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Alja [10]
4 years ago
13

Suppose you have been hired to mine a Manganese nodules off the bottom of the Pacific Ocean. Identify the problems you might hav

e, and design equipment that would enable you to get the nodules to the surface.
Chemistry
1 answer:
stepan [7]4 years ago
3 0

Answer:

The Deeper these nodules are, the harder it becomes. To use robotic submersibles eliminates the need for oxygen for people, but they would need an especially designed loader or mechanical “hands” to gather, and Balloons could Raise the mined Mn to the surface for collection.

Explanation:

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S_A_V [24]
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hope this helps 
comment me if you have any questions 


5 0
3 years ago
Which of the following correctly describes solid, liquid, and gas states of matter?
Alenkinab [10]

Explanation:

In solids, the particles are tightly packed together. In liquids, the particles have more movement, while in gases, they are spread out. Particles in chemistry can be atoms, ions or molecules. It is important to understand the particle nature of matter.

4 0
4 years ago
What is the mass in grams of 1.00*10^12 lead atoms
pav-90 [236]
34.5 X 10^-11 grams of lead
5 0
3 years ago
A force has to have which two Factors
rjkz [21]
Hello!
The answer is D Magnitude and Direction.
Hope This Helps! :)
4 0
3 years ago
The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample o
adoni [48]

Answer:

550 m/s

Explanation:

The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.

v = \sqrt{\frac{3 \times R \times T}{M} }

where,

  • R: ideal gas constant
  • T: absolute temperature
  • M: molar mass of the gas

We can use the info of argon to calculate the temperature for both samples.

T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K

Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.

v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times  10^{5}K }{20.18g/mol} } = 550 m/s

7 0
3 years ago
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