Answer:
An atom is a particle of matter that uniquely defines achemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons. ... The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons. A proton is positively charged.
The equilibrium constant of the reaction is 1.21 * 10^6 while the change in free energy is -34.7 kJ.
<h3>What is equilirium constant?</h3>
The equilibrium constant shows the extent of conversion of reactants to products.
Now we know from the Nernst equation that;
Ecell = E°cell - 0.0592/n logQ
E°cell = 0.52−0.16=0.36 V
Since Ecell = 0 V at equilibrium,
0 = 0.36 - 0.0592/1 log K
0.36 = 0.0592/1 log K
log K = 0.36/ 0.0592
K = antiog (0.36/ 0.0592)
K = 1.21 * 10^6
ΔG = -RT lnK
ΔG =-(8.314 * 298 * ln1.21 * 10^6)
ΔG =-34.7 kJ
Learn more about equilibrium constant:brainly.com/question/10038290
I think the correct answer would be the third option. An example of a pure research would be creating synthetic elements to study their properties. Pure research is also known as fundamental or basic research. It has an exploratory nature wherein it is done without any end use in mind. It is mostly out of the curiosity or intuition of a researcher. Most of the study under this type is aiming to advance an existing knowledge by explaining or identifying the relationship of the variables being studied. It is aimed at improving theories in order to the understanding of a specific phenomenon.
Chlorobenzene would essentially be a benzene ring that is substituted with chlorine. Thus it would have 6 carbon atoms, in its molecular formula.
Answer:
a) kc = 0,25
b) [A] = 0,41 M
c) [A] = <em>0,8 M</em>
[B] =<em>0,2 M</em>
[C] = <em>0,2M</em>
Explanation:
The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.
Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.
If global reaction is:
A(g) + B(g) ⇋ 2 C(g) + D(s)
The kc = ![\frac{[C]^2}{[A][B]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BC%5D%5E2%7D%7B%5BA%5D%5BB%5D%7D)
a) The concentrations of each compound are:
[A] = 
= <em>0,4 M</em>
[B] = 
= <em>0,1 M</em>
[C] = = <em>0,1 M</em>
<em>kc = </em>![\frac{[0,1]^2}{[0,4][0,1]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0%2C1%5D%5E2%7D%7B%5B0%2C4%5D%5B0%2C1%5D%7D)
= 0,25
b) The addition of B and D in the same amount will, in equilibrium, produce these changes:
[A] = 
[B] = 
[C] = 
0,25 = ![\frac{[0,60+2x]^2}{[1,60-x][0,60-x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0%2C60%2B2x%5D%5E2%7D%7B%5B1%2C60-x%5D%5B0%2C60-x%5D%7D)
You will obtain
3,75x² +2,95x +0,12 = 0
Solving
x =-0,74363479081119 → No physical sense
x =-0,043031875855476
Thus, concentration of A is:
= <em>0,41 M</em>
c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:
[A] = 
= <em>0,8 M</em>
[B] = 
= <em>0,2 M</em>
[C] = = <em>0,2M</em>
I hope it helps!
