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marissa [1.9K]
3 years ago
14

A sample of an ideal gas at 1.00 atm and a volume of 1.50 L was placed in a weighted balloon and dropped into the ocean. As the

sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 25.0 atm, what was the volume of the sample? Assume that the temperature was held constant.
Chemistry
1 answer:
Lady bird [3.3K]3 years ago
6 0

Answer:

Explanation:

Using Boyle's law

P₁ V₁ = P₂V₂ and temperature is constant

where P₁(pressure)  = 1.00atm, P₂ = 25 atm, V₁( volume) = 1.50L V₂ =

V₂ = ( P₁ V₁ ) / P₂ = ( 1 atm × 1.50L ) / 25 atm = 0.06 L

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Anuclear power plant is used to generate electricity

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What is the molarity of a solution containing 7.0 moles of solute in 569 mL of solution?
yawa3891 [41]
The formula of the molarity, M is: M = number of moles of solute / Volume of solution in liters.

Here number of moles of solute  is 7.0 and volume of solution in liters = 569 mL / 1000 mL/L = 0.569 L

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How is a series circuit different from a parallel circuit?
beks73 [17]

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Will a precipitate (ppt) form when 300. mL of 2.0 × 10 –5 M AgNO 3 are added to 200. mL of 2.5 × 10 –9 M NaI? Answer yes or no,
Vsevolod [243]

Answer:

A precipitate will form, AgI

Explanation:

When Ag⁺ and I⁻ ions are in an aqueous media, AgI(s), a precipitate, is produced or not based on its Ksp expression:

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<em>Where the concentrations of the ions are the concentrations in equilibrium</em>

For actual concentrations of a solution, you can define Q, <em>reaction quotient, </em>as:

Q = [Ag⁺] [I⁻]

<em>If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form</em>.

Actual concentrations of Ag⁺ and I⁻ are:

[Ag⁺] = [AgNO₃] = 2.0x10⁻⁵ × (300mL / 500.0mL) = 1.2x10⁻⁵M

[I⁻] = [NaI] = 2.5x10⁻⁹ × (200mL / 500.0mL) = 1.0x10⁻⁹M

<em>500.0mL is the volume of the mixture of the solutions</em>

Replacing in Q expression:

Q = [Ag⁺] [I⁻]

Q = [1.2x10⁻⁵M] [1.0x10⁻⁹M]

Q = 1.2x10⁻¹⁴

As Q > Ksp

<h3>A precipitate will form, AgI</h3>

8 0
3 years ago
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