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GrogVix [38]
3 years ago
7

1. Water has many unique properties that contribute to making Earth habitable. Describe all four of these unique properties of w

ater. In your answer be sure to dearly state the property of water and indicate how that property helps make Earth habitable.​
Biology
1 answer:
marysya [2.9K]3 years ago
5 0

The properties are:-

1. Water is polar. Water molecules are polar, with partial positive charges on the hydrogens, a partial negative charge on the oxygen, and a bent overall structure. This is because oxygen is more electronegative, meaning that it is better than hydrogen at attracting electrons.

2. Water is an excellent solvent. Water has the unique ability to dissolve many polar and ionic substances. This is important to all living things because, as water travels through the water cycle, it takes many valuable nutrients along with it!

3. Water has high heat capacity. It takes a lot of energy to raise the temperature of a certain amount of water by a degree, so water helps with regulating temperature in the environment. For example, this property allows the temperature of water in a pond to stay relatively constant from day to night, regardless of the changing atmospheric temperature.

4. Water has high heat of vaporization. Humans (and other animals that sweat) use water’s high heat of vaporization to cool off. Water is converted from its liquid form to steam when the heat of vaporization is reached. Since sweat is made mostly of water, the evaporating water absorbs excess body heat, which is released into the atmosphere. This is known as evaporative cooling.

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Gold has a specific heat of 0.129 J/(g×°C). How many joules of heat energy are required to raise the temperature of 15 grams of
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Answer:

\boxed {\boxed {\sf 121.905 \ J  }}

Explanation:

We are asked to find the energy given mass, specific heat, and change in temperature. Therefore, we must use this formula;

q= mc \Delta T

The mass is 15 grams and the specific heat is 0.129 J/(g×°C). Let's calculate the change in temperature.

  • ΔT= final temperature - initial temperature
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Now we know all the values:

m= 15 \ g \ \\c= 0.129 \ J / (g* \textdegree C) \\\Delta T= 63 \ \textdegree C

Substitue the values into the formula.

q= (15 \ g)( 0.129 \ J / (g* \textdegree C)) ( 63 \ \textdegree C)

Multiply the first numbers together. The grams will cancel.

q= (1.935 \ J/ \textdegree C) ( 63 \ \textdegree C)

Multiply again, this time the degrees Celsius cancels.

q= 121.905 \ J

<u>121.905 Joules</u> are required.

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