At sunset the light from the sun grazes across the surface of the earth, passing through a long column or pathlength of atmosphere. During its passage through the atmosphere the blue components of the light are reduced in intensity, making the transmitted beam of sunlight more yellow or even red in color.
hope this helps ^^
Answer:
The atomic theory has changed over time as new technologies have become available. Scientific knowledge builds on past research and experimentation. 400 B.C. Democritus, a philosopher in ancient Greece, began the search for a description of matter.
Explanation:
To calculate the pH of a solution that has a [H3O+] of 7.22x10^-7. You would do the following
pH=-log[H3O+]
pH=-log[7.22x10^-7]
pH=?
Answer:
34 gram of FeO produced 8 gram of oxygen.
Explanation:
Given data:
Mass of FeO = 34 g
Mass of oxygen = ?
Solution;
Chemical equation:
2FeO → 2Fe + O₂
Number of moles of FeO:
Number of moles = mass/ molar mass
Number of moles = 34 g /71.8 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of FeO with oxygen:
FeO : O₂
2 : 1
0.5 : 1/2 × 0.5 = 0.25
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 0.25 mol × 32 g/mol
Mass = 8 g
So 34 gram of FeO produced 8 gram of oxygen.
For the reaction of deprotonation of formic acid, the concentration of HCOO⁻ at equilibrium is 0.0151 M if the initial concentration of formic acid is 1.35 M.
The reaction of deprotonation of formic acid is the following:
CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺
At the equilibrium we have:
CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺ (1)
1.35 - x x x
The acid <em>equilibrium </em>constant for this reaction is:
![K_{a} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = 1.7 \cdot 10^{-4}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BCHOOH%5D%7D%20%3D%201.7%20%5Ccdot%2010%5E%7B-4%7D%20)
(2)
Entering the values of [CHOOH] = 1.35-x, [HCOO⁻] = [H₃O⁺] = x, into equation (2) we have:
![1.7 \cdot 10^{-4} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = \frac{x^{2}}{(1.35 - x)}](https://tex.z-dn.net/?f=%201.7%20%5Ccdot%2010%5E%7B-4%7D%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BCHOOH%5D%7D%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B%281.35%20-%20x%29%7D%20)
After solving the above <em>quadratic equation</em> and taking the positive value for x (<u>concentrations cannot be negative</u>), we have:
![x = [HCOO^{-}] = [H_{3}O^{+}] = 0.0151 M](https://tex.z-dn.net/?f=%20x%20%3D%20%5BHCOO%5E%7B-%7D%5D%20%3D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%200.0151%20M%20)
Therefore, the concentration of HCOO⁻ at equilibrium is 0.0151 M.
Learn more about the equilibrium constant here:
I hope it helps you!
