Answer: |x+2|=xF2+4
Explanation: not sure if this is right im just guessing
EVERYDAY MANS ON THE BLOCK. Lol its 4. You got this from Mans Not Hot, didn't you? xD
alumunum and oxygon with 2 Al atoms and 3 oxygen atoms make alumunum trioxide hope i went full god mode right there
Answer:
The value of Q must be less than that of K.
Explanation:
The difference of K and Q can be understood with the help of an example as follows
A ⇄ B
In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]
In this case ratio of [B] and [A] that is
K = [B] / [A] which is called equilibrium constant.
But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.
Q = [B] / [A] . It is clear that Q < K before equilibrium.
If Q < K , reaction will proceed towards equilibrium or forward reaction will
proceed .
The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.
The mass percent of lithium hydroxide can be calculated with the following equation:
(1)
Where:
(2)
We need to find the mass of LiOH.
From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.
Since mol = m/M, where M: is the molar mass and m is the mass, we have:
(3)
Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:
Solving for 
, we have:
Hence, the percent lithium hydroxide is (eq 1):
Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.
Learn more about mass percent here:
I hope it helps you!
