What is the controlled variable of your experiment? Your answer
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- The bond vibrations that can be observed in the IR spectrum of caffeine :
i) asymmetric, ii) symmetric, iii) wagging, iv) twisting, v) scissoring, vi) rocking
- while the observable stretching frequencies are ; 1700, 1300, 2900, 1500 and 3000.
Caffeine is a polyatomic molecule ( contains more than two atoms bonded via covalent bonds ) therefore its atoms can vibrate in three dimensions ( x, y, z ).therefore the bond vibrations that can be observed in the IR spectrum of caffeine are : asymmetric, symmetric, wagging, twisting, scissoring, and rocking.
The observable stretching frequencies that are in the IR spectrum of caffeine ; C = O is 1700 , C - H is 1300, C -H is 2900, C -N is 1500, and N -H = 3000.
Hence we can conclude The bond vibrations that can be observed in the IR spectrum of caffeine : i) asymmetric, ii) symmetric, iii) wagging, iv) twisting, v) scissoring, vi) rocking and the observable stretching frequencies are ; 1700, 1300, 2900, 1500 and 3000.
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Answer:
one atom of an element = 6.02 \times {10}^{23} atom
The mass of one atom of sulphur = 32g
The mass of one atom of aluminium = 27g
so one atom of aluminium = 6.02 \times {10}^{23}
27g of AL = 6.02 \times {10}^{23} atom
2.70g of AL = X atoms
Then you cross multiply ........
and get the answer
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