What is generally true about biomass and energy as we move from one trophic level to the next?

Strike441 [17]

Answer:

CH2 is the chemical name for methylene and belongs to the general compound category of Alkenes. It comprises of a single carbon atom bonded with 2 hydrogen atoms . The general formula for alkenes is C(n)H(2n) with n denoting the number of carbon atoms and thus deciding the name of the compound.

Explanation:

4 0

2 years ago

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According to the Kinetic Energy Theory, the more energy a particle has, the _______ it moves.

MissTica

B and temp is related to kinetic energy by how they both effect the speed of the particles. If the temp is hot the particles speed up and if cold the particles slow down

3 0

3 years ago

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In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form

erastovalidia [21]

Answer:

The time taken for the cross mark to disappear decreases steadily down the column.

Explanation:

Now if we look at the data provided, we will discover that the volume of the HCl was held constant while the volume of the thiosulphate was increased steadily and the volume of water decreased steadily.

Recall that a system is more concentrated when it contains less volume of water and more volume of reactants. Hence as the volume of water in the system is being reduced, the concentration of reactants is increased.

It has been established that an increase in the concentration of reactants lead to an increase in the rate of reaction. The disappearance of the cross shows the completion of the reaction between HCl and thiosulphate. The faster or slower the cross disappears, the faster or slower the rate of reaction.

Since increase in concentration of reactants increases the rate of reaction, it is observed that as the volume of the thiosulphate increases (reactant concentration increases) the cross disappears faster (rate of reactant increases). Hence as the volume of thiosulphate increases, it takes a shorter time for the cross to disappear. This implies that the time column in the table (refer to the question) will decrease steadily as the volume of thiosulphate increases.

5 0

3 years ago

Which of these can be described as a system of stars, gases, dust, and other matter that orbits a common center of gravity?

velikii [3]

It’s c galaxy ((: so yeah

8 0

2 years ago

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A. 14.8

Scilla [17]

Answer:

Explanation:

1)

Given data:

Initial volume = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

2)

Given data:

Initial volume = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1 mL

3)

Given data:

Initial volume = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75 L

T₂ = 720.78 L.K / 2.75 L

T₂ = 262.1 K

4)

Given data:

Initial volume = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89 L

5)

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8 L

6)

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79 mL

6 0

3 years ago