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3 years ago

Describe some ways that industry and agriculture use physical properties to separate substances.

2 answers:

3 0

This is more chemistry. But it is a process called fractional distillation, and it basically separates the long chained hydrocarbons from the short chained hydrocarbons through separation dependant on the boiling point of the crude oil.

Oliga [24]3 years ago

3 0

Some ways that industry and agriculture use physical properties to separate substances are as follows-

a) Separation based on the difference in DENSITY – In the process of sedimentation, the heavier particle settles at the bottom while the lighter particle remains dissolved or float in the solvent. The settled heavier substance is then removed periodically. For example mud and sand impurities from water are removed through this process.

b) Separation based on the difference in SOLUBILITY – Some substances that remain un dissolved in any other substance can be easily removed through the process of filtration. For example – Extraction of coffee from grounds

c) Separation based on the difference in Special metal Property – Iron possesses magnetism and thus it can be easily separated from nonmagnetic substances

d) Separation based on the difference in Vapour pressure/Boiling Point – It is the concept behind the process of separation through Distillation. In distillation, liquid with the lower boiling point boils first which is collected through condensation while the liquid with higher boiling point remain in the flask.

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A ___ is when too much water moves into an area

Marina86 [1]

A. Flood.
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6 0

4 years ago

Read 2 more answers

Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of coppe

alexandr402 [8]

Answer:

0.699 L of the fluid will overflow

Explanation:

We know that the change in volume ΔV = V₀β(T₂ - T₁) where V₀ = volume of radiator = 21.1 L, β = coefficient of volume expansion of fluid = 400 × 10⁻⁶/°C

and T₁ = initial temperature of radiator = 12.2°C and T₂ = final temperature of radiator = 95.0°C

Substituting these values into the equation, we have

ΔV = V₀β(T₂ - T₁)

= 21.1 L × 400 × 10⁻⁶/°C × (95.0°C - 12.2°C)

= 21.1 L × 400 × 10⁻⁶/°C × 82.8°C = 698832 × 10⁻⁶ L

= 0.698832 L

≅ 0.699 L = 0.7 L to the nearest tenth litre

So, 0.699 L of the fluid will overflow

6 0

3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

If there is a break at any point in a series circuit the current will

Tomtit [17]

Answer:

not work

Explanation:

in a series circuit, everything meaning the electrons are flowing on one path, therefore, it wouldn continue to work.

8 0

3 years ago

Find the speed of light in each of the following materials. (a) gallium phosphide m/s (b) carbon disulfide m/s (c) benzene

Oksanka [162]

Explanation:

We need to calculate the speed of light in each materials

(I). Gallium phosphide,

The index of refraction of Gallium phosphide is 3.50

Using formula of speed of light

$v=\dfrac{c}{\mu}$ ....(I)

Where, $\mu$ = index of refraction

c = speed of light

Put the value into the formula

$v=\dfrac{3\times10^{8}}{3.50}$

$v=8.6\times10^{7}\ m/s$

(II) Carbon disulfide,

The index of refraction of Gallium phosphide is 1.63

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.63}$

$v=1.8\times10^{8}\ m/s$

(III). Benzene,

The index of refraction of Gallium phosphide is 1.50

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.50}$

$v=2\times10^{8}\ m/s$

Hence, This is the required solution.

7 0

4 years ago