Answer: it is part of newtons second law. so the object to your question would accelerate
Explanation: newtons second law
Orbital diagram:

<h3>Explanation</h3>
Fluorine F is found in the second column from the right end of a modern periodic table. Fluorine is next to and on the left of the noble gas element neon. A neutral fluorine atom is one electron short of neon, which contains 8 electrons in the outermost shell when neutral. As a result, there are 7 electrons in the outermost shell of a fluorine atom.
Fluorine is in period 2. Its electrons occupy two main shells. The second main shell is the outermost shell of F. There are two subshells in the second main shell:
- 2s, which holds up to two electrons, and
- 2p, which holds up to six electrons.
A 2s electron carries less energy than a 2p electron. By Aufbau principle, the seven electrons will fill the two spaces in 2s before moving on the 2p. Among the 7 outermost shell electrons,
will end up going to 2p.
The only 2s orbital is filled with two electrons. The two 2s electrons will pair up with opposite spins, as seen with the two arrows. Two of the 2p orbitals will contain two electrons. Those electrons will also pair up. The third 2p orbital will contain only one electron. That electron can spin either
or
. Here that electron is shown as an upward arrow.
Answer:
D. Separation by melting point
Answer:
CaBr₂, CaO, FeBr₂, FeO
Explanation:
Per binary compound, there must be 1 cation (Ca²⁺, Fe²⁺) and 1 anion (Br⁻, O²⁻). The compound should have a charge of 0 (neutral). For this to occur, there sometimes needs to be more than one cation/anion present in the compound. Below I have included the math that displays why there are more ions necessary in some compounds.
CaBr₂
-----> Ca = +2 and Br = -1
-----> +2 + (-1) + (-1) = 0
CaO
-----> Ca = +2 and O = -2
-----> +2 + (-2) = 0
FeBr₂
-----> Fe = +2 and Br = -1
-----> +2 + (-1) + (-1) = 0
FeO
-----> Fe = +2 and O = -2
-----> +2 + (-2) = 0
Answer:
Air-drying clothes conserves more water than using an electric clothes dryer.