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3 years ago

How is the movement of particles in diffusion determined?

1 answer:

3 0

The particles move from an area of high concentration, to an area of low concentration, until the distribution of particles is equal

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The decomposition of nitrogen dioxide is described by the following chemical equation:

kow [346]

Answer:

The balanced chemical equation will be " $NO_{2}+O\rightarrow NO+O_{2}$ ".

Explanation:

The given equation is:

$2NO_{2}\rightarrow 2NO+O_{2}$

<u></u> $2NO_{2}\rightarrow NO+O$ ...(equation 1)

<u></u> $NO_{2}+O\rightarrow NO+O_{2}$ ...(equation 2)

On adding "equation 1" and "equation 2", we get

⇒ $NO_{2}+NO_{2}+O\rightarrow NO+O+NO+O_{2}$

⇒ $2NO_{2}\rightarrow 2NO+O_{2}$

The second step:

⇒ $NO_{2}+O\rightarrow NO+O_{2}$

3 0

2 years ago

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

11111nata11111 [884]

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

3 0

11 months ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: