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4 years ago

If a solution of hf (ka = 6.8 10-4) has a ph of 3.67, calculate the total concentration of hydrofluoric acid.

1 answer:

3 0

When PH = ㏒(H^+)
[H+] = 10^-3.67
[H+] = 2.14x 10^-4

so according to the reaction equation:
HF ↔ H^+ + F^-
at equilibrium X-(2.14x10^-4) (2.14x10^-4) (2.14x10^-4)

by substitution in Ka formula:
Ka = [H+][F]-/[HF]
6.8x10^-4 = (2.14x10^-4)*(2.14x10^-4) /(X-2.4x10^-4)
X-2.4x10^-4 = (4.58x10^-8)* (6.8x10^-4)
∴X = 2.4x10^-4
∴[HF] = X- (2.14x10^-4)= (2.4x10^-4)-(2.14x10^-4) = 2.6 X 10^-5

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What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?

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The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency $1.0 \times {10^{15}}\;{\text{Hz}}$ is $\boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}$

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When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

${\mathbf{E=h\nu }}$ ......(1)

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${\text{E}}$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

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Substitute these values in equation (1)

$\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}$

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is ${\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}}$ .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

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3 years ago

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