Answer:
Less than 0.033 M:
![[Z]_{eq}=2.4x10^{-3}M](https://tex.z-dn.net/?f=%5BZ%5D_%7Beq%7D%3D2.4x10%5E%7B-3%7DM)
Explanation:
Hello,
In this case, the described equilibrium is:

Thus, the law of mass action is:
![K=\frac{[Z]^2}{[A]^2[B]}=0.43](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BZ%5D%5E2%7D%7B%5BA%5D%5E2%5BB%5D%7D%3D0.43)
Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:
![\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BK%7D%3D%5Cfrac%7B%5BA%5D%5E2%5BB%5D%7D%7B%5BZ%5D%5E2%7D%3D%5Cfrac%7B1%7D%7B0.43%7D%3D2.33)
Know, by introducing the change
due to the reaction extent, we can write:

Which has the following solution:

But the correct solution is
since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:
![[Z]_{eq}=0.033M-2(0.0153M)](https://tex.z-dn.net/?f=%5BZ%5D_%7Beq%7D%3D0.033M-2%280.0153M%29)
![[Z]_{eq}=2.4x10^{-3}M](https://tex.z-dn.net/?f=%5BZ%5D_%7Beq%7D%3D2.4x10%5E%7B-3%7DM)
Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).
Regards.
<span>Petroleum and biomass are burned in combustion reactions, which liberate energy stored in chemical bonds. This is chemical energy. In contrast, nuclear energy comes from the conversion of mass into energy when an nuclear reaction occurs. Geothermal energy comes directly from heat sources underground, with no chemical or nuclear reactions.</span>
Answer:
Know this
1mole of a substance contains 6.022x10²³molecules
Now
8.55x10^-15moles x 6.022x10²³molecules/1mole
=5.15x10^9molecules.
Answer:
obtain something from (a specified source).
Explanation:
160 g of SO3 are needed to make 400 g of 49% H2SO4.
<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>
The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:

1 mole of SO3 produces 1 mole of H2SO4
Molar mass of SO3 = 80 g/mol
Molar mass of H2SO4 = 98 g/mol
80 g of SO3 are required to produce 98 og 100%H2SO4
mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3
Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g
Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.
Learn more about mass and moles at: brainly.com/question/15374113
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