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Yuri [45]
3 years ago
7

How much co2 is realsed by burning trash

Chemistry
1 answer:
V125BC [204]3 years ago
3 0
A lot ...................
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Write the formulas for the following compounds :
Vanyuwa [196]
Sodium phosphate = Na₃PO₄

phosphorus triiodide = PI3

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What is the term for photosynthesis<br>​
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The process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water.
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There is no overall change in reactants and products whenever a chemical reaction a. goes in just one direction. b. goes in two
rodikova [14]

There is no overall change in reactants and products whenever a chemical reaction reaches equilibrium. It is a fundamental point in a reaction.

<h3>Chemical reactions</h3>

A chemical reaction is a process where one or more reactants interact with an enzyme to generate one or more products.

The equilibrium refers to the state where reactants and products are found in the same concentrations.

During the equilibrium, there is no change in the properties of the reaction and the amounts of substances involved remain the same.

Learn more about chemical reactions:

brainly.com/question/6876669

5 0
2 years ago
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

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3 years ago
Consider the energy diagram below.
Kazeer [188]

Answer:

A) The catalyzed reaction passes through C.

Explanation:

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