Answer: all other conditions equal, the rate evaporation of a contained liquid will be slower than the rate of evaporation of an uncontained liquid.
Justification:
1) The rate of evaporation increases as the surface area of the liquid (relative to the whole content) increases. This is, the greater the surface is the faster the evaporation.
2) That is so because the higher the surface of the liquid the more the number of particles in the liquid that are in contact with the surrounding air and so the more the particles will escape from the liquid to the air (which is what evaporation is).
3) A liquid contained will take the form of the container, so part of the liquid wil remain below the surface, while an uncontained liquid will spread all over the surface and so pratically all the liquid is in contact witht the air surrounding it.
The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.
<h3>What are the required properties of the logarithm?</h3>
The required logarithm properties are
logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);
Where a is the base of the logarithm.
<h3>Calculation:</h3>
It is given that,
log₄(x) = 12;
On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;
So,
log₄(x) = 12 ⇒ 4¹² = x
⇒ x = (2²)¹² = 2²⁴
Then, calculating log₂(x/4):
log₂(x/4) = log₂(2²⁴/4)
= log₂(2²⁴/2²)
= log₂(2²⁴ ⁻ ²)
= log₂(2²²)
On applying the property logₐ(xⁿ) = n logₐ(x);
log₂(x/4) = 22 log₂2
We know that logₐa = 1;
So,
log₂(x/4) = 22(1)
∴ log₂(x/4) = 22.
Learn more about the properties of logarithm here:
brainly.com/question/12049968
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Answer:
The system is not in equilibrium and will evolve left to right to reach equilibrium.
Explanation:
The reaction quotient Qc is defined for a generic reaction:
aA + bB → cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where the concentrations are not those of equilibrium, but other given concentrations
Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where the concentrations are those of equilibrium.
This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
Comparing Qc with Kc allows to find out the status and evolution of the system:
- If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
- If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
- If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.
In this case:
![Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BSo_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BSO_%7B2%7D%20%5D%5E%7B2%7D%2A%20%5BO_%7B2%7D%5D%20%7D)
Q=100,000
100,000 < 4,300,000 (4.3*10⁶)
Q < Kc
<u><em>
The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.
