The scientist that was the first to use the telescope in astronomy was Newton
2Ca(OH)2(aq) + 2FeCl3(aq) on the dead locs
Answer:
<h3>A neutral (no charge) particle found in the nucleus of an atom. ... The atomic number of an atom is equal to the number of protons or neutrons? false. The atomic number is always equal to the atomic mass.</h3>
Answer:
0.0119
Explanation:
There was a part missing. I think this is the whole question:
<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>
A (aq) ⇌ 2B (aq) + C(aq)
<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>
First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.
A (aq) ⇌ 2B (aq) + C (aq)
I 0.0510 0 0
C -x +2x +x
E 0.0510-x 2x x
Since the concentration at equilibrium of A is 0.0153 M, we get

We can use the value of x to calculate the concentrations at equilibrium.
![[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\](https://tex.z-dn.net/?f=%5BA%5De%20%3D%200.0153%20M%20%5C%5C%5BB%5De%20%3D%202x%20%3D%202%280.0357%29%20%3D%200.0714%20M%20%5C%5C%5BC%5De%20%3D%20x%20%3D%200.0357%20M%20%5C%5C)
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
![Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BB%5D%5E%7B2%7D%20%20%5Ctimes%20%5BC%5D%7D%7B%5BA%5D%7D%20%3D%20%5Cfrac%7B0.0714%5E%7B2%7D%20%20%5Ctimes%200.0357%7D%7B0.0153%7D%20%3D%200.0119)
The equilibrium constant for this reaction at equilibrium is 0.0119.
You can learn more about equilibrium here: brainly.com/question/4289021
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