So in this task first I think you should use the formula of the: Density/percentage
68.8% divided 26.2%= 2.625954198473282
2.625954198473282 X =
13.12977099236641
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
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Answer:
Volume of the solutions
This is the most important factor for her to control.
Answer:
1, C: BB, Bb, Bb, BB
2. C: Hybrid
Explanation:
1. If u do a punnet square for BB and Bb you will get: BB, Bb, Bb, Bb
B| B|
B| BB. BB
b| Bb Bb
2. You do a punnet square for BB and bb and you'll get: Bb, Bb, Bb, Bb, which means all the children are hybrids of Dominant alleles ans recessive alleles
<u>B </u> <u>B</u>
b| Bb Bb
b| Bb Bb