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N76 [4]
3 years ago
7

From your knowledge about the distribution of electrons in the levels and from the atomic number (in parentheses), indicate the

most likely charge on the ion when this atom forms an ion. (Remember the 2, 8, 18 level distribution.)
Answer Choices:
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Chemistry
1 answer:
Simora [160]3 years ago
7 0
Hydrogen is a special case (there are a lot of special cases in chemistry)

Atomic number: 1

Electron configuration 1s1.

If hydrogen gains 1 electron it will form the ionn H- . This is feasible and likely to happen because with on additional electron the electron configuration will be 1s2, and it will complete the las shell (same confiugration of He) which is a stable confirguration.

Then the answer is 1-. But you will find that H can also loose its electron and form the ion H+.

A more representative analysis can be done with Oxygen, whose atomic number is 8 and the electron configuration is 2s2 2p6, then by gaining two electrons it will acquire the stable electron configuration of Ne: 2s2 2p8


By gaining two electrons, the ion has two negative charges, this is O 2-.

Whith this I have ilustrated the method: 1) use the atomic number to make the electron confirguration, 2) look at the valence electron shell and determine if it is easier to loose electrons or to gain electrons (gain 1 or two electrons is easier than loosing 7 or 6 electrons) to acquire the electron configuration of the closest Noble gas (full valence electron shell)
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For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in
Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>

<em>C(s) + 2H2(g) <--> CH4(g). </em>

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With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

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We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

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*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

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The relation between moles and mass is given by the formula:


# of moles = (mass in grams) / (molar mass)


So, given the molar mass of 18.02 and 0.025 mol, you can solve the formula for the mass of water in grams:


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