The boiling point of HF is higher than the boiling point of 
, and it is higher than the boiling point of 
.
<h3>What is the boiling point?</h3>
The boiling point is the temperature at which the pressure exerted by the surroundings upon a liquid is equalled by the pressure exerted by the vapour of the liquid.
has weak dispersion force attractions between its molecules, whereas liquid HF has strong ionic interactions between 
and 
ions.
Only London Forces are formed - Therefore more energy is required to break the intermolecular forces in HF than in the other hydrogen halides and so HF has a higher boiling point.
and will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.
Hence, the boiling point of HF is higher than the boiling point of , and it is higher than the boiling point of .
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Answer:
pH = 3.65
Explanation:
given data
pKa of HNO2 = 3.40
nitrous acid (HNO2) = 0.110 M
NaNO2 = 0.200 M
to find out
What is the pH
solution
we get here ph for acidic buffer that is express as
pH = pKa + log(salt÷acid) ........................1
put here value and we get
pH = 3.40 + log(0.200÷0.110)
pH = 3.65
If you change the subscripts it would change the reactants or products and then you would be solving a different formula, you would change what the chemical is
Answer:
ΔH° = -186.2 kJ
Explanation:
Hello,
This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:
(1) it is changed as:
SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ
That is why the enthalpy of reaction sign is inverted.
(2) remains the same:
Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ
Therefore, by adding them, we obtain the requested chemical reaction:
(3) SnCl2(s) + Cl2(g) --> SnCl4(l)
For which the enthalpy change is:
ΔH° = 325.1 kJ - 511.3 kJ
ΔH° = -186.2 kJ
Best regards.
