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3 years ago

Give 7 and example of how each wave in the EM spectrum in used in our daily lives

Chemistry

1 answer:

bogdanovich [222]3 years ago

8 0

1. it’s use for our food in like microwaves and ovens
2. airplanes are guided by radar waves,
3.the tv is used by electromagnetic waves
4. they are used in heaters,
5.infrared cameras which detect people in the dark
6.Allows airport security to observe the internal contents of objects and luggage using airport scanners
7.doctors use electromagnetic waves in x-rays

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Give the name and formula of the compound formed from the following elements: (a) sodium and nitrogen; (b) oxygen and strontium;

lisov135 [29]

Answer:

For a: The chemical name and chemical formula formed is sodium nitride and $Na_3N$

For b: The chemical name and chemical formula formed is strontium oxide and $SrO$

For c: The chemical name and chemical formula formed is aluminium chloride and $AlCl_3$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

For a: Sodium and nitrogen:

Sodium is the 11th element of periodic table having electronic configuration of $[Ne]3s^1$

To form $Na^{+}$ ion, this element will loose 1 electron.

Nitrogen is the 7th element of periodic table having electronic configuration of $[He]2s^22p^3$ .

To form $N^{3-}$ ion, this element will gain 3 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for sodium nitride is $Na_3N$

For b: Oxygen and strontium

Strontium is the 38th element of periodic table having electronic configuration of $[Kr]5s^2$ .

To form $Sr^{2+}$ ion, this element will loose 2 electrons.

Oxygen is the 8th element of periodic table having electronic configuration of $[He]2s^22p^4$ .

To form $O^{2-}$ ion, this element will gain 2 electrons.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for strontium oxide is $SrO$

For c: Aluminium and chlorine

Aluminium is the 13th element of periodic table having electronic configuration of $[Ne]3s^23p^1$

To form $Al^{3+}$ ion, this element will loose 3 electrons.

Chlorine is the 17th element of the periodic table having electronic configuration of $[Ne]3s^23p^5$

To form $Cl^{-}$ ion, this element will gain 1 electron.

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

So, the chemical formula for aluminium chloride is $AlCl_3$

4 0

4 years ago

Fe2(SO4)3 +__BaCl2 → __BaSO4 +_ FeCl3

Elenna [48]

Answer:

Fe2(SO4)3 + 3BaCl2 → 2FeCl3 + 3BaSO4

8 0

3 years ago

When there is a chemical reaction between hydrochloric acid and potassium hydroxide. Which substances are the reactants? Which s

Tasya [4]

Answer:

Good question i really dont know sorry

Explanation:

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2 years ago

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Which of the following is an indicator of a chemical reaction?

yuradex [85]

Answer:

Explanation:

B is the best showing of a chemical reaction out of the choices

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2 years ago

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Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).

wolverine [178]

Answer: Li is the reducing agentg and O is the oxidizing agent.

Explanation:

1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.

2) The given reaction is:

4Li(s) + O₂ (g) → 2 Li₂O(s)

3) Determine the oxidation states of each atom:

Li(s): oxidation state = 0 (since it is alone)

O₂ (g): oxidation state = 0 (since it is alone)

Li in Li₂O (s) +1

O in Li₂O -2

That because 2× (+1) - 2 = 0.

4) Determine the changes:

Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.

O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.

7 0

4 years ago

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