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Viefleur [7K]
3 years ago
12

What's the frequency of 506 nanometers

Chemistry
1 answer:
Dominik [7]3 years ago
7 0
It would be in the visible light wavelength, specifically in the color green.
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Typically, a solid chloride unknown (not the AgCl product) for analysis is dried in an oven to remove absorbed water before the
tatyana61 [14]

Answer:

%Cl would be too high if not dried.

Explanation:

Cl (chlorine) is a liquid. During the process of drying AgCl, the chlorine particles evaporate and chlorine is lost.

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3.87 moles of iron bromide yielded10.4 moles of sodium bromide
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Answer: 89.57 %

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3 years ago
How many grams of potassium chloride, KCl, must be dissolved in 500.0 mL of solution to produce a 1.5 M solution?
denis23 [38]

Answer:

How many grams of potassium chloride, KCl, must be dissolved in 500.0 mL of solution to produce a 1.5 M solution? Answer: g 4. What is the molarity of a solution in which 84.0 grams of sodium chloride, NaCl, is dissolved in 1.25 liters of solution? Answer: M 5.

Explanation:

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2 years ago
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Zina [86]
I really cant read it sorry i tried
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3 years ago
Read 2 more answers
Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

3 0
3 years ago
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