Answer:
this is because the van der waals forces or also known as induced dipole-dipole interactions increase because more electrons are present within an atom as you go down
Explanation:
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Answer:
Here's what I get
Explanation:
CH₃CH₂CH₂CH₂CH₂CH₃ — hexane
CH₂=CHCH₂CH₂CH₂CH₃ — hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted
CH₃C≡CCH₃ — but-2-yne (PIN); 2-butyne is accepted
CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane
CH₃CH₂CHCICH₂CH₃ — 3-chloropentane
Answer:
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Explanation: I think this is what you are looking for. Hope this helps.
Answer:
![[OH^-]=9.24x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D9.24x10%5E%7B-3%7DM)
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Explanation:
Hello,
In this case, since the ionization of methylamine is:
The equilibrium expression is:
![Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
And in terms of the reaction extent 
which is equal to the concentration of OH⁻ as well as that of CH₃NH₃⁺ via ice procedure we can write:
Whose solution for via quadratic equation is 9.24x10⁻³ M since the other solution is negative so it is avoided. Therefore, the concentration of OH⁻ is:
![[OH^-]=x=9.24x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dx%3D9.24x10%5E%7B-3%7DM)
With which we can compute the pOH at first:
![pOH=-log([OH^-])=-log(9.24x10^{-3})=2.034](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%289.24x10%5E%7B-3%7D%29%3D2.034)
Then, since pH and pOH are related via:
The pH turns out:
Best regards.
