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Marina CMI [18]
3 years ago
6

A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What vol

ume of water evaporated from the 0.24 M solution?
Physics
1 answer:
Gwar [14]3 years ago
4 0

Answer:

The volume of water evaporated is 199mL

Explanation:

Concentration is calculated with the following formula

C=\frac{n}{V}

where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.

So we isolate the variable n to know the amount of moles, using the volume given in liters

230mL=0.23L

n=C*V=0.240 M*0.23L=0.055 mol

Now, we isolate the variable V to know the new volume with the new concentration given.

V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL

Finally, the volume of water evaporated is the difference between initial and final volume.

V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL

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Describe an experiment to determine how the frequency of a vibrating string depends on the length of the string
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Answer:

For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length. ... The fundamental frequency of a vibrating string is inversely proportional to its length.

Explanation:

Sounds of a single pure frequency are produced only by tuning forks and electronic devices called oscillators; most sounds are a mixture of tones of different frequencies and amplitudes. The tones produced by musical instruments have one important characteristic in common: they are periodic, that is, the vibrations occur in repeating patterns. The oscilloscope trace of a trumpet's sound shows such a pattern. For most non-musical sounds, such as those of a bursting balloon or a person coughing, an oscilloscope trace would show a jagged, irregular pattern, indicating a jumble of frequencies and amplitudes.

A column of air, as that in a trumpet, and a piano string both have a fundamental frequency—the frequency at which they vibrate most readily when set in motion. For a vibrating column of air, that frequency is determined principally by the length of the column. (The trumpet's valves are used to change the effective length of the column.) For a vibrating string, the fundamental frequency depends on the string's length, its tension, and its mass per unit length.

In addition to its fundamental frequency, a string or vibrating column of air also produces overtones with frequencies that are whole-number multiples of the fundamental frequency. It is the number of overtones produced and their relative strength that gives a musical tone from a given source its distinctive quality, or timbre. The addition of further overtones would produce a complicated pattern, such as that of the oscilloscope trace of the trumpet's sound.

How the fundamental frequency of a vibrating string depends on the string's length, tension, and mass per unit length is described by three laws:

1. The fundamental frequency of a vibrating string is inversely proportional to its length.

Reducing the length of a vibrating string by one-half will double its frequency, raising the pitch by one octave, if the tension remains the same.

2. The fundamental frequency of a vibrating string is directly proportional to the square root of the tension.

Increasing the tension of a vibrating string raises the frequency; if the tension is made four times as great, the frequency is doubled, and the pitch is raised by one octave.

3. The fundamental frequency of a vibrating string is inversely proportional to the square root of the mass per unit length.

This means that of two strings of the same material and with the same length and tension, the thicker string has the lower fundamental frequency. If the mass per unit length of one string is four times that of the other, the thicker string has a fundamental frequency one-half that of the thinner string and produces a tone one octave lower.

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Place /our live /music / important / has / in / an​
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(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is  7.61 x 10⁻⁶  

(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

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  • g is acceleration due to gravity
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σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

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strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

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ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

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