Without solving for the dipole moment, we can easily determine which among the common gases has the smallest dipole moment just by determining the differences in their electronegativity. The greater the difference in the electronegativity, the higher is the value of the dipole moment.
From the given above, there are obvious differences between the electronegativity between the atoms composing LiF, ClF, and HF. For Cl2, since this is the same molecule then, the difference in the electronegativity is zero.
Answer: Cl2.
The experimental method for measuring the change in concentration with time for the given reaction is by measuring the amount of gas a reaction releases over time.
2NO(g) + Cl₂(g) → 2NOCl(g)
<h3>What is reaction rate?</h3>
- The reaction rate is the rate at which a chemical reaction proceeds.
- Which is proportional to both the increase in a product's concentration per unit time and the decrease in a reactant's concentration per unit time.
- There is a wide range in reaction times.
- The general definition is that the term "rate of a reaction" refers to the pace at which a reaction occurs.
- As an illustration, iron rusting has a low reaction rate since the process is slow but wood burning has a high reaction rate because the process is quick.
Learn more about reaction rate here:
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Answer: The pygmy shrew
Explanation: which weighs in at less than an ounce, has the fastest heartbeat of any mammal at 1,200 beats per minute, according to the National Wildlife Federation.
Answer:
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one could compute the pH considering that the pKa of hydrofluoric acid, HF, is 3.14:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base and acid are computed by considering the mixing process with a total volume of 300 mL (0.3 L):
![[HF]=\frac{0.005molHF}{0.3L}=0.017M](https://tex.z-dn.net/?f=%5BHF%5D%3D%5Cfrac%7B0.005molHF%7D%7B0.3L%7D%3D0.017M)
![[NaF]=\frac{0.02molHF}{0.3L}=0.067M](https://tex.z-dn.net/?f=%5BNaF%5D%3D%5Cfrac%7B0.02molHF%7D%7B0.3L%7D%3D0.067M)
Therefore, the pH turns out:
Regards.
