Is bubble chamber one of your choices? Bubble chamber sounds like a good fit for the question.
The correct answer is option 1. Carbon dioxide is nonpolar because the shape of the molecule is symmetrical. It is a linear molecule where the oxygen atoms are symmetrical on each end. CO2 molecule do not have a region of unequal sharing.
Answer:
Pottasium reacts with water vigorously and the reation is exothermic. The heat released causes the hydrogen released to ignite
Explanation:
<u>The troposphere: </u>
H. This layer can have thunderstorms or clear, sunny skies.
A. The biosphere interacts most with this layer.
<u>The stratosphere:</u>
B. It is the second layer from Earth's surface.
G. Winds are strong and steady in this layer.
<u>The mesosphere:</u>
E. It is heated by the ozone layer beneath it.
D. This layer is where most meteor showers occur.
<u>The thermosphere :</u>
F. It contains the ionosphere and exosphere.
C. It contains layers of single, unmixed gas.
<u>Explanation:</u>
Depending on the Earth's temperature the atmosphere can be separated into layers. The troposphere, the stratosphere, the mesosphere and the thermosphere are those layers. The lowest layer is named as Troposphere (0-10 km from the Earth outer surface), it comprises about 75% of the atmosphere's total air and nearly most the water vapor.
Stratosphere (10-30) includes much of the surface ozone. The change in height temperature arises as this ozone absorbs ultraviolet (UV) radiation from the sun. The temperature in Mesosphere (30-50 Km) declines again with height, hitting a minimum of about -90 ° C at the "mesopause." Above this thermosphere (50-400 Km) is settled which is a area where temperatures rise with height once again. The penetration of intense UV and X-ray radiation from the sun induces this temperature rise.
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.
The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:
The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.
Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.
The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, 
A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.
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