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Vitek1552 [10]
2 years ago
5

A sample of N2O3(g) has a pressure of 0.046 atm . The temperature (in K) is then doubled and the N2O3 undergoes complete decompo

sition to NO2(g) and NO(g). Find the total pressure of the mixture of gases assuming constant volume and no additional temperature change. Enter your answer numerically, in terms of atm.
Chemistry
1 answer:
iogann1982 [59]2 years ago
5 0

Answer:

0.184 atm

Explanation:

The ideal gas equation is:

PV = nRT

Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.

So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.

The decomposition occurs:

N₂O₃(g) ⇄ NO₂(g) + NO(g)

So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.

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Two isotopes of hypothetical element X exist with abundances of 30.00% 100X and 70.00% 101X. What is the approximate atomic mass
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C. 100.7 amu

Explanation:

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4 0
2 years ago
Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital
Thepotemich [5.8K]

Answer:

<u>Number of electrons</u> = 118

<u>Electronic configuration</u>: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the <u>valence shell of such an element is 7p.</u>

⇒ the principal quantum number <u>(n) for the valence shell is 7.</u>

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

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Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

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<u>Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.</u>

7 0
3 years ago
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