A model of an atom is shown above. The blue dots represent electrons, and the red dot represents the nucleus. This model accurat
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The masses of the components are obtained as;
- Sodium hydrogen carbonate = 3.51 g
- Sodium carbonate = 8.708 g
The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.
Now putting down the equation of the reaction, we have;
Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g
Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol
= 3.51 g
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Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the <u>vibration of the bonds</u> by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is <em>a specific energy that generates a specific vibration</em>. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the <u>lower the wavenumber we will have less energy</u>. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have <u>heteroatoms</u> (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of <u>resonance structures</u> which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the <u>cyclohexanone.</u>
See figure 1
I hope it helps!
