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3 years ago

Describe how freezing could be used to remove sugar from a mixture of sugar and water?

2 answers:

7 0

When the mixture (the sugar and water) is frozen, it separates. The water molecules get closer together, separating and pushing the sugar crystals to the top.<span />

NikAS [45]3 years ago

6 0

Answer

Freezing can be used to separate sugar from a mixture of sugar and water if the solution is slowly cooled.

Explanation

Depending on the concentration of the sugar in the mixture, a separation can be attained if a slowly cooling process is performed. For example, if the solution is highly concentrated for example in a syrup solution, the mixture could be slowly cooled to allow the sugar to crystallize out. If the concentration is dilute but cooled slowly, pure water can crystallize first to leave a concentrated sugar solution until it freezes too.

You might be interested in

How much pressure is applied to the ground

statuscvo [17]

Answer:

Pressure applied by the man= 285103.125 $Pa$ or 41.35 $lb/in^{2}$

Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e. $Pressure=\frac{Force}{Area}$

Now, $Force= mg$

where, $m$ = mass of the body(man) = 93 kg

$g$ = acceleration due to gravity of Earth = 9.81 $m/{s^{2}}$

$Area$ covered is equal to the area of both stilts(a man generally stands on two feet)

therefore $Area=2(0.04)^{2}$ $m^{2}$

and putting in the values, we get,

$Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}$

Now we need to convert to our required units:

$1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}$

(We can get the above result by individually converting kg to lb and meters to inches respectively)

Using the above relations we get,

$Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}$

7 0

3 years ago

5. Is it likely that light with frequencies higher than ultraviolet was

Vinvika [58]

Answer: YES

Explanation:

If Tia Ana exposes her eyes to light ray in which it frequency is higher than ultraviolet ray, it may result to a gradual eyes issue such as cataracts as the radiation damages the cornea and the lens in front of the eyes.

3 0

3 years ago

A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2

il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

$f_o = \frac{v - v_o}{v}f$

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

$f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz$

The frequency heard by the observer is 993.52 Hz.

4 0

3 years ago

yaroslaw [1]

Answer:

12 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 7.6 kg

Distance (d) = 6 m

Velocity (v) = 5 m/s

Force (F) = 2 N

Workdone (Wd) =.?

Workdone can be defined as the product of force and distance moved in the direction of the force. Mathematically, it is expressed as:

Workdone = Force × distance

Wd = F × d

With the above formula, we can obtain the workdone as follow:

Distance (d) = 6 m

Force (F) = 2 N

Workdone (Wd) =.?

Wd = F × d

Wd = 2 × 6

Wd = 12 J

Thus, the workdone is 12 J

6 0

2 years ago

When are you most aware of your motion in a moving vehicle: when it is moving steadily in a straight line or when it is accelera

jenyasd209 [6]

Answer:

A) Accelerating. You would not be aware of the motion if you did not look outside the car

Explanation:

Since human cannot sense the motion (unless visually), but the inertia force caused due to the acceleration of the motion. So if you are in a car with constant velocity, there's no acceleration and no inertia force, you would not be able to sense the motion at all unless you look outside.

3 0

2 years ago