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3 years ago

In a graph, what effect would adding a catalyst have on D?

Chemistry

2 answers:

Anika [276]3 years ago

7 0

It would decrease hope this helps

CaHeK987 [17]3 years ago

3 0

Answer:

C. It would stay the same

Explanation:

PLATO answer

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irga5000 [103]

Answer:

Explanation:

mirrors reflect light

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3 years ago

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3 years ago

Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i

anygoal [31]

Answer:

A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

Explanation:

A. The equilibrium constant K is defined as

$K=\frac{Products}{reagents}$

In any case

aA +Bb equilibrium Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so $K_{2}$ will be less than $K_{1}$ the forward reactions is not favored.

${e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

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3 years ago

jolli1 [7]

Answer:

B. first level consumers.

Explanation:

The first level consumers are also the primary consumers. They are the animals that depends or rely on plants to transform chemical energy into carbohydrates and sugar.

Primary consumers are mostly herbivores. They feed on plant matter by which they derive their nutrition. They have special enzymes that helps to convert plant materials into substances needed for their body.

So, the first level consumer takes plant matter and convert them into useful food materials for their own nutrition.

The second level consumers depend on the first level consumers for nutrition.

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3 years ago

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Flura [38]

Answer:

Glucose and oxygen are required for cellular respiration. As the law of conversation states, in a biochemical reaction, mass is conserved. The mass of hydrogen in the glucose is therefore conserved in the water molecules products.

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3 years ago