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lesya [120]
3 years ago
10

The reaction 2 NO(g) + Br2(g) equilibrium reaction arrow 2 NOBr(g) has Kp = 109 at 25°C. If the equilibrium partial pressure of

Br2 is 0.0155 atm and the equilibrium partial pressure of NOBr is 0.0763 atm, calculate the partial pressure of NO at equilibrium.
Chemistry
1 answer:
anyanavicka [17]3 years ago
3 0

Answer:

See explanation below

Explanation:

To know this, we need to write again the reaction, and then, write the expression for the equilibrium constant Kp:

2NO + Br₂ <------> 2NOBr    Kp = 109

The Kp expression for this reaction is:

Kp = PpNOBr² / PpNO² * PpBr₂

Solving for the NO we have:

PpNO² = PpNOBr² / PpBr₂ * Kp

All we have to do now, is replace the given values:

PpNO² = (0.0763)² / (0.0155) * 109

PpNO² = 3.45x10⁻³

PpNO = √3.45x10⁻³

PpNO = 0.0587 atm

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"Write the electron Configuration expected for element 113 and the configurations for the two cations it is most likely to form"
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Answer:

Element:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Cations:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

Explanation:

The electron configuration is the distribution of the electrons in the sublevels in order of the crescent energy of them. The crescent energy of the sublevels follows the Linus Pauling's diagram, which is attached below. The sublevel "s" comports until 2 electrons, the sublevel "p" until 6 electrons, the sublevel "d" until 10 electrons, and sublevel "f" until 14 electrons.

So, for the element with an atomic number of 113, the neutral atom will have 113 electrons:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Thus the element is at the 7 period (the highest level), and group 13 (most energic sublevel p with 1 electron), the group of the aluminum. It needs to lose 3 electrons to be stable and follow the octet rule, but the subshells of the last shell are too far away in energetic order, thus, it most probably to lose the electron of 7p and form a monovalent cation, and can lose the two electrons of 7s to form a trivalent cation:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

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