The Average atomic weight of X is 28.7amu
Isotopes are atoms with the same number of protons but differing numbers of neutrons.
Different isotopes have various atomic masses.
The proportion of atoms with a particular atomic mass that can be found in a naturally occurring sample of an element is known as the relative abundance of an isotope.
An element's average atomic mass is computed as a weighted average by multiplying the relative abundances of its isotopes by their respective atomic masses, then adding the resulting products.
Using mass spectrometry, it is possible to determine the relative abundance of each isotope.
The atomic weight of the element will be a weighted average of the isotopes based on the relative abundance:
(27.730 x 0.6058) + (28.841 x 0.1835) + (31.321 x 0.2107) = 16.7988 + 5.2923+ 6.599 = 28.690 = 28.7 amu.
Average atomic weight of X is 28.7amu
Learn more about Average Atomic Weight here
brainly.com/question/6200158
#SPJ4
Answer:
(a) r = 6.26 * 10⁻⁷cm
(b) r₂ = 6.05 * 10⁻⁷cm
Explanation:
Using the sedimentation coefficient formula;
s = M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle
s = M ( 1 - Vρ) / N*6πnr
making r sbjct of formula, r = M (1 - Vρ) / N*6πnrs
Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s
r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)
r = 6.26 * 10⁻⁷cm
b. Using the formula r₂/r₁ = s₁/s₂
s₂ = 0.035 + 1s₁ = 1.035s₁
making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁
r₂ = 6.3 * 10⁻⁷cm / 1.035
r₂ = 6.05 * 10⁻⁷cm
Answer:
See explanation
Explanation:
All molecules possess the London dispersion forces. However London dispersion forces is the only kind of intermolecular interaction that exists in nonpolar substances.
So, the only kind of intermolecular interaction that exists in dimethyl ether is London dispersion forces.
As for ethyl alcohol, the molecule is polar due to the presence of polar O-H bond. In addition to London dispersion forces, dipole-dipole interactions and specifically hydrogen bonding also occurs between the molecules.
Because ethyl alcohol is polar, it is more soluble in water than dimethyl ether.