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3 years ago

The partial electron configuration of an atom with 14 electrons is shown. 1s22s22p63s2X Which of the following does X represent?

Chemistry

2 answers:

andrew-mc [135]3 years ago

7 0

<u>Answer:</u> The X represents $3p^2$

<u>Explanation:</u>

We are given:

An element having 14 electrons

The distribution of electrons around the nucleus of an atom in defined orbits is defined as the electronic configuration of an atom.

The electronic configuration given to us is $1s^22s^22p^63s^2X$

The next sub-shell that is filled after '3s' sub-shell is '3p' Till now, 12 electrons have been filled up. The remaining electrons are filled in '3p' sub-shell.

Now, the electronic configuration becomes: $1s^22s^22p^63s^23p^2$

Hence, the X represents $3p^2$

victus00 [196]3 years ago

4 0

3p2

The second subshell is p and there are still 2 electrons to be placed

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2 years ago

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide.

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Answer:

$[H_2]_{eq}=0.183M$

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Explanation:

Hello.

In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

$2HI\rightleftharpoons H_2+I_2$

Whereas the law of mass action (equilibrium expression) is:

$Kc=\frac{[H_2][I_2]}{[HI]^2}$

That in terms of initial concentrations and reaction extent or change $x$ turns out:

$Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}$

And the solution via solver or quadratic equation is:

$x_1=0.183M\\\\x_2=0.210M$

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:

$[H_2]_{eq}=0.183M$

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Regards.

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3 years ago