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:
<h2>Oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.</h2>
<h3>Valence electrons are the electrons in the outermost shell, or energy level, of an atom. </h3>
<h3>Configuration of oxygen's valence electrons as 2s²2p⁴.</h3>
Explanation:
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<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Answer:
ya, it's gravity. What is the problem??
Explanation:
Answer:
bowl 5, bowl 3, bowl 4, bowl 1, bowl 2
Explanation:
