"The other halogens are not as electronegative and so other hydrogen halides cannot form hydrogen bonds between molecules. Only London Forces are formed. - Therefore more energy is required to break the intermolecular forces in HF than the other hydrogen halides and so it has a higher boiling point."
not a hack link, just stating where i got your answer from! -
https://www.mytutor.co.uk/answers/17558/A-Level/Chemistry/Explain-the-unusually-high-boiling-point-of-HF/
Answer: Option C. p-dichlorobenzene and 1,4-dichlorobenzene.
Explanation:
A line-angle formula with six vertices and a circle inscribed corresponds to the compound known as benzene.
Further, according to the IUPAC standards for naming benzene derivatives, you must first number the position of the substituent. In this case, the substituents (chloros) are located at the positions 1 and 4; also, for the benzene derivatives when they have 2 substituents and the positions are 1 and 4, this configuration is known as <em>para </em>or <em>p </em>configuration.
Additionally, this compound has 2 substituents (chloros) so you have to indicate this number (di).
Therefore, the correct answer is C. p-dichlorobenzene and 1,4-dichlorobenzene.
Answer:
9 moles of ions
Explanation:
Our compound is: CaCl₂(s)
We dissociate it:
CaCl₂(aq) → Ca²⁺ (aq) + 2Cl⁻(aq)
Per 1 mol of chloride, we have 1 mol of calcium cation and 2moles of chlorides, so in total we have 3 moles of ions.
Therefore in 3 moles of chloride, we would have 9 moles of ions (3 . 3)
Answer:
The answer is 1 and 4.
Explanation:
Mass is most concentrated in the nucelus of an atom. Therefore, if you are looking to find the area with the least mass, go outside of the nucelus. Points one and four are the furthest outside of the nucleus.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
