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andrew11 [14]
3 years ago
8

during a rising tide, ocean waves often become larger. if the amplitude of a wave increases by a factor of 1.1, by how much does

the energy increase?
Chemistry
1 answer:
Sergio039 [100]3 years ago
8 0

Answer:

1.21 times

Explanation:

The energy of a wave is proportional to the square of the amplitude of the wave.

Mathematically:

E\propto A^2

where

E is the energy of the wave

A is its amplitude

In this problem, the amplitude of the wave increases by a factor of 1.1; it means that the new amplitude can be written as

A'=1.1 A

Therefore, this means that the energy of the wave increases by a factor:

E'\propto A'^2=(1.1 A)^2 = 1.1^2 A^2 =1.21 A^2 = 1.21 E

Therefore, the energy of the wave increases by a factor 1.21.

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Yield for this reaction?<br> Reaction: N2(g) + 3 H2(g) → 2 NH3 (g)
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Answer:

Yes, yield.

Explanation:

N2(g) + 3 H2(g) → 2 NH3 (g) balanced equation

First, find limiting reactant:  

Moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2

Moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2

The mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)

Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3

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11) Predict the products of this reaction.
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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

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Answer:

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Explanation:

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