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4 years ago

Please Help I need this to be done by today. There is a big reward if you get it right.

1 answer:

5 0

1)Mid Ocean Ridge
2) Sea Floor Spreading; New crust is formed
3) It gets subducted and melts back into the mantle
4) 1. Molten Material (funny looking rocks)
2. Drilling Samples (age of rocks)
3. Magnetic stripes
5) Subduction: Old oceanic crust is more dense and sinks
6) Sonar
7) Trench
8) Sea floor spreading
9) Subduction
10) Mid ocean ridge

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sesenic [268]

Answer:

For number 2, your answer is C.

Explanation:

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3 years ago

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

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Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}$

Hence, the value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

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