The given statement is true .
<h3>What is Rutherford’s gold foil experiment?</h3>
- A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.
- The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.
- For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.
To learn more about Rutherford’s gold foil experiment, refer to:
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Answer:
The glowing splint test is a test for an oxidizing gas, such as oxygen. In this test, a splint is lit, allowed to burn for a few seconds, then blown out by mouth or by shaking. Whilst the ember at the tip is still glowing hot, the splint is introduced to the gas sample that has been trapped in a vessel. Oxygen supports combustion so a good method of testing for oxygen is to take a glowing splint and place it in a sample of gas, if it re-ignites the gas is oxygen. This is a simple but effective test for oxygen.
Answer:
directly proportional to one another
Explanation:
Volatility refers to how quickly a substance changes from liquid to gas.
If a substance has a high vapour pressure, the substance is highly volatile. Similarly, if a substance has a low vapour pressure, then the substance is much less volatile.
This implies that volatility and vapour pressure gives a direct proportionality.
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!
