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omeli [17]
2 years ago
8

If you have a mixture that contains the solid benzoic acid mixed with a volatile alkyl halide liquid. what is the most efficient

technique to separate these two components?
Physics
1 answer:
Goryan [66]2 years ago
6 0

Fractional distillation is the most efficient technique to separate solid benzoic acid mixed with a volatile alkyl halide liquid components.

<h3>Fractional distillation :</h3>

Separating a mixture into its constituent parts, or fractions, is known as fractional distillation. Chemical compounds are separated by heating chemical mixtures to a temperature at which one or more fractions of the mixture evaporate. Distillation is used to fractionate.

Boiling a mixture of liquids results in vapors rising up a glass tube known as a "fractionating column," which is followed by separation. The fractionating column, which is placed between the flask holding the mixture and the "Y" adaptor, improves the separation of the liquids being distilled.

<h3>Uses of fractional distillation :</h3>

Alcohol purification, desalination, crude oil refinement, and the creation of liquefied gases from air are all applications of distillation.

To know more about purification visit :

brainly.com/question/3130146

#SPJ4

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4. An object is moving with an initial velocity of 9 m/s. It accelerates at a rate of 1.5
Nata [24]

Answer:

11.9 m/s

*<em>But</em><em> </em><em>if</em><em> </em><em>you</em><em> </em><em>need</em><em> </em><em>the</em><em> </em><em>answer</em><em> </em><em>with</em><em> </em><em>sig</em><em> </em><em>figs</em><em> </em><em>it</em><em> </em><em>should</em><em> </em><em>be</em><em> </em><em>10m</em><em>/</em><em>s</em><em> </em><em>(</em><em>1sf</em><em>)</em><em> </em><em>or</em><em> </em><em>12m</em><em>/</em><em>s</em><em> </em><em>(</em><em>2sf</em><em>)</em>

Explanation:

{vf}^{2}  -  {vi}^{2}  = 2ad

{vf}^{2}  = 2ad +  {vi}^{2}

{vf}^{2}  = 2 \times 1.5 \times 20 +  {9}^{2}

{vf}^{2}  = 60 + 81

{vf}^{2}  = 141

vf =  \sqrt{141}

vf = 11.874

best answer if it worked

3 0
3 years ago
Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add
jok3333 [9.3K]

Answer:

When x = 2.8 cm, B_{x1} = 0.0265 T

When x = 5.5 cm, B_{x2} = 0.0209 T

when x = 7.3 cm, B_{x3} = 0.0169 T

When x = 11.0 cm, B_{x4} = 0.0103 T

Explanation:

According to Biot-Savart law,

B_{x} = \frac{N \mu_{o}IR^{2}  }{2(x^{2} +R^{2}  )^{3/2} }\\.......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

\mu_{o}  = 4\pi  * 10^{-7} N/A^{2}

When x₁ = 2.8 cm = 0.028 m

B_{x1} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.028^{2} +0.11^{2}  )^{3/2} }\\B_{x1} = 0.0265 T

When x₂ = 5.5cm = 0.055 m

B_{x2} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.055^{2} +0.11^{2}  )^{3/2} }\\B_{x2} = 0.0209 T

When x₃ = 7.3 cm = 0.073 m

B_{x3} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.073^{2} +0.11^{2}  )^{3/2} }\\B_{x3} = 0.0169 T

When X₄ = 11.0 cm = 0.11 m

B_{x4} = \frac{300 *(4\pi * 10^{-7} ) *  17 *0.11^{2}  }{2(0.11^{2} +0.11^{2}  )^{3/2} }\\B_{x4} = 0.0103 T

4 0
4 years ago
A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in
creativ13 [48]

Answer:

The power require to accelerate the car is, P = 299700 watts

Explanation:

Give data,

The mass of the car, m = 2000 kg

The initial velocity of the sports car, u = 30 m/s

The final velocity of the sports car, v = 60 m/s

The time period of acceleration, t = 9 s

The acceleration of the car, a =  (v-u) / t

                                                  = (60 - 30) / 9

                                                  = 3.33 m/s²

The displacement of the car,

                                               S = ut + ½ at²

                                                  = 30 x 9 + ½ x 3.33 x 9²

                                                  = 405 m

The force acting on the car, F = m x a

                                                  = 2000 x 3.33

                                                  = 6660 N

The work done by the car, W = F  S

                                                  = 6660 x 405

                                                  = 2697300 J

The power of the car,           P = W / t

                                                  = 2697300 / 9

                                                  = 299700 watts

Hence, the power require to accelerate the car is, P = 299700 watts

4 0
4 years ago
HELP THIS IS URGENT! IT WOULD BE GREATLY APPRECIATED IF YOU ANSWER ALL BUT ONE IS FINE!
sukhopar [10]

Answer:

1 both increase

2. highest to lowest A C D B

3 the kinetic energy is higher in the larger vehicle as they have the same speed

Explanation:

6 0
3 years ago
A mass with a charge of 4.60 x 10-7 C rests on a frictionless surface. A compressed spring exerts a force on the mass on the lef
stira [4]

Answer:

The compression of the spring is 24.6 cm

Explanation:

magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C

magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C

distance between the two charges, r = 3 cm = 0.03 m

spring constant, k = 14 N/m

The attractive force between the two charges is calculated using Coulomb's law;

F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(4.6\times 10^{-7})(7.5\times 10^{-7})}{(0.03)^2} \\\\F= 3.45 \ N

The extension of the spring is calculated as follows;

F = kx

x = F/k

x = 3.45 / 14

x = 0.246 m

x = 24.6 cm

The compression of the spring is 24.6 cm

4 0
3 years ago
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