True, an atoms nucleus has both protons and electrons.
Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The first or initial ionization energy or Ei of an atom or molecule is the energy required to remove one mole of electrons from one mole of isolated gaseous atoms or ions
Answer: Option (B) is the correct answer.
Explanation:
As the given reaction is as follows.
Equilibrium constant for this reaction will be as follows.
![K_{c} = \frac{[CO_{2}]}{[CO]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCO_%7B2%7D%5D%7D%7B%5BCO%5D%5E%7B2%7D%7D)
According to Le Chatelier's principle, when we increase the temperature then the equilibrium will shift towards the right hand side.
As a result, concentration of carbon dioxide will decrease whereas concentration of carbon monoxide will increase.
Thus, we can conclude that in the given reaction equilibrium constant for this reaction will decrease with increasing temperature.
Answer:
Scientific experiments need to be repeated so that scientists can make sure the results are as accurate as possible.
Explanation:
The point of a scientific experiment is to find the answer to a question. If a scientist performs an experiment only one time, they might find an answer to their question. However, they might have messed something up, so the answer is completely wrong. Doing an experiment momre than once can help a scientist make sure that their answer is correct. I hope this helps!
Answer:
The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>
Explanation:
Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M
Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M
<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)
The equilibrium constant (
) for the given chemical reaction, is given by the equation:
![K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5C%3A%20%5BI_%7B2%7D%5D%7D)
<u><em>At the original equilibrium state:</em></u>

<u><em>Therefore, at the new equilibrium state:</em></u>
![\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%5E%7B2%7D%20%3D%20713.59%20%5Ctimes%200.01805%20%3D%2012.88)
![\Rightarrow [HI] = \sqrt {12.88} = 3.589 M](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%20%3D%20%5Csqrt%20%7B12.88%7D%20%3D%203.589%20M)
<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>