I really don’t know the answer, please help this kiddo with his question
Magnesium potassium phosphate hexahydrate and a number of its deuterated counterparts were studied for their Fourier transform infrared and Raman spectra. By contrasting the spectra obtained at the boiling point of liquid nitrogen with those obtained at ambient temperature and by examining the spectra of a succession of partially deuterated MgKPO4
<h3>What is Infrared and raman spectroscopy ?</h3>
Raman and infrared spectroscopy both investigate how radiation interacts with molecular vibrations, but they do so in different ways depending on how the photon energy is delivered to the molecule by altering its vibrational state.
- While IR spectroscopy depends on a change in the dipole moment, Raman spectroscopy depends on a change in the polarizability of a molecule. In contrast to IR spectroscopy, which measures the absolute frequencies at which a sample absorbs light, Raman spectroscopy measures the relative frequencies at which a sample scatters light.
Learn more about Spectroscopy here
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The answer is (CH₃)₂CH-CH(CH₃)₂.
That is the free radical generated in bromination of propane is (CH₃)₂CH° and CH₃CH₂CH₂° , but due to more stability of secondary free radical , (CH₃)₂CH° is more stable than that of CH₃CH₂CH₂°, so reaction proceeds via (CH₃)₂CH° radical. and than at termination the the product formed will be (CH₃)₂CH-CH(CH₃)₂.
The answer is (3) 0.200 kJ
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