Answer:

Explanation:
Hello there!
In this case, according to the given information, it will be possible for us to solve this problem by using the Boyle's law as an inversely proportional relationship between pressure and volume:

In such a way, we solve for the final volume, V2, and plug in the initial volume and pressure and final pressure to obtain:

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Answer:
B. Conclusions are based on observations
Explanation:
"A <span>gas condensing to a liquid" is the one system among the following choices given in the question where the entropy is decreasing. The correct option among all the options that are given in the question is the fourth option or option "D". I hope that this is the answer that has come to your desired help.</span>
Answer:
In He2 molecule,
Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.
Molecular Orbitals thus formed are:€1s2€*1s2
It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .
Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2
Bond Order =Nb-Na/2
Bond Order =2-2/2=0
Since the bond order is zero so that He2 molecule does not exist.
Explanation:
1. Cycloalkane
2. Alkene
3. Saturated Hydrocarbon
4. Aromatic Hydrocarbon
5. Alkane
6. Alkyne