Answer:
In this phenomenon we talk about ideal gases, that is why in these equations the constant is the number of moles and the constant R, which has a value of 0.082
Explanation:
The complete equation would have to be P x V = n x R x T
where n is the number of moles, and if it is not clarified it is because they remain constant, as the question was worded.
On the other hand, the symbol R refers to the ideal gas constant, which declares that a gas behaves like an ideal gas during the reaction, and its value will always be the same, which is why it is called a constant. The value of R = 0.082.
The ideal gas model assumes that the volume of the molecule is zero and the particles do not interact with each other. Most real gases approach this constant within two significant figures, under pressure and temperature conditions sufficiently far from the liquefaction or sublimation point. The real gas equations of state are, in many cases, corrections to the previous one.
The universal constant of ideal gases is not a fundamental constant (therefore, choosing the temperature scale appropriately and using the number of particles, we can have R = 1, although this system of units is not very practical)
Fluorine.
Because:- Atoms want to become stable, for an atom to become stable, they need 8 valence electrons. Since Fluorine has 7 valence electrons, it only needs one more electron to become stable and have an octet. An octet is when an atom/element has 8 valence electrons. Since Fluorine will need to gain an electron, it will have a negative charge, and become an anion.
Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:
Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
The concentration of 
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Therefore, the pH of the solution is, 2.67
Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.
