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:
1. Rubidium metal reacts very rapidly with water to form a colorless basic solution of rubidium hydroxide (RbOH) and hydrogen gas (H2).
2. Rubidium sinks because it is less dense than water. It reacts violently and immediately, with everything leaving the container. Rubidium hydroxide solution and hydrogen are formed.
Use Raoult's Law:
Psolution = (χsolvent) (P°solvent)
24.90 = (x) (25.756)
x = 0.966765 (this is the solvent mole fraction)
χsolute = 1 - 0.966765 = 0.033235
χsolute = 0.03324 (to four sig figs)
Radioactive decay => C = Co { e ^ (- kt) |
Data:
Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min
Time conversion: 4 hr 39 min = 4.65 hr
1) Replace the data in the equation to find k
C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}
=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719
2) Use C / Co = 1/2 to find the hallf-life
C / Co = e ^ (-kt) => -kt = ln (C / Co)
=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k
t = ln(2) / 0.44719 = 1.55 hr.
Answer: 1.55 hr
