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Alenkasestr [34]
4 years ago
10

1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False

Physics
2 answers:
Studentka2010 [4]4 years ago
4 0
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is? 
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
ohaa [14]4 years ago
3 0

Answer:

1.True

2.6

3.All of the above

Explanation:

1.In a single atom,  no more than 2 electrons can occupy a single orbital.

By pauli exclusion principle, an orbital can contain a maximum of only 2 electrons, the two electrons must have opposite spins.

Therefore, it is true.

2.We know that maximum number of electrons in p sublevel is 6.

Hence, the maximum number of electrons allowed in a p sublevel of the 3rd principle level is 6.

3.A neutral atom has a ground state electronic configuration of 1s^2 2s^2.

Therefore, Total number of electrons=4

Atomic number=4

Number of protons=Atomic number=4

The atom has  total number of orbital is 2.

The element has no unpaired electrons.

Hence, all of the above is correct option.

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A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t
tresset_1 [31]

Answer:

Part a)

Reading = 2.00 kg

Part b)

Reading = 2.00 kg

Part c)

Reading = 4.04 kg

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part b)

When elevator is decending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

F_{net} = ma

T - mg = ma

T = mg + ma

Reading is given as

Reading = \frac{mg + ma}{g}

Reading = 2.00\frac{9.81 + 10}{9.81}

Reading = 4.04 kg

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0
3 years ago
a wall of glass 2cm thickhas inside temperature of 30°C,outside temperature of15°C.how much heat is flowing through the glass(k=
kenny6666 [7]

Answer:46.5

Explanation:is the topical formula of F= 30x+15

5 0
4 years ago
Maria rides a bicycle at a velocity of 8 m sl. She brakes suddenly and stops after a distance of 2 m. What is the acceleration o
Svetach [21]

But I can help you if you need to answer this question

Dony

8 0
2 years ago
Serena drops a 5-kg book down the stairwell. What is the book's final velocity vf after it falls
elena55 [62]

Answer:

39.2m/s

Explanation:

Given parameters:

Mass of book  = 5kg

Time taken for fall  = 4s

Unknown:

Final velocity of the book  = ?

Solution:

Serena dropped the book from rest therefore, the initial velocity of the book is 0.

 Let us find the appropriate motion equation to solve this problem;

        V  = U + gt

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity  = 9.8m/s²

t is the time taken

  Insert the given parameters and solve;

       V = 0 + 9.8 x 4  = 39.2m/s

3 0
3 years ago
In the diagram, 91, 92, and q3 are in a straight line. Each of these particles has a charge of -2.35 x 10-6 C. Particles q₁ and
Strike441 [17]

The net force on particle q₃ is  6.2128125 N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/d²

where k = 9 x 10⁹ N.m²/C²

Given is the diagram in which each of the particles has a charge of -2.35 x 10⁻⁶ C. Particles q₁ and q₂ are separated by 0.100 m and particles q₂ and q₃ are separated by 0.100 m.

Force acting on q₃ due to q₁

F₃₁  = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1)²

F₃₁ = 4.97025 N(in right direction)

Force acting on q₃ due to q₂

F₃₂ = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1+0.1)²

F₃₂ = 1.2425 N (in right direction)

Net force acting on particle q₃ is

F₃ = F₃₁ +F₃₂

F₃ = 4.97025 N + 1.2425 N

F₃ = 6.2128125 N

Thus the net force on a charged particle is  6.2128125 N to the right.

Learn more about electrostatic force.

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3 0
2 years ago
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