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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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sukhopar [10]

Answer:

4) True. The change of direction needs an unbalanced force

Explanation:

Let us propose the resolution of the problem using Newton's second law.

    F = m a

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1) and 3) False, even when the speed changes, the direction changes

2) False with the speed change can be determined

4) True. The change of direction needs an unbalanced force

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6 0
3 years ago
Creativity is always discouraged in scientific endeavors.
shepuryov [24]
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In Short, Your Answer would be "False"

Hope this helps!
3 0
3 years ago
Read 2 more answers
Two cars are traveling in the same direction down a highway at 65 miles per hour. What is the relative velocity of the second ca
Levart [38]

Answer:

5 hours

Explanation:

Let the required time be x hours. The time will be the same for both cars.

The cars will cover different distances because they are travelling at different speeds.

<em>D=S×T </em>

The distance travelled by the slower car = 50×x miles.

The distance travelled by the faster car = 58×x miles.

The two distances differ by 40 miles.

58x−50x=40

8x=40

x=5 hours

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A second method:

The difference in the distances is 40 miles

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The time to make up the 40 miles= \frac{40}{8}=5 hours

8 0
3 years ago
Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

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5 0
4 years ago
A phonograph record 0.15 m in its radius rotates 18 times per 90 seconds what is the frequency?
ioda

Answer:

The frequency of the phonograph record is 0.2 Hz

Explanation:

The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period

The given parameters of the phonograph record are;

The radius of the record = 0.15 m

The number of times the phonograph record rotates, n = 18 times

The time it takes the phonograph record to rotate the 18 times, t = 90 seconds

The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)

∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz

The frequency of the phonograph record = 0.2 Hz.

6 0
3 years ago
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