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3 years ago

Which structure is located between the auricle and the eardrum? O stirrup O ear canal O auditory nerve Ocochlea

Chemistry

1 answer:

Damm [24]3 years ago

3 0

Answer:

The structure that is located between the auricle and the eardrum is the ear canal.

Explanation:

The ear canal, or external ear canal, is a a tubular hole about 30 mm long that runs from the auricle to the eardrum, forming part of the external ear.

Its function is to conduct sound, in the form of vibrations, from the outside to the eardrum. It also has the function of producing a viscous secretion called cerumen, capable of trapping dust particles and small foreign bodies.

Other options are not correct because:

<em>Stirrup is located in the middle ear, along with the anvil and hammer. </em>
<em>Cochlea is in the inner ear and continues with the auditory nerve.</em>

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2 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

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The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

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$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

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Which structure is located between the auricle and the eardrum? O stirrup O ear canal O auditory nerve Ocochlea​

Which structure is located between the auricle and the eardrum? O stirrup O ear canal O auditory nerve Ocochlea