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satela [25.4K]
2 years ago
9

Sophia was fascinated while studying the role of oceans in the hydrosphere. She decided to illustrate the features of the ocean

floor on a poster for her part in a group project. Identify and describe the kinds of features found on the ocean floor.
Chemistry
1 answer:
ANEK [815]2 years ago
7 0

Answer:

The Major features of the ocean floor are:

  1. Continental Shelf
  2. Continental Slope
  3. Continental Rise
  4. Abyssal Plain
  5. Oceanic Trench
  6. Mid-Ocean Ridge

Explanation:

1. Continental Shelf: This refers to the part of the land on every continent that is covered with water that is not too deep. The types of animals that can be found on the continental shelf are:

Crab, Tuna, Lobster, Dungeness cod, etc. Within the Continental shelf, there are permanent rocks that house other organisms such as sponges, anemones, clams, sponges, oysters. The continental shelf also contains the route of migration for bigger animals such as sea turtles dolphins and even whales.

2. Continental Slope: This spans from the shelf break to the continental rise.  It can slope up to 4 degrees. Slopes can be created by faulting, slumping of huge boulders of sediments, rifting, etc.

Some of the aquatic animals that can be found in this region include but are not limited to:

Sablefish, Dover sole rockfish, etc.

3. Continental rise

This part of the ocean floor usually has a very steep gradient or angle slope. It slopes very steeply into the abyssal plain of the ocean.

The following can help form continental rise:

  • Mass wasting;
  • deposition from contour currents and
  • the longitudinal settling of biogenic and clastic particles

4. Abyssal Plain.

This is the real bottom of the ocean. There is a very high probability that one would find animals such as nematodes, polychaetes, etc which are all types of worms down there. The Abyssal plain is also home to molluscs,  and echinoderms.

5. Oceanic Trench

Sometimes there is a long and narrow indenture or depression along the seafloor. These are called Trenches. Trenches are sometimes formed by the boundaries between one lithospheric plate and another. The deepest trench on earth is found in the Pacific Ocean. It has been nick-named the Challenger Deep and said to be the deepest point known on earth reaching almost 11 kilometers.

6. Mid-Ocean Ridge

This is a mountain range underneath the ocean. It is formed when there is an upward push by convection currents of the mantle beneath the oceanic crust. When this happens and molten magma is ejected or created at the boundary between the plates, the result is a Mid-Oceanic Ridge.

Cheers

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netineya [11]
B. Precipitates are formed when two solutions combine to form another solution and a solid.
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A student is asked to standardize a solution of potassium hydroxide. He weighs out 1.08 g potassium hydrogen phthalate (KHC8H4O4
natka813 [3]

Answer:

A. 0.143 M

B. 0.0523 M

Explanation:

A.

Let's consider the neutralization reaction between potassium hydroxide and potassium hydrogen phthalate (KHP).

KOH + KHC₈H₄O₄ → H₂O + K₂C₈H₄O₄

The molar mass of KHP is 204.22 g/mol. The moles corresponding to 1.08 g are:

1.08 g × (1 mol/204.22 g) = 5.28 × 10⁻³ mol

The molar ratio of KOH to KHC₈H₄O₄ is 1:1. The reacting moles of KOH are 5.28 × 10⁻³ moles.

5.28 × 10⁻³ moles of KOH occupy a volume of 36.8 mL. The molarity of the KOH solution is:

M = 5.28 × 10⁻³ mol / 0.0368 L = 0.143 M

B.

Let's consider the neutralization of potassium hydroxide and perchloric acid.

KOH + HClO₄ → KClO₄ + H₂O

When the molar ratio of acid (A) to base (B) is 1:1, we can use the following expression.

M_{A} \times V_{A} = M_{B} \times V_{B}\\M_{A} = \frac{M_{B} \times V_{B}}{V_{A}} \\M_{A} = \frac{0.143 M \times 10.1mL}{27.6mL}\\M_{A} =0.0523 M

8 0
3 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
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How many moles are contained in 2.0 L of N2 at standard temperature and pressure.
GuDViN [60]

0.091 moles are contained in 2.0 L of N2 at standard temperature and pressure.

Explanation:

Data given:

volume of the nitrogen gas = 2 litres

Standard temperature = 273 K

Standard pressure = 1 atm

number of moles =?

R (gas constant) = 0.08201 L atm/mole K

Assuming nitrogen to be an ideal gas at STP, we will use Ideal Gas law

PV = nRT

rearranging the  equation to calculate number of moles:

PV = nRT

n = \frac{PV}{RT}

putting the values in the equation:

n = \frac{1X2}{0.08201 X 273}

n = 0.091 moles

0.091 moles of nitrogen gas is contained in a container at STP.

6 0
2 years ago
Water is dripping into a bucket at a rate of 51 drops per minute. if each drop has a volume of 0.050 milliliters, what volume of
serious [3.7K]

112.2 milliliters volume of water will drip into the bucket in 44 minutes.

<h3>What is volume?</h3>

How much space an object or substance takes up. • Measured in cubic meters (m3), liters (L) & milliliters (mL).

Total drop = Drops per minute X time

= 51 drops per minute X 44 minutes

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Volume of water will drip into the bucket in 44 minutes

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Hence, 112.2 milliliters volume of water will drip into the bucket in 44 minutes.

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8 0
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