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jekas [21]
3 years ago
8

A 0.025 g sample of o is injected into a 1.00 l flask at 30.0c. what forms of water will be present at equilibrium? vapor pressu

re of water at 30.0c is 31.8 mm hg.
Physics
1 answer:
marysya [2.9K]3 years ago
5 0
The question asked is only the forms of water that will be present at equilibrium. Since the temperature of the system is 30 C, water will be in a form of liquid. Assuming that equilibrium will be achieved in the container, some liquid water will vaporize and mix with the sample. So, the forms of water that will be present are liquid and water vapor. 
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It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy whe
boyakko [2]

Answer:

10.4L of water is expended when 1L of crude oil is burned.

Explanation:

This problem requires us to calculate the volume of water that must be expended to absorb the amount of energy released from the burning of 1.00L of crude oil.

In going from water at 18.5°C to steam at 285°C, the water undergoes 3 stages:

Absorption of heat from 18.5°C to 100°C >> phase change to vapour at 100°C >> heating from 100°C to steam at 285°C

Given Cw = 4186 J/kg°C and Cs = 2020 J/kg°C. The latent heat of of vaporization of water Lv = 2.256 ×10⁶ J/kg

The total heat absorbed in the process per kilogram

H = Cw × (100 – 18.5) + Lv + Cs × (285 – 100)

H = 4186 × 81.5 + 2. 256 ×10⁶ + 2020× 185

= 2.696 ×10⁶ J/kg

The amount of heat in J/L of water needed can be calculated as follows:

Density of water = 1000kg/m³ =

1000 kg/m³ × 1m³/1000L = 1kg/L (Basically conversion of density in kg/m³ to kg/L)

Let the volume of water needed be V litres.

Then the mass of water that must be expended = Density × Volume

= 1kg/L × V L = Vkg

The heat that would be absorbed by the water when 1L of crude oil is burned is V×H

= 2.696×10⁶ × V

This is also equal to 2.80×10⁷ J of energy (given).

So,

2.696×10⁶V = 2.8×10⁷

V = (2.80×10⁷)/(2.696×10⁶) = 10.4L of water.

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