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Alenkasestr [34]
3 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]3 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]3 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 bag of apples has a weight of 22 newtons. find its approximate mass in kilograms.
Leya [2.2K]
<span>Force = mass x acceleration

F = 22 N
m = ?
a = 10 m/s²

22 = 10m

m = 22 / 10

m = 2,2 kg</span>
3 0
3 years ago
What are the three key points to seismic waves​
leonid [27]

Earthquakes produce three types of seismic waves: primary waves, secondary waves, and surface waves. Each type moves through materials differently. In addition, the waves can reflect, or bounce, off boundaries between different layers.

three key points:

1. they all change the earth by shifting plates in different ways.

2. these waves lead to our plates form today, which are always changing.

3. even the smallest earthquakes make the slightest changes to our ground. and many happen in just a day! we just never feel it.

6 0
3 years ago
Two blocks collide on a frictionless surface. After the collision, the blocks sticktogether. Block A has a mass M and is initial
jasenka [17]

Answer:

ΔPa = –2/3Mv

ΔPb = 2/3Mv

ΔPc = Mv.

Explanation:

See attachment below please.

5 0
3 years ago
HELP PLSSS I HAVE AN EXAM MONDAY AND I THINK THIS IS GONNA BE ON ITTTT
il63 [147K]
<h2>Answer:</h2>

(a) 3.2 x 10²s

(b) 0.9 m/s (S 13 E)

(c) 2.9 x 10²m

<h2>Explanation:</h2>

The sketch illustrating the scenario has been attached to this response.

As shown;

The fish swims due east with a velocity V_{x} = 0.2m/s

The river current has a velocity V_{y} due South = 0.9m/s

The resultant of the velocity is V

The width of the river is x = 64m

(a) To calculate how long it took the fish to get across the river, we know that velocity is the rate of change in distance, therefore we can use the relation;

V = \frac{d}{t}      -------------(i)

Where;

V = velocity of the fish = V_{x} = 0.2m/s

d = distance from the start to the end = width of the river = x = 64m

t = time taken to move for that distance

Make t subject of the formula in equation (i);

t = \frac{d}{V}

Substitute the values of d and V into the equation;

t = \frac{64m}{0.2m/s}

t = 320 s

t = 3.20 x 10²s

Therefore, the time taken for the fish to get across the river is 3.20 x 10²s

(b) The resulting vector of the fish is V whose magnitude is the algebraic sum of vectors  V_{x} and  V_{y}, and direction is given by θ. i.e

<em>The magnitude of the resulting vector is;</em>

|V| = \sqrt{(V_x)^2 + (V_y)^2}

|V| = \sqrt{(0.2)^2 + (0.9)^2}

|V| = \sqrt{(0.04) + (0.81)}

|V| = \sqrt{(0.85)}

|V| = 0.92m/s

|V| ≅ 0.9m/s

<em>The direction of the resulting vector θ and is given by;</em>

tan θ = \frac{V_y}{V_x}

tan θ = \frac{0.9}{0.2}

tan θ = 4.5

θ = tan⁻¹ ( 4.5)

θ = 77.47° South of East.

θ  ≈ 77.5° South of East.

Subtracting θ = 77.5° from 90° gives its value East of South

i.e

90 - 77.5 = 12.5° East of South

<em>This can also be written as S12.5°E</em>

<em>Approximating to the nearest whole number gives </em>S 13 E

Therefore, the resulting velocity of the fish is 0.9m/s in the direction S13°E

(c) When the fish arrives on the opposite bank, its distance from being at the point directly across from where it started is the product of the velocity of the river current and the time taken by the fish to get across the river. This point is equivalent to k as shown in the diagram.

Therefore;

distance = velocity of river current x time taken

distance = 0.9m/s x 3.20 x 10²s

distance = 2.88 x 10²m

distance ≅ 2.9 x 10²m

<em>Notice that the velocity of the river current is used since that's the velocity of the fish on the y-axis.</em>

<em />

<em />

7 0
2 years ago
When a boat is placed in liquid, two forces act on the boat. Gravity pulls the boat down with a force equal to the weight of the
wel

Answer:

the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

Explanation:

This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law

                  B- W = 0

                  B = W

           

the thrust force is equal to the weight of the liquid that is dislodged

                  B = ρ g V

we substitute

             ρ g V = m g

             V = m /ρ_fluid          1

we can write the mass of the pot as a function of its density

             ρ_body = m / V_body

            m = ρ_body  V_body

             V_fluid / V_body = ρ_body / ρ _fluid         2

Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

The effect appears the pot as if it had a lower apparent weight

3 0
3 years ago
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