9.54 The phase change between graphite and diamond is difficult to observe directly. Both substances can be hurned, however. From these equations, calculate Δ H ° for the conversion of diamond into graphite. C ( s , graphite ) + O 2 ( g ) → CO 2 ( g ) Δ H ° = -393.51 kJ C ( s , diamond ) + O 2 ( g ) → CO 2 ( g ) Δ H ° = -395.94 kJ
9.54 The phase change between graphite and diamond is difficult to observe directly. Both substances can be hurned, however. From these equations, calculate Δ H ° for the conversion of diamond into graphite. C ( s , graphite ) + O 2 ( g ) → CO 2 ( g ) Δ H ° = -393.51 kJ C ( s , diamond ) + O 2 ( g ) → CO 2 ( g ) Δ H ° = -395.94 kJ
Solution Summary: The author explains that Hess's law allows us to calculate the heat of reaction for any reaction that occurs at standard conditions.
9.54 The phase change between graphite and diamond is difficult to observe directly. Both substances can be hurned, however. From these equations, calculate
Δ
H
°
for the conversion of diamond into graphite.
How many arrangements are there of 15 indistinguishable lattice gas particles distributed on:
a.V = 15 sites
b.V = 16 sites
c.V = 20 sites
For which element is the 3d subshell higher in energy than that 4s subshell?
Group of answer choices
Zr
Ca
V
Ni
ii) Molecular ion peak
:the peak corresponding to the intact molecule (with a positive charge)
What would the base peak and Molecular ion peaks when isobutane is subjected
to Mass spectrometry? Draw the structures and write the molecular weights of
the fragments.
Circle most stable cation
a) tert-butyl cation
b) Isopropyl cation c) Ethyl cation. d) Methyl cation
6. What does a loss of 15 represent in Mass spectrum?
a fragment of the molecule with a mass of 15 atomic mass units has been lost during
the ionization Process
7. Write the isotopes and their % abundance of isotopes of
i) Cl
Chapter 9 Solutions
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