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AlladinOne [14]
3 years ago
14

Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled wit

h some standard solution,such as aqueous potassium chloride. The conductivity of water is 76 mS m^(-1) at 25 C and the conductivity of 0.100 mol dm^(-3) KCl (aq) is 1.1639 S m^(-1) A cell has a resistance of 33.21ohm when filled with 0.100 mol dm^(-3) KCl (aq) and 300.0 ohm when filled with 0.100 mol dm CHaCOOH (aq). What is the molar conductivity of acetic acid at that concentration and temperature?
Physics
1 answer:
Bezzdna [24]3 years ago
3 0

Answer:

1200 Sm^2mol^-1

Explanation:

Given data :

conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1

conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1

Kkcl = 1.1639 - 0.076 = 1.0879  Sm^-1

Resistance = 33.21 Ω

where conductivity can be expressed as = \frac{Cell constant}{Resistance }

hence cell constant = conductivity * Resistance

                                 = 1.0879 * 33.21 = 36.13m^-1

conductivity of  CH3COOH ( kCH3COOH ) =  36.13 / 300

                                                                       = 0.120 Sm^-1

<u>Determine the molar conductivity of acetic acid</u>

= ( kCH3COOH * 1000 ) / C

C = 0.1 mol dm

=  (0.120 * 1000) / 0.1  =  1200 Sm^2mol^-1

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This makes defecation very difficult.

So therefore, changes in daily routine or lifestyle like traveling can lead to constipation

Learn more about changes in life style as one of the causes of constipation:

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2 years ago
A certain organ pipe, open at both ends, produces a fundamental frequency of 288 Hz in air. Part A If the pipe is filled with he
alina1380 [7]

Answer:

773.25 Hz

Explanation:

Concept : In an open organ pipe in fundamental mode of vibration

wave length of wave λ = 2L

where L  is length of  the pipe

frequency   = velocity of sound / λ

Given values: fundamental frequency = 288 Hz

fluid is air. velocity of sound = 340 m/s

⇒ 288 = 340/2L

⇒L = 59.02 cm

The point to be noted is if the pipe is filled with helium initially at the same temperature, there would be change in the sound velocity .Then,  frequency of note produced will also be changed .

We know that velocity of sound is inversely proportional to  square root of molar mass of gas

velocity of sound in air / velocity of sound in helium = Square root of (Molar mass of Helium/ molar mass of air)

\frac{V_a}{V_{He}} = \sqrt{\frac{4}{28.8} } \\\frac{340}{V_{He}} =0.3725\\V_{He} =912.5 m/s

Now, frequency  = velocity of sound / λ

= 912.75 / (2 x 0.5902)

= 773.25 Hz

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3 years ago
Why do we believe that most of the mass of the milky way is in the form of dark matter?
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Dark matter is a type of matter, whose composition is unknown and which corresponds to 80% of the matter in the universe. Its name refers to the fact it does not emit or interact with any type of electromagnetic radiation, being completely transparent throughout the electromagnetic spectrum.

However, it interacts with the known matter through <u>gravity</u>.

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4 years ago
The 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined pl
Scorpion4ik [409]

<u>Given</u><u> </u><u>:</u><u>-</u>

  • A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
  • \mu_s = 0.3 \ \& \ \mu_k = 0.2

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • Would the block move ?
  • If it moves what is its speed after it has descended a distance of 5m down the plane .

<u>Solution</u><u> </u><u>:</u><u>-</u>

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the <u>angle</u><u> of</u><u> </u><u>repose</u><u> </u>. We can find it as ,

\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)

Substitute ,

\longrightarrow \theta_{repose}= tan^{-1}( 0.2)

Solve ,

\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}

Hence ,

\longrightarrow\theta_{plane}>\theta_{repose}

<u>Hence</u><u> the</u><u> </u><u>block</u><u> will</u><u> slide</u><u> down</u><u> </u><u>.</u>

Now assuming that block is released from the reset , it's <u>initial</u><u> </u><u>velocity </u> will be 0m/s .

And the net force will be ,

\longrightarrow F_n = mgsin53^o - \mu_k N

Substitute, N = mgcos53⁰ ( see attachment)

\longrightarrow ma_n  = mgsin53^o - \mu_k mgcos53^o

Take m as common,

\longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)

Simplify ,

\longrightarrow a_n = gsin53^o - \mu_k g cos53^o

Substitute the values of sin , cos and g ,

\longrightarrow a_n = 10( 0.79 - 0.2 (0.6))

Simplify ,

\longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}

Now using the <u>Third </u><u>equation</u><u> </u><u>of</u><u> motion</u><u> </u>namely,

\longrightarrow2as = v^2-u^2

Substituting the respective values,

\longrightarrow2(6.7)(5) = v^2-(0)^2

Simplify and solve for v ,

\longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }}

<u>Hence</u><u> the</u><u> </u><u>velocity</u><u> after</u><u> </u><u>covering</u><u> </u><u>5</u><u>m</u><u> </u><u>is </u><u>8</u><u>.</u><u>1</u><u>8</u><u> </u><u>m/</u><u>s </u><u>.</u>

7 0
3 years ago
Using 15 percent as machine energy efficiency, what is the actual work output if the total work input is 7500 kilojoules​?
g100num [7]

Answer:

The actual work output is 1,125 kilojoules.

Explanation:

Energy efficiency is defined as the efficient use of energy. An appliance, process, or facility is energy efficient when it consumes less than the average amount of energy to perform an activity. Then, energy efficiency is the ratio between the amount of energy used in an activity and the amount expected to be carried out.

Efficiency is calculated as:

efficiency = output / input

Where output is the amount of mechanical work (in watts) or energy consumed by the process (in joules), and input (input) is the amount of work or energy that is used as input to carry out the process.

In this case:

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  • output=?
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Replacing:

0.15=output/7500 kilojoules

Solving:

Output=0.15* 7500 kilojoules

output=1,125 kilojoules

<u><em>The actual work output is 1,125 kilojoules.</em></u>

7 0
3 years ago
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