[CO] = 1 mol / 2L = 0.5 M
[
According to the equation:
and by using the ICE table:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
initial 0.5 0.5 0 0
change -X -X +X +X
Equ (0.5-X) (0.5-X) X X
when Kc = X^2 * (0.5-X)^2
by substitution:
1.845 = X^2 * (0.5-X)^2 by solving for X
∴X = 0.26
∴ [CO2] = X = 0.26
Answer:
3.6 × 10⁻⁵ M
Explanation:
Ergosterol has a maximum absorbance at λ = 282 nm. The absorbance of an analyte is related to its concentration through the Beer-Lambert's law.
A = ε × <em>l</em> × c
where,
A: absorbance
ε: molar absorptivity
<em>l</em>: optical path length
c; molar concentration
c = A / ε × <em>l </em>= 0.43 / (11,900 M⁻¹cm⁻¹) × 1.00 cm = 3.6 × 10⁻⁵ M
Answer:
Explanation:
A. The charge on an element is determined by the differences between the number of protons and electrons in an atom.
An atom will have no charges if the number of protons and electrons are the same.
- When an atom loses or gains electrons, the number of electrons will either decrease or increase
- if the number of electrons is more than the number of protons, the excess electrons is the charge on the atom. And this makes the atom become a negatively charged ion.
- if the number of electrons is lesser than the number of protons, the deficient electrons makes the atom a positively charged ion. The number of electrons by which the atom is deficient makes the atom a positively charged ion.
Charge = number of protons - number of electrons
B. Electrons form the charges they do because with the charge, they become stable like the noble gases.
The desire of every atom is to have stable electronic configuration like those of the noble gases.
A potassium atom with a configuration 2 8 8 1 will prefer to lose an electron to become an Argon atom making the ion stable.
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>
The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
Learn more here:brainly.com/question/15962928
HI.3H2O. This is the answer
