Answer:
The reaction will continue in the forward direction until all the NO or all the NO₂ is used up.
Explanation:
- <em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
- So, according to Le Chatelier's principle, removing the product (N₂O₃) from the system means decreasing the concentration of the products; thus, the reaction will proceed forward to produce more product to minimize the stress of removing N₂O₃ from the system.
- <em>So, the reaction will continue in the forward direction until all the NO or all the NO₂ is used up.
</em>
<em></em>
Answer:
0.718L of 0.81M HCl are required
Explanation:
Based on the reaction:
Cd(s)+2HCI(aq) → H2(g)+CdCl2(aq)
<em>1 mol of Cd reacts with 2 moles of HCl</em>
<em />
To solve this question we must, as first, find the moles of Cd. With the moles of Cd we can find the moles of HCl needed to react completely with the Cd. With the moles and the molarity we can find the volume:
<em>Moles Cd -Molar mass: 112.411g/mol-:</em>
32.71g * (1mol / 112.411g) = 0.2910 moles Cd
<em>Moles HCl:</em>
0.2910 moles Cd * (2 moles HCl / 1mol Cd) =
0.5820 moles HCl
<em>Volume:</em>
0.5820 moles HCl * (1L / 0.81moles) =
<h3>0.718L of 0.81M HCl are required</h3>
Answer:
8.279
Explanation:
The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.
At the equivalence point, we have
= 25.00 x 0.200
= 5.00 m-mol
= 0.005 mol
Volume of the base that is added to reach the equivalence point is
Number of moles of 
= 0.005 mol
Volume at the equivalence point is 25 + 5 = 30.00 mL
Therefore, concentration of 
= 0.167 M
Now the ICE table :
I (M) 0.167 0 0
C (M) -x +x +x
E (M) 0.167-x x x
Now, the value of the base dissociation constant is ,
= 
Base ionization constant, ![$K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$](https://tex.z-dn.net/?f=%24K_b%20%3D%20%5Cfrac%7B%5Cleft%5BHNO_2%5Cright%5D%20%5Cleft%5BOH%5E-%20%5Cright%5D%7D%7B%5Cleft%5BNO%5E-_2%20%5Cright%5D%7D%24)
So, ![$[OH^-]=1.9054 \times 10^{-6 } \ M$](https://tex.z-dn.net/?f=%24%5BOH%5E-%5D%3D1.9054%20%5Ctimes%2010%5E%7B-6%20%7D%20%5C%20M%24)
pOH =- ![$\log[OH^-]$](https://tex.z-dn.net/?f=%24%5Clog%5BOH%5E-%5D%24)
= 
=5.72
Now, since pH + pOH = 14
pH = 14.00 - 5.72
= 8.279
Therefore the ph is 8.279 at the end of the titration.
