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)
Answer:
The answer is one
Explanation:
I will just type any rubbish here bcuz my answer should be more than 20 words..... Just know the answer is 1
Answer:
One
Explanation:
An element is a pure substance in which there are only one kind of atom. Elements are distinct substances that cannot be split up into simpler substances.
Such substances consists of only one kind of atom. There are over a hundred known elements to date.
Generally, as a pure substance, the composition of an element is definite and they are homogenous in all parts.
Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
Iodine has an electronegativity of 2.5, and potassium has an electronegativity of 0.8, so the difference is:
2.5 - 0.8 = 1.7
