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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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b) The projectile was 21.3 s in the air.

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r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

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v = velocity vector at time t

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vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

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