Hund’s rule of maximum multiplicity specifies that _____. A. electrons will fill orbitals of lower energy first, pairing up only
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<h3>Answer: C) 142 degrees</h3>
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Explanation:
Extend segment MN such that it intersects side ST. Mark the intersection as point A. See the diagram below.
We're given that angle MNT is 72 degrees. The angle TNA is equal to 180-(angle MNT) = 180 - 72 = 108 degrees, since angles MNT and TNA add to 180.
For now, focus entirely on triangle TNA. We see from the diagram that T = 34 and we just found that N = 108. Let's find angle A
A+N+T = 180
A+108+34 = 180
A+142 = 180
A = 180-142
A = 38
So angle NAT is 38 degrees.
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Since segment MA is an extension of MN, and because MN || SQ, this means MA is also parallel to SQ.
We found at the conclusion of the last section that angle NAT was 38 degrees. Angles QST and NAT are corresponding angles. They are congruent since MA || SQ. This makes angle QST to also be 38 degrees
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The angles QSR and QST are a linear pair, so they are supplementary
(angle QSR) + (angle QST) = 180
angle QSR = 180 - (angle QST)
angle QSR = 180 - 38
angle QSR = 142 degrees
