Answer:
b- 4.4 * 10^-12.
Explanation:
Hello.
In this case, as the reaction:
A + 2B → 3C
Has an equilibrium expression of:
![K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BC%5D%5E3%7D%7B%5BA%5D%5BB%5D%5E2%7D%3D2.1x10%5E%7B-6%7D)
If we analyze the reaction:
2A + 4B → 6C
Which is twice the initial one, the equilibrium expression is:
![K_2=\frac{[C]^6}{[A]^2[B]^4}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BC%5D%5E6%7D%7B%5BA%5D%5E2%5BB%5D%5E4%7D)
It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:
Thus, the equilibrium constant of the second reaction turns out:
Therefore, the answer is b- 4.4 * 10^-12.
Best regards.
Answer: A
Explanation: There is a lot of sand that will blow and form dunes unlike option b
63/29 Cu (copper-63)
65/29 Cu (copper-65)
24/12 Mg (magnesium 24)
25/12 Mg (magnesium 25)
26/12 Mg (magnesium 26)
6/3 Li (lithium 6)
7/3 Li (lithium 7)
The top number is the mass number and you find that by adding the number of neutrons and atomic number
The bottom number is the atomic number and it’s the number or protons
3 ethyl, 4 methylheptane. The compound is named by first identifying the longest carbon chain in the structure. in this case the chain has seven carbon atoms thus the prefix hept-.
Next you identify the substituent groups attached to the long carbon chain and name them from the lowest value of the integer assigned to the carbon atoms from either side. From the right, the ethyl group is attached to carbon number 3 while from the left, the methyl group is attached to carbon number 4. We therefore start with the right and name the attached groups first, including the carbon atoms to which they are attached.
Then we also take into consideration the highest number of bonds between the carbon atoms which is one from the question. Thus the suffix -ane is added if a maximum of one bond, -ene,if two bonds and -yne if three bonds.
The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
