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Troyanec [42]
3 years ago
12

What is the order in which electrons start filling the orbitals?

Chemistry
2 answers:
earnstyle [38]3 years ago
5 0
<span>Take a look at this periodic table. You start in the left upper corner (1s) then you go to the right untill you can't go further, then you go 1 row down and start at the left again. So the order will be 1s,2s,2p,3s,3p,4s,3d,4p... etc</span>
Leokris [45]3 years ago
4 0

i believe the answer would be the first option if im not mistaken


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Answer

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Explanation

<em>Given:</em>

pH = 9.9

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pH + pOH = 14

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Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi
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<u>Answer:</u> The value of K_c for the net reaction is \frac{K_b}{K_w}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b

<u>Equation 2:</u>  H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}

The net equation follows:

B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c

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

The value of equilibrium constant for net reaction is:

K_c=K_1\times K_2

We are given:  

K_1=K_b

K_2=\frac{1}{K_w}

Putting values in above equation, we get:

K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}

Hence, the value of K_c for the net reaction is \frac{K_b}{K_w}

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state of matter

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prohojiy [21]
A. Water can dissolve a wide variety of substances
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