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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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At STP, which substance has metallic bonding?
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Answer:

Explanation:

From the given information:

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Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

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The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

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3 years ago
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