Which of the following elements is NOT involved in the cycling of energy and matter on Earth? O Phosphorus O Gold O Nitrogen O Carbon ​

Which Of The Following Elements Is NOT Involved In The Cycling Of Energy And Matter On Earth? O Phosphorus

Answers

Answer 1

Gold is not involved in the cycling of energy and matter on Earth.

option B.

What is Biogeochemical cycles?

Biogeochemical cycles refer to the pathways by which essential elements such as carbon, nitrogen, and phosphorus move through the Earth's biosphere, atmosphere, geosphere, and hydrosphere. These cycles are crucial for maintaining the balance of elements required for life on Earth, and they involve both biotic and abiotic processes.

The carbon cycle involves the movement of carbon through the atmosphere, the biosphere, and the oceans, with processes such as photosynthesis, respiration, and decomposition contributing to carbon uptake and release. Carbon is also stored in rocks and sediments as fossil fuels.

The other three elements listed - phosphorus, nitrogen, and carbon - are all essential elements involved in biogeochemical cycles on Earth, including the phosphorus cycle, nitrogen cycle, and carbon cycle. Gold, on the other hand, is a relatively inert element and does not play a significant role in the cycling of energy and matter on Earth.

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Related Questions

Sodium and Lithium have similar chemical properties. What characteristic of these elements explains why they are chemically similar?

F. Their atoms both have 1 valence electron

G. Their atoms both have more neutrons than protons

H. Their atoms have the same number of energy levels

J. Their atoms contain equal numbers of protons and electrons


I am not in college...

Answers

These elements' shared possession of a single valence electron in their atoms explains why they are chemically related. As a result, F is the right response.

An atom's valence shell electrons control how it interacts with its neighbours and, consequently, its chemical characteristics. Lithium and sodium both contain one valence electron, hence they have comparable chemical characteristics. Elements from this family include lithium, sodium, potassium, rubidium, cesium, and francium (Li, Na, K, Rb, Cs, and Fr, respectively). The components of Group 1 have similar traits. Each of them is a delicate silver metal. These metals have low melting temperatures and are extremely reactive because of their low ionisation energy.

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The beach in Antibes, France, is composed of small, smooth rocks that have been worn
down by being tumbled against each other by the energy of the sea water
Physical or Chemical?
Specific type of physical or chemical

Answers

The beach at Antibes, France, is made mostly of small, smooth rocks that have been worn down by being pushed against one another by the energy of the sea water.

What exactly is erosion?

The geological process of erosion occurs when earthen materials are pushed and worn away by natural forces such as wind or water. Although it does not necessitate movement, a similar process known as weathering dissolves or disintegrates rock.

Weathering occurs when rocks and minerals on the Earth's surface disintegrate or dissolve. Weathering is caused by a variety of factors such as water, ice, acids, salts, plants, animals, and temperature swings. After a rock has been broken down, a process known as erosion transports the shards of rock and mineral away. The Earth's rocks are all too soft to endure weathering and erosion.

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PLEASE HELPPPPP I WILL GIVE A BRAINLIST PLEASE


Introduction (6 points)
1. What do you think will happen when the balloon is released? (2 points)



2. What caused the balloon to move? (2 points)




3. How do you think Newton's third law of motion applies to the motion of a rocket? Hint: Think about what happened to the balloon when it was released. (2 points)





Part 1: Rocket Cars (10 points)
4. Describe the action-reaction force pair that occurs when the air in the balloon is allowed to escape. (2 points)





5. Collect data during the experiment with the rocket car. In the data table below, record the amount of air in the balloon during each test run and the distance traveled by the car. (4 points)
Test run Amount of air Distance traveled (cm)
1 Small
2
3
4
6. What relationship do you observe between the amount of air in the balloon, the size of the action-reaction force pair, and the distance traveled? (4 points)




Part 2: Car Collisions (14 points)
7. In the data table below, record the force applied to car 1 during each test run and the distance traveled by car 2 after the collision. Hint: Distance traveled by car 2 = final position of car 2 – 20 cm. (4 points)
Test run Force applied to car 1 (N) Final position of car 2 (cm) Distance traveled by car 2 (cm) Description of change in motion of car 1
1
2
3
8. How did the action and reaction forces affect the motion of each car? (4 points)



9. What is the relationship between the force of the collision and the distance traveled by car 2? Which of Newton's laws of motion explains this? (2 points)



10. The formula F = ma describes how force, mass, and acceleration are related. Explain what the formula means. How could you increase the size of the action force an object exerts and the size of the reaction force it experiences during a collision? (4 points)





Part 3: Car and Brick Collisions (10 points)
11. Describe the strengths and directions of the action and reaction forces that occur when a moving car collides with a stationary brick. How do these forces explain the changes in motion of the car and brick? (2 points)








12. What happened when the car collided with the brick with twice as much force? How does Newton's third law of motion explain this? (2 points)








13. How could you change the test to make the brick move even more? (2 point)




14. What happened when the car with the sponge bumper hit the brick with the same amount of force as the car without the bumper? (1 point)




15. Why did the car with the sponge bumper rebound farther after its collision with the brick? (1 point)





16. A rocket moving through space at a high velocity can collide with small objects, such as meteors or space junk. If that happens, the rocket may be damaged, destroyed, or pushed off course. Based on what you just observed, what do you think could be done to help reduce the effect of a collision between a rocket and space junk? (2 points)

Answers

1) Gas escapes to equalize pressure inside and outside of balloon.

2) Pressure pushes gases out of the balloon when released.

3) The gas pushes against the outside air, pushing it back.

What are gas balloons called?

For its creator, Frenchman Jacques Charles, a gas balloon may also be referred to as a Charlière. Large blimps and tiny latex celebration balloons are two common types of gas balloons today. Before hot-air balloons took over, human balloon flying used gas balloons for almost 200 years, well into the 20th century.

Fill your balloon with two teaspoons of baking soda, and fill the container with half a cup of acetic vinegar. Avoid using too much cola! The baking soda in the balloon will descend into the container and mix with the vinegar when the bottleneck is inserted into the balloon's neck and the balloon is straightened.

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a physician prescribes an ophthalmic suspension to contain 100mg of cortisone acetate in 8 ml of normal saline solution. the pharmacist has on hand a 2.5% suspension of cortisone acetate in normal saline solution. how many milliliters of this and how many milliliters of normal saline solution should be used in preparing the prescribed suspension?
the answer is 4ml and 4ml but i don't know how you get that?

Answers

2 mL οf 2.5% cοrtisοne acetate suspensiοn and 6 mL οf nοrmal saline sοlutiοn shοuld be used tο prepare the prescribed suspensiοn.

Hοw much οf 2.5% suspensiοn and nοrmal saline sοlutiοn shοuld be used?

- Cοrtisοne acetate is required in 100mg

- 2.5% suspensiοn οf cοrtisοne acetate in nοrmal saline sοlutiοn is available

- Cοncentratiοn οf cοrtisοne acetate in 2.5% suspensiοn is 25mg/mL

- Thus, 4mL οf 2.5% suspensiοn is required tο get 100mg οf cοrtisοne acetate

- Thus, 4mL οf 2.5% suspensiοn and 4mL οf nοrmal saline sοlutiοn tοtal 8mL is required tο make the prescribed suspensiοn

- But, 2.5% suspensiοn already cοntains nοrmal saline sοlutiοn

- Sο, final requirement οf 2.5% suspensiοn and nοrmal saline sοlutiοn is 2mL and 6mL respectively

2 mL οf 2.5% cοrtisοne acetate suspensiοn and 6 mL οf nοrmal saline sοlutiοn shοuld be used tο prepare the prescribed suspensiοn.

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The phase diagram of ammonia.

Answers

Explanation:

Ammonia is a gas,

however at low temperature or high pressure the gas become liquid and at extreme conditions,a solid

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What is the density of a sample of argon gas at 69 ∘C
and 768 mmHg
?

Answers

The density of the sample of argon gas at 69 °C and 768 mmHg is 1.44 g/L (1st option)

How do i determine the density of the gas?  

From the question given above, the following data were obtained:

Temperature of gas (T) = 69 °C = 69 + 273 = 342KPressure of gas (P) = 768 mmHg = 768 / 760 =  1.01 atmGas constant (R) = 0.0821 atm.L/Kmol Molar mass of argon (M) = 40 g/molDensity of gas (D) = ?

The density of the sample of argon gas can be obtained as illustrated below:

D = MP / RT

D = (40 × 1.01)  / (0.0821 × 342)

D = 1.44 g/L

Thus, from the above calculation, we can conclude that the density of the gas is 1.44 g/L (1st option)

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Hi, I need help with this please.

Answers

The final temperature of the water when thermal equilibrium is reached is approximately 24.36°C.

What is the final temperature of the water?

The principle of conservation of energy which states that the heat lost by the metal cubes must be equal to the heat gained by the water is used to determine the final temperature of the water.

The formula to calculate the heat lost by a metal cube is:

Q = m × c × ΔT

where;

Q is the heat lost,m is the mass of the cube,c is the specific heat capacity of the metal, and ΔT is the change in temperature of the metal.

For the silver cube:

m = density × volume

m = 10.49 g/cm³ × (2.31 cm)³

m = 58.48 g

c = 0.235 J/g°C (specific heat capacity of silver)

ΔT = (82.2°C - 19.6°C)

ΔT = 62.6°C

Q = 58.48 g × 0.235 J/g°C × 62.6°C

Q = 877.4 J

For the gold cube:

m = density × volume

m = 19.32 g/cm³ × (2.78 cm)³

m = 170.5 g

c = 0.129 J/g°C (specific heat capacity of gold)

ΔT = (82.2°C - 19.6°C)

ΔT  = 62.6°C

Q = 170.5 g × 0.129 J/g°C × 62.6°C = 1354.4 J

The total heat lost by both cubes is:

Qtotal = 877.4 J + 1354.4 J

Qtotal= 2231.8 J

To calculate the final temperature of the water, we can use the formula:

Q = mwater × cwater × ΔT

where;

mwater is the mass of water,cwater is the specific heat capacity of water, and ΔT is the change in temperature of the water.

Solving for ΔT:

ΔT = Qtotal / (mwater × cwater)

ΔT = 2231.8 J / (113.0 g × 4.184 J/g°C)

ΔT = 4.76°C

Final temperature = 19.6°C + 4.76°C

Final temperature= 24.36°C

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