Answer:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwjkwv-cqrjnAhVCheAKHWaFBBgQFjAAegQICBAB&url=https%3A%2F%2Fwww.acs.org%2Fcontent%2Fdam%2Facsorg%2Feducation%2Fresources%2Fk-8%2Finquiryinaction%2Fstudent-activity-sheets%2Fgrade-5%2Fchapter-3%2Flesson-3.3-forming-a-precipitate.pdf&usg=AOvVaw1fT7fpXG9PNWroM87puvgQ
Explanation:
that has the answers copy and paste it in your google
Answer:
Hey there!
CS2) Carbon Disulfide.
PBr3) Phosphorus Tribromide
NO) Nitric Oxide
CF4) Carbon Tetrafluoride
P2O5) Phosphorus Pentoxide
Let me know if this helps :)
Answer:
1st Question: A
2nd Question: B
Explanation:
The 1st answer would be A because if a sample is at absolute zero then the sample is at its lowest temperature none of the molecules would be able to move, this is because lower temperature= lower kinetic energy.
The 2nd answer would be B because if a sample has more temperature it speeds up it has more temperature and more kinetic energy, meaning it would move faster because there is more temperature.
Communication will be the answer
Answer:
Here's what I get.
Explanation:
The MO diagrams of KrBr, XeCl, and XeBr are shown below.
They are similar, except for the numbering of the valence shell orbitals.
Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.
However, the MO diagrams are approximately correct.
The ground state electron configuration of KrF is
KrF⁺ will have one less electron than KrF.
You remove the antibonding electron from the highest energy orbital, so the bond order increases.
The KrF bond will be stronger.
