Blank 1: polar
The difference in electronegativity between N and H causes electrons to preferentially orbit N, making the bond polar.
Blank 2: trigonal pyramidal
There are four “things” attached to N - 3 H’s and 1 lone pair of electrons. The four things together are arranged into a tetrahedral formation. However, the lone pairs don’t actually contribute to the shape of the molecule per se; it’s only the actual atoms that do. The lone pair creates a bit of repulsion that pushes the 3 H’s down, creating a trigonal pyramidal shape (as opposed to a trigonal planar one).
Blank 3: polar
The molecule as a whole is also polar because the “things” around it, though arranged in a tetrahedral pattern, are not all the same. The side of the molecule with the lone pair is slightly negative, while the side with the 3 H’s is slightly positive due to the differences in electronegativity described above.
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
Answer:
0.962 atm.
97.4 kPa.
731 torr.
14.1 psi.
97,434.6 Pa.
Explanation:
Hello.
In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:
- Atmospheres: 1 atm = 760 mmHg:
- Kilopascals:: 101.3 kPa = 760 mmHg:
- Torrs: 760 torr = 760 mmHg:
- Pounds per square inch: 14.69 psi = 760 mmHg:
- Pascals: 101300 Pa = 760 mmHg:
Best regards.
Answer:
I'm thinking cooper but not sure
Hello!
I believe the correct answer to this question is H+ and H2O.
I hope you found this helpful! :)
