3-Methylpentane is the IUPAC name for the substance.
whether in a continuous chain or a ring, the longest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature. According to a precise set of priorities, all deviations—whether they involve numerous bonds or atoms other than carbon and hydrogen—are denoted by prefixes or suffixes.
+3-Methylpentane is the IUPAC name for the substance in question. It has a lengthy chain of 5 carbon atoms, which gives it the prefix pent-, and a single bond is what gives it the postfix -ane (alkane). Given that the methyl group is present at the third carbon, it is 3-methylpentane.
Learn more about IUPAC Nomenclature here-
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Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
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
The atomic structure of the atom contains 9 positively charged particles (protons) and 10 neutrally charged particles (neutrons) in the center of the atom in a clump called the nucleus. Those 9 negatively charged particles (electrons) are moving around outside of the nucleus.
There are 10 neutral charges, because the mass of 19 comes from the number of neutral charges plus the number of positive charges.
To calculate the number of neutral charges, subtract the positive charges from the mass (19 - 9), and you get the number of neutral charges (10).
