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lilavasa [31]
3 years ago
9

What is the energy of a 4.66 x 1014 Hz wave?

Chemistry
1 answer:
docker41 [41]3 years ago
4 0
I attached the working and the answer to both questions below.

Please note that E = energy, ν = frequency, h = Plank's Constant

<span>The energy of a 4.66 x 10</span>¹⁴<span> Hz wave is 3.088 </span>× <span>10</span>⁻¹⁷ J

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The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
4 years ago
All of the following are solutions except?
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A/an bonds allows metals to conduct electricity
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Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha
Aleks04 [339]

Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

pH = pKa + log \frac{[Base]}{[Acid]}

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

7.5 = 6.3 + log \frac{[base]}{[acid]}

1.2 = log \frac{[base]}{[acid]}

\frac{[Base]}{[Acid]} = 10^{1.2}

\frac{[Base]}{[Acid]} = 15.8

[Base] = 15.8 \times [Acid]

The total of mole fraction of acid and base is 1. Therefore we have,

[Acid] + [Base] = 1

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

[Acid] + 15.8 \times [Acid] = 1

16.8 [Acid] = 1

[Acid] = \frac{1}{16.8}

[Acid] = 0.0595

[Acid] = 0.0595 x 100 = 5.95 %

The fraction of carbonic acid present in the blood is 5.95%

4 0
4 years ago
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