Answer 1) When a strong acid like

reacts with

usually the equilibrium shifts to the right because
As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change." So, in the given reaction when

reacts with

it generates carbon dioxide and water as a by product, if we are adding

it will remove some of the

molecule from the reaction mixture, which then tends to shift the equilibrium towards right.
Answer 2) The same would be observed in this case, if we replace

with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.
As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing

and

ions from the solution.
The compound
is formed only by sharing of electrons between the atoms. The structure of the compound is shown in the image.
Each line between two atoms represents the sharing of an electron pair which results in the formation of a single bond. Since, carbon has 4 electrons in its valence shell and hydrogen has 1 electron in its valence shell so in order to complete the octet ( to have 8 electrons in their valence shell, noble gas configuration) to attain stability carbon needs 4 more electrons and hydrogen needs 1 electron. So, sharing of electron will occur as shown in the image and the formed compound is stable in nature.
Since, the bond that is formed by sharing of electrons between atoms is known as covalent bond. So, covalent bonding is most important in
.
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
![Ksp=[Ag^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%2B%5D%5BCl%5E-%5D)
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!
To calculate the <span>δ h, we must balance first the reaction:
NO + 0.5O2 -----> NO2
Then we write all the reactions,
2O3 -----> 3O2 </span><span>δ h = -426 kj eq. (1)
O2 -----> 2O </span><span>δ h = 490 kj eq. (2)
NO + O3 -----> NO2 + O2 </span><span>δ h = -200 kj eq. (3)
We divide eq. (1) by 2, we get
</span>O3 -----> 1.5O2 δ h = -213 kj eq. (4)
Then, we subtract eq. (3) by eq. (4)
NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)
eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)
O -----> 0.5O2 <span>δ h = -245 kj eq. (6)
</span>
Add eq. (6) to eq. (5), we get
NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj
<em>ANSWER:</em> <em>NO + O ----> NO2 δ h = -232 kj</em>