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2 years ago

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This characteristic of a sound wave changes as the volume of the sound wave changes

2 answers:

Ket [755]2 years ago

8 0

<span>When a sound wave reaches near the ear, the outer part of the ear acts like if its a funnel, so it catches the wave and funnels it into the ear. They the wave hits the eardrum. Then the bones in the eardrum cause each other to vibrate. Then they are sensed by nerves and sends signals to the brain to interpret sound. Hope This Helps!</span>

natima [27]2 years ago

5 0

AMPLITUDE of a sound wave changes as the volume of the sound wave changes

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user100 [1]

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2 years ago

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Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

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