32.85$

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$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{25}{1000}kW=0.025 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.025 kW\times 2190 h=54.75 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

$Cost=\$ (54.75\times 0.15)=\$ 8.21$

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7 0

3 years ago

Consider the chemical equations shown here.

vichka [17]

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Explanation:

The overall equation for the reaction that produces P₄0₁₀ is :

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

Now let us derive this equation:

Given equations:

P₄ $_{s}$ + 30₂ $_{g}$ ⇒ P₄0₆ $_{s}$ equation 1;

P₄ $_{s}$ + 50₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$ equation 2;

To get the overall combined equation, the equation 1 must be reversed and added to equation 2:

P₄0₆ $_{s}$ ⇒ P₄ $_{s}$ + 30₂ $_{g}$ equation 3

equation 2:

P₄ $_{s}$ + 50₂ $_{g}$ + P₄0₆ $_{s}$ ⇒ P₄0₁₀ $_{s}$ + P₄ $_{s}$ + 30₂ $_{g}$

cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;

P₄0₆ $_{s}$ + 20₂ $_{g}$ ⇒ P₄0₁₀ $_{s}$

learn more:

Net equation brainly.com/question/2947744

#learnwithBrainly

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3 years ago

What are some of the uses of halogens?​

What are some of the uses of halogens?