During the process of condensation, water vapor:____________.

Chemistry

1 answer:

monitta3 years ago

7 0

Answer: (2) releases 2260 J/g of heat energy

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Latent heat of condensation is energy released when 1 mole of vapor condenses to form liquid droplets.

The temperature does not change during this process, so heat released goes into changing the state of the substance, thus it is called latent which means hidden. The energy released in this process is same in magnitude as latent heat of vaporization. The heat of condensation of water vapour is about 2,260 J/g.

You might be interested in

What is an animals life in a desert?

rodikova [14]

An animals life in a desert is to survive, depending on the type of desert though, but I assume the most common desert, the hot/dry desert. Most animals are nocturnal, because it becomes cooler at night, and live underground during the day.

8 0

3 years ago

The enzyme urease catalyzes the breakdown of urea in the body. Urease breaks urea down to 2NH3+CO2. This is an example of a hydr

Alika [10]

Answer:

Look at the picture.

Explanation:

On stage one binding of a substrate occurs (and also the geometry of active site may change) and water comes to the site. On stage two the hydrolisis takes place and on stage 3 products deabsorb from the enzyme.

8 0

3 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$