25.2 / 34.6 x 100
which would get you 73%
if you found my answer helpful pls give brainliest
Answer:
Oxidation occurs at the anode: Fe(s) | Fe2+(aq) half cell. (ii) Reduction occurs at the cathode: Ag(s) | Ag+(aq) half cell. Oxidation occurs at the anode: Pt | Sn2+(aq), Sn4+(aq) half cell. (iii) Electrons flow from the anode to the cathode: from the Pt(s) → Ag(s) electrode.
Following the key in the diagram (see the attached image), the only particle diagram that represents a mixture of three substances is diagram 2.
To simplify it, let us replace the key in the diagram as follows;
- atom of one element = A
- atom of different element = B
Diagram 1 consists of only AA and AB
Diagram 2 consists of AA, BB, and AB.
Diagram 3 consists of AA and ABA
Diagram 4 consists of AA and BAB
Thus, only diagram 2 has a mixture of 3 substances.
More on mixtures can be found here: brainly.com/question/6594631
Answer:
atoms tend to react in order to gain 8 valence electrons
Explanation:
The octet rule describes the tendency of atoms of elements to react in order to have eight electrons in their valence shell. This is because having eight valence electrons confers stability to the atoms of these elements in the compounds they form.
The octet rule only does not apply to the transition elements or the inner transition elements as only the s and p electrons are involved. the electronic configuration in atoms having an octet is s²p⁶.
For example, sodium atom has one valence electron in its valence shell but a complete octet in the inner shell; it will react with chlorine atom which has seven valence electrons to form a stable compound, sodium chloride by donating its one valence electron in order to have an octet. Similarly, the chlorine atom will then have an octet by accepting the one electron from sodium atom.
Answer:
Frequency = 6.16 ×10¹⁴ Hz
λ = 4.87×10² nm
Explanation:
In case of hydrogen atom energy associated with nth state is,
En = -13.6/n²
For n = 2
E₂ = -13.6 / 2²
E₂ = -13.6/4
E₂ = -3.4 ev
Kinetic energy of electron = -E₂ = 3.4 ev
For n = 4
E₄ = -13.6 / 4²
E₄ = -13.6/16
E₄ = -0.85 ev
Kinetic energy of electron = -E₄ = 0.85 ev
Wavelength of radiation emitted:
E = hc/λ = E₄ - E₂
hc/λ = E₄ - E₂
by putting values,
6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev - (-3.4ev )
6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J
λ = 4.87×10⁻⁷ m
m to nm:
4.87×10⁻⁷ m ×10⁹nm/1 m
4.87×10² nm
Frequency:
Frequency = speed of electron / wavelength
by putting values,
Frequency = 3×10⁸m/s /4.87×10⁻⁷ m
Frequency = 6.16 ×10¹⁴ s⁻¹
s⁻¹ = Hz
Frequency = 6.16 ×10¹⁴ Hz
