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Sunny_sXe [5.5K]
2 years ago
11

A methane molecule, CH4, in the stratosphere or (2) a CH3 molecule and a hydrogen atom formed from breaking one of the carbon‐hy

drogen bonds in a CH4 molecule.
Chemistry
1 answer:
MA_775_DIABLO [31]2 years ago
5 0

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

#SPJ1

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\qquad ☀️\pink{\bf{ {Answer  = \: \:   85.57g }}}

Molar mass of \bf Cu_2O

\qquad \twoheadrightarrow\sf 63.546 \times 2 +16

\qquad \pink{\twoheadrightarrow\bf 143.092 g}

<u>As we know</u>–

1 mol =\bf 6.02×10^{23} formula units

1 mol\bf Cu_2O = 143.092 g = \bf 6.02×10^{23}formula units

Henceforth –

\bf 3.60×10^{23} formula units \bf Cu_2O–

\qquad \sf :\implies \dfrac{143.092 \times3.60×10^{23  }}{6.02×10^{23}}

\qquad \sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23  }}}{6.02×\cancel{10^{23}}}

\qquad \pink{:\implies\bf 85.57 g}

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Refer to attachment.

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