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Free_Kalibri [48]
3 years ago
5

(True or False) A single particle has NO temperature. True False

Chemistry
1 answer:
Elenna [48]3 years ago
6 0

Answer:

True

Explanation:

No, a single particle cannot have a temperature at all, because temperature is a property of a large number of particles.- this is not my own words the website I got it from is https://www.quora.com/Can-a-single-particle-have-any-temperature

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Rank these real gases according to how closely they resemble an ideal gas. CO, N2, Ne, He, NH3
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Formula for Hydrocloric Acid.
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Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a
horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound            Amount

     Ni                        12.7 g

   CO                        1.98 g

Ni(CO)_4                  0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of Ni(CO)_4 = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M

\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M

\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M

For the given chemical reaction:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

The expression of equilibrium constant for the reaction:

K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1

Hence, the value of equilibrium constant for the reaction is 2448.1

4 0
3 years ago
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