Answer:
D
Explanation:
The answer is D- There are more collisions of air molecules against the wall of the balloon.
Squeezing an air-filled balloon increases the number of collisions of air molecules with the wall of the balloon, leading to more pressure inside and potentially a higher temperature.
Explanation:When you squeeze an air-filled balloon, the correct answer is D. There are more collisions of air molecules against the wall of the balloon. This is because squeezing the balloon increases the pressure inside it as the volume available to the gas molecules is reduced. Given a constant amount of gas, compressing it makes the molecules collide more frequently with the walls of the balloon. According to the principles of kinetic molecular theory, increased pressure results from an increased number of collisions. This can raise the temperature inside the balloon, as suggested by the relationship described in Charles's Law, which states that volume and temperature are directly proportional when the pressure is held constant. However, if the pressure increases due to compression, this does not necessarily indicate a change in temperature, as that depends on energy transfer, which is not specified in the scenario. Instead, it is the frequency of molecular collisions that increases.
Johnny is very careless in the lab. He purified two white solids (A and B) by re-crystallization, but forgot to label the vials. Both A and B have a melting point range of 10² - 10⁴ °C, so he is not sure which solid is which. He has a small amount of an authentic sample of pure A available in the lab. What should he do?
Answer:
The solution are in the explanation below
Explanation:
- Find the melting point of authentic sample of pure A.
- Mix sample A into both vials.
- Use melting point depression
- Lower melting component will liquefy first, and melting point will lower/broaden the range. (Determine which vial holds sample B mixed with A)
- Vial with the same sample A melting point range would remain consistent to 102-104˚C.
Johnny can identify the unknown samples A and B by conducting a melting point analysis with a mixture of the unknown and a known authentic sample of A. A unchanged melting point indicates the unknown is A, while a changed range suggests it is B. Proper technique is necessary for accurate results.
Johnny should perform a melting point analysis to determine which vial contains which substance. He can take a small amount of the authentic sample of pure A and mix it with a small amount of the unknown substance. He should then measure the melting point range of the mixture. If the mixture has a sharp, unchanged melting point range similar to the authentic sample's known value, it can be concluded that the unknown substance is also substance A. If the melting point range is depressed or broadened, it indicates the unknown substance is not A, but rather substance B.
It is important to note that if the expected melting point is not known, the substance should be heated at a medium rate to determine an approximate melting point. A second attempt should be made with a fresh sample for accuracy. Additionally, ensuring that the amount of solvent is suitable for the volume of the flask, and that heat is properly trapped can guarantee optimal re-crystallization conditions.