Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)
<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
Answer:
A
Explanation:
hail will fall push it back up making it bigger comes back down
<h3>
Answer:</h3>
Chlorine gas (Cl₂)
<h3>
Explanation:</h3>
- According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
- Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
- Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.
In this case we are given gases, CI₂
, H₂,He and Ne.
- We are required to identify the gas that will diffuse at the slowest rate.
- In other words we are required to determine the heaviest gas.
Looking at the molar mass of the gases given;
Cl₂- 70.91 g/mol
H₂- 2.02 g/mol
He - 4.00 g/mol
Ne- 20.18 g/mol
Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.
Answer:
B and BBr3
Explanation:
1) 3HBr + B = BBr3 + H2 (double all equation because H2)
2) 6HBr + <em>2B </em>= <em>2BBr3</em> + 3H2
