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kaheart [24]
3 years ago
8

What is the mass of 1.00 mile of NE

Chemistry
1 answer:
sasho [114]3 years ago
3 0
If you mean moles of Ne, 1 mole of any compound of element is equal to its molecular weight in grams . In this case, you’re dealing with 1 mole of an element, so it’s mass is going to be equivalent to it’s mass in grams from the period table of elements .
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Methane reacts with oxygen to produce carbon dioxide and water,What are the products? A. methane and water. B. oxygen and carbon
EleoNora [17]
D, carbon dioxide and water. Methane and oxygen PRODUCE carbon dioxide and water in their reaction. 
3 0
3 years ago
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Please help.This is due tomorrow.It's worth 2 grades.Please help.God bless u.Please and thankyou so much.
Daniel [21]

Answer:

1. False - compression

2. True

3. False - transform faults

4. False - horizontally

5. True

6. False- perpendicular

7. False - away from

8. False - increase

9. True

10. True

Explanation:

1. Mountains, oceanic trenches, and rift valleys are created by tension and compression stress. They are formed by divergent and convergent boundaries. Compression stress occurs when plates are pushing against each other, while tension stress occurs when the plates are pulling away from each other.

**Shear stress happens when the plates grind against each other. Often found in transform boundaries.

2. Transform faults happen when two plates glide or slide against each other. These areas are called transform boundaries. Transform faults occur in the ocean. When these boundaries are formed on land, they are called strike-slip faults.

3. Shear stress that occur in transform boundaries produce transform faults. These faults are usually identified by long faults and ridges. Sometimes small ponds form in the cracks due to deposition.

*** Rift valleys are produced by divergent boundaries or tension stress, when the plates are pulled apart.

4. Transform boundaries are formed when two plates slides against each other. Transform faults are formed in these boundaries and the movement of the plates are horizontal.

*** They do not move vertically.

5-6. Mid-oceanic ridges are segmented or divided by transform faults. The transform faults in the mid-oceanic ridges are perpendicular to the oceanic ridges. They separate them into distinct segments and can run across for hundreds of kilometers

7. New faults form as they move away from the ridges. Mid oceanic ridges are formed when the plates move apart, pushing the seafloor outwards and along with that, the transform faults. When new crust however overlaps the transform fault, they stop moving against each other, and start moving side by side, creating a crack.

8. Transform faults increase in size as long as the plates continue to move. The areas of transform faults, especially in the surface create earthquake faults.

9. Faults at the surface can be part of a larger underground system. Some faults can cut across continental crusts. These faults are created by different geological processes, like compression stress from convergent boundaries, tension stress from divergent boundaries, and shear stress from transform boundaries.

10. Fault zones are areas where you can find different faults formed, relatively close to each other. The faults in fault zones can be shallow or deeper like the fault zone Sierra Madre.

6 0
3 years ago
Question 1 (1 point)
Marizza181 [45]
Both ehhevshahahbsbdvhshs
4 0
2 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
What is meant by atomic radius
Kaylis [27]

Answer:

The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons. ... The value of the radius may depend on the atom's state and context.

Explanation:

8 0
3 years ago
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