The diagram shows the movement of particles from one end of the container to the opposite end of the container.

1 answer:

3 0

The event which is most likely occurring in this scenario is effusion because there is a movement of a gas through a small opening into a larger volume and is denoted as option C.

<h3>What is Effusion?</h3>

This is referred to as the process in which a gas or a substance escapes from a container through a hole of diameter which is usually smaller.

The type of event which is most likely occurring is effusion because of the presence of the small holes in which the balls are made to pass through the center which is why option C was chosen.

Read more about Effusion here brainly.com/question/2097955

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The options are:

diffusion because particles move from regions of high concentration to regions of low concentration.
diffusion because particles move from regions of low concentration to regions of high concentration.
effusion because there is a movement of a gas through a small opening into a larger volume.
effusion because there is a movement of a gas through a large opening into a smaller volume

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

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2 years ago

Why are the densities of corn syrup and gasoline expressed as a range of values

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<span> because gasoline changes volume as a function of temperature or because there are different grades of gasoline or because the values are given in different units of measure .</span>

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2 years ago

SOLVE The density of a gas is 0.68 g/mL. What amount of space would 23.8 g of this gas occupy?

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Answer:

<h2>The answer is 35 mL</h2>

Explanation:

Density of a substance can be found by using the formula

$Density(\rho) = \frac{mass}{volume}$