Answer:
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Explanation:
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<h3>
Answer:</h3>
12 neutrons
<h3>
Explanation:</h3>
- A neutral atom is an atom that has no charge, which means it has no additional electrons or electrons removed from its energy levels.
- An atom is made up of sub-atomic particles neutrons and protons found in the nucleus and electrons found in the energy levels.
- The number of protons in the nucleus of a neutral atom is known as the atomic number.
- In a neutral atom, the atomic number is equal to the number of electrons as protons are equal to the number of electrons.
- The sum of protons and neutrons in the nucleus of an atom is known as the mass number of an atom.
Therefore;
Protons = Atomic number
Protons + neutrons = Mass number
Thus;
Neutrons = Mass number - protons or
= Mass number - atomic number
In this case;
Atomic number = 11
Mass number = 23
Thus;
Neutrons = mass number - atomic number
= 23 - 11
= 12 neutrons
Answer:
Explanation:
For a general equilibrium
aA +bB ⇔ cC + dD ,
the equilibrium constant is K = [C]^c [D]^d / [A]^a[B]^b.
Our reasoning here should be based on the fact that Q has the same expression as K, but is used when the system is not at equilibrium, and the system will react to make Q = K to attain it ( Le Chatelier´s principle ).
So with this in mind, lets answer this question.
1. False: Q can large or small but is not the value of the equilibrium constant, it will predict the side towards the equilibrium will shift to attain it.
2. False: Given the expression for the equilibrium constant, we know if K is small the concentrations of the reactants will be large compared to the equilibrium concentrations of the products.
3. False: when the value of K is large, the equilibrium concentrations of the products will be large and it will lie on the product side.
4. True: From our previous reasongs this is the true one.
5. False: If K is small, the equilibrium lies on the reactants side.
Answer:
The concentration c is equal to Ka
Explanation:
The acid will ionize as observed in the following reaction:
HA = H+ + A-
H+ is the proton of the acid and A- is the conjugate base
. The equation to calculate the Ka is as follows:
Ka = ([H+]*[A -])/[HA]
Initially we have to:
[H+] = 0
[A-] = 0
[HA] = c
During the change we have:
[H+] = +x
[A-] = +x
[HA] = -x
During balance we have:
[H+] = 0 + x
[A-] = 0 + x
[HA] = c - x
Substituting the Ka equation we have:
Ka = ([H+]*[A-])/[HA]
Ka = (x * x)/(c-x)
x^2 + Kax - (c * Ka) = 0
We must find c, having as [H+] = 1/2c. Replacing we have:
(1/2c)^2 + (Ka * 1/2 * c) - (c * Ka) = 0
(c^2)/2 + Ka(c / 2 - c) = 0
(c^2)/2 + (-Ka * c/2) = 0
c^2 -(c*Ka) = 0
c-Ka = 0
Ka = c
