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The methane molecule in the stratosphere has a higher potential energy than the CH₃ molecule and the hydrogen atom formed from breaking one of the carbon‐hydrogen bonds in a CH₄ molecule.
The complete question is:
<em>For each of the following situations, you are asked which of two objects or substances has the higher energy. Explain your answer with reference to the capacity of each to do work and say whether the energy that distinguishes them is kinetic energy or potential energy.</em>
<em>a. (1) A methane molecule, CH4, in the stratosphere or (2) a CH3 molecule and a hydrogen atom formed from breaking one of the carbon-hydrogen bonds in a CH4 molecule.</em>
<h3>Which have a higher energy?</h3>
The methane molecule in the stratosphere is a stable molecule and possesses chemical potential energy.
The CH₃ molecule and the hydrogen atom formed from breaking one of the carbon‐hydrogen bonds in a CH₄ molecule are unstable molecules and possesses kinetic energy. However, some of their energy has been used in breaking the bond.
Thus, the methane molecule in the stratosphere has a higher potential energy than the CH₃ molecule and the hydrogen atom formed from breaking one of the carbon‐hydrogen bonds in a CH₄ molecule.
In conclusion, the energy in the methane molecule is higher.
Learn more about potential energy at: brainly.com/question/14427111
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Answer:
The most abundant isotope is 1.007 amu.
Explanation:
Given data:
Average atomic mass = 1.008 amu
Mass of first isotope = 1.007 amu
Mass of 2nd isotope = 2.014 amu
Most abundant isotope = ?
Solution:
First of all we will set the fraction for both isotopes
X for the isotopes having mass 2.014 amu
1-x for isotopes having mass 1.007 amu
The average atomic mass is 1.008 amu
we will use the following equation,
2.014x + 1.007 (1-x) = 1.008
2.014x + 1.007 - 1.007 x = 1.008
2.014x - 1.007x = 1.008 - 1.007
1.007 x = 0.001
x= 0.001/ 1.007
x= 0.0009
0.0009 × 100 = 0.09 %
0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.
now we will calculate the abundance of second isotope.
(1-x)
1-0.0009 = 0.9991
0.9991 × 100= 99.91%
