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kondor19780726 [428]
3 years ago
6

When heat energy is removed, two changes are observed:

Chemistry
2 answers:
Softa [21]3 years ago
3 0

Answer: A is condensation and B is freezing.

The process in which the gas changes to the liquid phase due to the removal of heat is known as condensation. Rains are formed in this way. The water vapour rises. As it raises above it loses heat and then cools down and condenses and comes down as rainfall.

The process in which the liquid changes to a solid on removal of heat is known as freezing. Formation of ice is an example. When the heat is removed at -4 degree Celsius the water freezes to ice.

nikitadnepr [17]3 years ago
3 0

Answer:

A is condensation and B is freezing.

Explanation:

Condensation is the process in which a gas changes into a liquid. Evaporation is the process in which a liquid changes into a gas. Freezing is when liquid changes to solid/solid liquid. Melting is when a solid/solid liquid changes to  liquid.

In change A, a gas changes to a liquid; this is called condensation

In change B, a liquid changes into a solid; this is called freezing.

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Explain why raising the temperature of a liquid would speed up diffusion
nikitadnepr [17]
Because diffusion<span> is the process when molecules move to lower or higher concentration, so as the molecules move faster they are going to lower or high concentration faster.</span>
3 0
3 years ago
A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori
Nitella [24]

Answer : The specific heat of metal is 0.481J/g^oC.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of metal = 129.00 g

m_2 = mass of water = 45.00 g

T_f = final temperature = 39.6^oC

T_1 = initial temperature of metal = 97.8^oC

T_2 = initial temperature of water = 20.4^oC

Now put all the given values in the above formula, we get

129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC

c_1=0.481J/g^oC

Therefore, the specific heat of metal is 0.481J/g^oC.

4 0
3 years ago
Using the trend for chlorine, create a rule for naming halogen oxyacids.
Vlada [557]
All oxayacids have cations so no need to name the cation (H+) If name of polyatomic anion ends in -ate change to -ic for acid and if it ends with -ite change to -ous for acid

example:

ion nitrate is called nitric acid
ion nitrate is called nitrous acid
7 0
3 years ago
What is the longitude of the international date line
sveta [45]

Answer:

The International Date Line passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south 

line on the Earth. It is located halfway round the world from the prime meridian—the zero degrees longitude established in Greenwich

8 0
3 years ago
A glucose solution that is prepared for a patient should have a concentration of 180 g/L. A nurse has 18 g of glucose. How
Irina-Kira [14]
<h3>Answer:</h3>

0.10 L

<h3>Explanation:</h3>

The concentration of glucose  is given as 180 g/L

The mass of glucose is 18 g

  • Concentration in g/L is calculated by dividing mass of the solute by the volume of the solution.
  • When calculating molarity on the other hand, we divide number of moles of the solute by the volume  of the solution.
  • Concentration in g/L = Mass of solute ÷ Volume

Rearranging the formula,

Volume = Mass of the solute ÷ concentration

             = 18 g ÷ 180 g/L

            = 0.10 L

Therefore, volume of water is 0.10 L

6 0
3 years ago
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