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Hitman42 [59]
4 years ago
13

What is stratosphere

Chemistry
1 answer:
____ [38]4 years ago
4 0
The stratosphere is above the earth's troposphere. It is about 32 miles above the earth's surface.

Have a nice day! :)
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Is potassium fluoride a covalent bond or ionic bond
Ivahew [28]

It's an ionic bond! Potassium is a cation, or a metal with a positive charge, and fluoride is an anion, or a nonmetal with a negative charge.

A covalent bond is the bond between two nonmetals.

Hope this helped!

7 0
4 years ago
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Define a table salt in salty water<br><br>A). Solution<br>B). Solute<br>C). Solvent<br>D). Salute​
ozzi

Answer:

Table salt in salty water is Solute

Solvent in this solution is Water

3 0
3 years ago
What attractive force draws in surrounding electrons for chemical bonds?
Dmitrij [34]

The formation of chemical bonds occurs due to the attractive forces between oppositely charged ions (ionic bonds) or by sharing of electrons (covalent bonds).

An atom having tendency of attracting a shared pair of electrons towards itself and this chemical property is said to Electronegativity .

Thus, the attractive forces which draws in surrounding electrons for chemical bonds is electronegativity.

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4 years ago
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Particles of clay and silt eroded and deposited by the wind are called
Ivanshal [37]
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4 0
4 years ago
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Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib
Dvinal [7]

<u>Answer:</u> The value of K_c for the final reaction is 7.16\times 10^{25}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1

<u>Equation 2:</u>  HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2

The net equation follows:

S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=\frac{1}{K_1}\times \frac{1}{K_2}

We are given:

K_1=9.57\times 10^{-8}

K_2=1.46\times 10^{-19}

Putting values in above equation, we get:

K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}

Hence, the value of K_c for the final reaction is 7.16\times 10^{25}

5 0
3 years ago
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