Uses a special type of paper to separate liquids based on their different densities

What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?

xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

$K_{a}=1.8\times 10^{-5}$

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

$\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}$

The expression for dissociation constant of acid is:

$K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}$

$1.8\times 10^{-5}=\frac{x^2}{1.8-x}$

$1.8\left(1.8-x\right)=100000x^2$

Solving for x, we get:

x = 0.00568 M

Percentage ionization = $\frac{0.00568}{1.8}\times 100=0.32 \%$

Option B is correct.

8 0

3 years ago

the normal human eye responds to visible light of wavelength range from about 390 to 710nm. Determine the frequency range of the

antoniya [11.8K]

Answer:

$\:$

A typical human eye will respond to wavelengths from 390 to 710 nm (in vacuum or air). In terms of frequency this corresponds to a band in the vicinity of 420–790 THz.

3 0

3 years ago

Read 2 more answers

I need help with this plz help

schepotkina [342]

Answer:

$\rm ^{103}_{\phantom{1}40}Zr$ , zirconium-103.

Explanation:

In a nuclear reaction, both the mass number and atomic number will conserve.

Let $^{A}_{Z}\mathrm{X}$ represent the unknown particle.

The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle $^{A}_{Z}\mathrm{X}$ , $A$ is the mass number while $Z$ while $Z$ is the atomic number.

Sum of mass numbers on the left-hand side of the equation:

$\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240$ .

Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:

$\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137$ .