Claude Monet is an artist for avoiding sorrowful subjects, and instead focusing on jovial scenes.
Who is Claude Monet?
Born on November 14, 1840, Claude Monet. The artist and founder of the French Impressionist movement, Claude Monet, literally gave the movement its name.
French painter and inventor of impressionist paintings was Claude Monet. He is considered as a key innovator of modernism, especially for his attempts to capture nature as he perceived it.
Claude Monet produced artwork that captured the enchanted details and nuances he noticed in wide seascapes, tranquil lily ponds, and other settings.
As a result, option (a) Claude Monet is correct.
Learn more about on Claude Monet, here:
brainly.com/question/1967226
#SPJ1
Answer:
Do the famous painting of the ''Royal Red and Blue'' by Mark Rothko
Explanation:
Just use 5 beautiful wardrobe objects , one royal red for the top, kinda dark pink for the middle, royal blue for the bottom, and than two light pinks at the sides.
<span>pH = pKa + log ([R-]/[RH])
Where pH is the pH of the buffer, [R-] is the concentration of the basic species, and [RH] is the concentration of the acidic species.
At pH 2.4, the amino group on glycine (pKa = 9.6) will be, for accounts and purposes, 100% protonated. This means our buffer will be dealing with the two ionic forms of the carboxyl group (pKa = 2.4).
When pH = pKa, the two species are in equilibrium. This can be seen using the HH equation:
2.4 = 2.4 + log ([R-]/[RH])
0 = log ([R-]/[RH])
1 = ([R-]/[RH])
[RH] = [R-]
Now we add in another equation, our conservation of mass.
M = [RH] + [R-]
where M is the molarity of the buffer
But since [RH] = [R-]:
M = 2 [RH]
0.2 = 2 [RH]
And we wind up with:
[RH] = [R-] = 0.1 M
Now to figure out the moles of each needed, we multiply by the volume of the buffer.
0.1 M * 0.1 L = 0.01 mol
This shows that to make 100 ml of 0.2 M glycine buffer, we'll need 0.01 mol of each species.
0.01 mol of 0.5 M HCl:
0.5 mol HCl / 1 L = 0.01 mol / v
solve for v
v/1 = 0.01 / 0.5 ==> v = 0.02 L or 20 mL
weight of glycine:
MW: 75.07 g/mol
0.01 mol glycine * (75.07g glycine / 1 mol) = 0.75 g glycine
And there's your answer
--------
To make this buffer you would add 0.75g glycine to 20 mL of 0.5 M HCl and fill with water until a 100mL volume was achieved.</span>
