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Pepsi [2]
3 years ago
11

What is the method of seperating inmisible liquid​

Chemistry
1 answer:
nikitadnepr [17]3 years ago
6 0
Decantation is a process for the separation of mixtures of immiscible liquids or of a liquid and a solid mixture such as a suspension.
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The answer is true. Gas will always be gas.
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9. A box is pushed 1.5 m to the right in 5 s. What is the box’s average speed to the nearest hundredth of a m/s? *
poizon [28]

Answer:

s = 0.30 m/s

Explanation:

Given data:

Distance travel = 1.5 m

Time taken = 5 s

Average speed of box = ?

Solution:

s = d/t

s = speed

d = distance

t = time

by putting values,

s = 1.5 m/ 5 s

s = 0.30 m/s

7 0
2 years ago
Stuck stupid science homework. anyone know answers?
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The charge of a Rb ion would be +1
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A sample of gas has a volume of 3.75 L at a temperature of 25°C and a pressure of 1.15 atm. What will the volume be at a tempera
omeli [17]

Answer: V = 5.1 L

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Solution attached.

5 0
3 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
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