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4 years ago

What method of separation would work best on a homogeneous mixture salt water?

2 answers:

5 0

The appropriate answer is c. evaporation. Evaporation is the process by which a liquid changes to a gas and in this case is the best way to remove the water from the homogenous mixture. The reaction can be sped up by heating the mixture. All the water will eventually evaporate leaving behind the sodium chloride crystals as a precipitate.
Filtration works best on mixtures that have precipitates and distillation is for separating liquids with different boiling points.

NNADVOKAT [17]4 years ago

5 0

Answer:

evaporation

Explanation:

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

Which of the following would be an example of basic research

abruzzese [7]

Basic research, also called pure research or fundamental research, is scientific research aimed to improve scientific theories for improved understanding or prediction of natural or other phenomena.

(source: wikipedia)

5 0

3 years ago

Read 2 more answers

A metal rod has a length of 123. cm at 200°C. At what temperature will the length be 92.6 cm if the coefficient of linear expans

mestny [16]

Answer:

$\theta_{2} = 15400^0 C$

Explanation:

The formula for linear expansivity, $\alpha = \frac{l_{2} - l_{1} }{l_{1} ( \theta_{2} - \theta_{1} )}$

original length, l₁ = 123 cm = 1.23 m

final length, l₁ = 92.6 cm =0.926 m

original temperature, θ₁ = 200°C

Linear expansivity, α = 2 * 10⁻⁵ °C⁻¹

Putting these values into the formula:

$2 * 10^{-5} = \frac{1.23 - 0.926 }{l_{1} ( \theta_{2} -200 )}\\ \theta_{2} -200 = \frac{0.304}{2 * 10^{-5} } \\\theta_{2} = 15200 + 200\\\theta_{2} = 15400^0 C$