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andriy [413]
3 years ago
13

In 1887, Albert A. Michelson and Edward W. Morley conducted an experiment to measure the speed at which the Earth moves through

the ether. Their experiment was so refined that it should have finally detected the luminiferous ether. But it didn't. Repeated trials verified the surprising result that the ether did not exist. How did the failure of the Michelson-Morley experiment change scientific theories of light? A. The Michelson-Morley experiment provided a new theory that fit the data better than the theory of the luminiferous ether did. B. The results of the Michelson-Morley experiment had to be rejected because the theory of the luminiferous ether was so well established. C. The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, so the theory had to be rejected. D. The Michelson-Morley experiment proved that the luminiferous ether moved at the exact same speed as the Earth during all seasons of the year.
Chemistry
2 answers:
vichka [17]3 years ago
7 0
<span> C.The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, so the theory had to be rejected. </span>
frutty [35]3 years ago
3 0

Answer:

The correct answer is option C. "The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, so the theory had to be rejected".

Explanation:

The notorious Michelson-Morley experiment changed the way empty spaces were perceived at their time. Before 1887, the scientific community believed the existence of the luminiferous ether, a substance that supposedly was in empty spaces and allowed electromagnetic waves propagation. The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, and the experiment was so refined that the theory had to be rejected.

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The pH of a 0.150 molar solution of a weak acid is 4.10. What is the pKa of the acid?
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Explanation:


1) Assume a general equation for the ionization of the weak acid:

Let HA be the weak acid, then the ionization equation is:

HA ⇄ H⁺ + A⁻

2) Then, the expression for the ionization constant is:

Ka = [H⁺][A⁻] / [HA]

There, [H⁺] = [A⁻], and [HA] = 0.150 M (data given)


3) So, you need to determine [H⁺] which you do from the pH.

By definition, pH = - log [H⁺]

And from the data given pH = 4.1


⇒ 4.10 = - log [H⁺] ⇒ [H⁺] = antilog (- 4.10) = 7.94×10⁻⁵

4) Now you have all the values to calculate the expression for Ka:

ka = 7.94×10⁻⁵ × 7.94×10⁻⁵ / 0.150 = 4.21×10⁻⁸
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