Answer:
Average atomic mass of uranium = 237.97 amu
Explanation:
Given data:
Abundance of U-234 = 0.01%
Abundance of U-235 = 0.71%
Abundance of U-238 = 99.28%
Average atomic mass of uranium = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (234×0.01)+(235×0.71)+(238+99.28) /100
Average atomic mass = 2.34 + 166.85+23628.64 / 100
Average atomic mass = 23797.83 / 100
Average atomic mass = 237.97 amu.
A element mixture (key word: mixture) due to the mixture containing 2 different types of elements it can be said that it’s highly likely that the mixture was not a compound but instead of a mixture due to mixture meaning more than 1 material. Hopes this help!
Water is made of only one molecule which can be chemically separated into hydrogen and oxygen. The answer is D.
The average atomic mass of iron is the sum of the products of the percentage abundance and masses of the isotopes. That is,
average atomic mass = 0.0582 x 53.940 amu + 0.9166 x 55.935 amu + 0.0219 x 56.935 amu + 0.0033 x 57.933 amu = 55.847 amu
Thus, the average atomic mass of iron is approximately 55.847 amu.
Answer: The correct answer is option E.
Explanation:
When (1R,2R)-1-chloro-2-methylcyclohexane is heated with a strong base such as NaOCH3, the predominant product will be an alkene.
It is important to note that when such compounds undergo elimination as a result of heat, the resultant product is usually an alkene.
The available options are:
A. 3-methylcyclohex-1-ene (racemic)
B. methylcyclohexene
C. (1S,2R)-1-methoxy-2-methylcyclohexane
D. (S)-3-methylcyclohex-1-ene
E. (R)-3-methylcyclohex-1-ene
F. (1R,2R)-1-methoxy-2-methylcyclohexane
The correct answer is E.
Therefore, the compound will undergo elimination reaction to give an alkene( . (R)-3-methylcyclohex-1-ene ).
