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Tema [17]
3 years ago
6

If a soccer ball is left on a hot blacktop driveway, predict what would happen to the pressure, volume and density of the air in

side the ball. Support your answers with an explanation of what is happening with the motion of the air particles as well.
Chemistry
2 answers:
hoa [83]3 years ago
7 0
Since the soccer ball is heating up the air in the ball will expand, which will cause the ball to grow in size and it will have more volume and pressure but less density. (if you put a soccer ball in water it'll float since it has less weight from the air, that's it's density) hope this helps.  <3
stira [4]3 years ago
7 0
Determining the effects of applying heat to a system requires analyzing the system itself. Assuming that the soccer ball can expand (like it does in real life), then applying heat will increase the volume, but NOT the pressure or the density. Density refers to how compressed the air is, but since the air can freely expand into a larger system, the air doesn’t get more dense. The same applies to pressure.

The air particles gain energy when heat is applied to the system, causing them to move faster and therefore collide with the walls of the ball more often. This either causes the pressure to increase (if the system is rigid), or the volume to increase (if the system isn’t rigid, which it isn’t). In applying heat, the ball expands, but the pressure and density remain constant.
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A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

3 0
3 years ago
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Nat2105 [25]

A good reason for a desert fox to show this pattern of behavior because hunting at night allows the fox to use its night vision.

<h3>What is Hunting?</h3>

Thi9s is commonly practised by predators such as fox in which they capture and kill other animals for food.

The fox has a good night vision which makes it able to hunt for animals during the night also. This is why option C is chosen as the most appropriate choice.

Read more about Hunting here  brainly.com/question/81175

6 0
2 years ago
What are the four most common problems that can arise in your motor design?
Nuetrik [128]

Electric motors are an essential part of our daily life as many systems, applications, and services depend on them. Motors today have a long service life and require a minimum level of maintenance to make sure that they perform efficiently. In large buildings, motors have to be maintained on a regular basis because they need to be in operation all the time; one small problem could cause a great loss to the organization.

Usually in large organizations, a motor maintenance program is carried out in which the causes of motor failures are identified and some necessary steps are taken to avoid them or lower their impact. Motors need to be inspected regularly, and other maintenance activities need to be performed to ensure efficient operation. Whenever a problem occurs, it should be corrected immediately to avoid further loss.

4 0
3 years ago
What is the charge on an atom after it gains two electrons during the formation of a bond?
vekshin1

Answer:

two negative charges is the answer of your question

8 0
3 years ago
What will the pressure be if 89.9 moles of argon are contained in a 12.0 L cylinder that is pressurized at a temperature of 300
irinina [24]
  • P=nRT/V
  • p=89.9(8.314)(12)/300
  • P=8969.14/300
  • P=29.89atm
  • P=29.9atm

Done!

7 0
2 years ago
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