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3 years ago

What is an example of a Low Energy-created depositional environment?

Chemistry

1 answer:

sweet-ann [11.9K]3 years ago

4 0

Answer:

River flood plains, swamps, lakes, lagoons, marshes, and offshore.

Explanation:

Slow-moving currents prevent coarse-grained sediment from migrating into low-energy depositional environments. Fine materials can be carried long distances before they can settle out in the absence of waves and currents.

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List earth and the gas giant planets from the hottest to the coldest planet

inessss [21]

Answer:

1. Venus

471°C

2. Mercury

(430°C) during the day, (-180°C) at night

3. Earth

16°C

4. Mars

-28°C

5. Jupiter

-108°C

6. Saturn

-138°C

7. Uranus

-195°C

8. Neptune

-201°C

Explanation:

8 0

2 years ago

ANSWER: b i hope i helped

Sergio [31]

Answer:

thanks

Explanation:

7 0

3 years ago

Read 2 more answers

Which describes Ernest Rutherford’s experiment?

wel

Rutherford's Gold Foil Experiment proved the existance of a small massive center to atoms, which would later be known as the nucleus of an atom. Ernest Rutherford, Hans Geiger and Ernest Marsden carried out their Gold FoilExperiment to observe the effect of alpha particles on matter.

heres the link if you need more information

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjjre2m_ovN....

7 0

4 years ago

On a hot summer day you and some friends decide you want to cool down your pool. Determine the mass of ice you would need to add

sergeinik [125]

Answer: The mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is $16.14 \times 10^{4}$ kg.

Explanation:

We know that relation between heat energy and specific heat is as follows.

q = $m \times S \times \Delta T$

As density of water is 1 kg/L and volume is given as 400,000 L. Therefore, mass of water is as follows.

Mass of water = Volume × Density

= $400,000 L \times 1 kg/L$

= 400,000 kg

or, = $400,000 \times 10^{3}$ g (as 1 kg = 1000 g)

Specific heat of water is 4.2 J/gm K. Therefore, change in temperature is as follows.

$\Delta T$ = 305 K - 273 K

= 32 K

Now, putting the given values into the above formula and calculate the heat energy as follows.

q = $m \times S \times \Delta T$

= $400,000 \times 10^{3} \times 4.2 \times 32 K$

= $5376 \times 10^{7}$ J

or, = $5376 \times 10^{4}$ kJ

According to the enthalpy of melting of ice 333 kJ/Kg of energy absorbed by by 1 kg of ice. Hence, mass required to absorb energy of $5376 \times 10^{4}$ kJ is calculated as follows.

Mass = $\frac{5376 \times 10^{4} kJ}{333 kJ/Kg} \times 1 kg$

= $16.14 \times 10^{4}$ kg

Thus, we can conclude that the mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is $16.14 \times 10^{4}$ kg.

3 0

3 years ago

Determine the number of miles of H in each sample;

Inessa [10]

1) Answer is: there are 9.04 moles of the hydrogen in 4.52 mol of the water.

n(H₂O) = 4.52 mol; amount of the water molecules.

In one molecule of the water, there are two hydrogen atoms.

n(H₂O) : n(H) = 1 : 2.

n(H) = 2 · n(H₂O).

n(H) = 2 · 4.52 mol.

n(H) = 9.04 mol; amount of the hydrogen atoms.

2) Answer is: there are 25.29 moles of the hydrogen in 8.43 mol of the ammonia (NH₃).

n(NH₃) = 8.43 mol; amount of the ammonia molecules.

In one molecule of the ammonia, there are three hydrogen atoms.

n(NH₃) : n(H) = 1 : 3.

n(H) = 3 · n(NH₃).

n(H) = 3 · 8.43 mol.

n(H) = 25.29 mol; amount of the hydrogen atoms.

3) Answer is: there are 0.408 moles of the hydrogen in 0.102 mol of the hydrazine (N₂H₄).

n(N₂H₄) = 0.102 mol; amount of the hydrazine molecules.

In one molecule of the hydrazine, there are four hydrogen atoms.

n(N₂H₄) : n(H) = 1 : 4.

n(H) = 4 · n(N₂H₄).

n(H) = 4 · 0.102 mol.

n(H) = 0.408 mol; amount of the hydrogen atoms.

4) Answer is: there are 770 moles of the hydrogen in 35.0 moles of the decane (C₁₀H₂₂).

n(C₁₀H₂₂) = 35.0 mol; amount of the decane molecules.

In one molecule of the decane, there are twenty two hydrogen atoms.

n(C₁₀H₂₂) : n(H) = 1 : 22.

n(H) = 22 · n(C₁₀H₂₂).

n(H) = 22 · 35.0 mol.

n(H) = 770 mol; amount of the hydrogen atoms.

6 0

4 years ago