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3 years ago

An atom of arsenic has how many electron-containing orbitals

Physics

1 answer:

irina [24]3 years ago

8 0

Answer:

The Arsenic has three electron-containing orbitals. The orbitals s, p and d.

Explanation:

Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:

$1s^{2}$

$2s^{2}$

$2p^{6}$

$3s^{2}$

$3p^{6}$

$3d^{10}$

$4s^{2}$

$4p^{3}$

Therefore, the Arsenic has three electron-containing orbitals (s, p d).

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A balloon is rubbed against a sweater. Which of the following describes the result of this interaction?(1 point)

Ghella [55]

Fibers of the sweater lose electrons because electrons are leave it.

One of the ways of charging a body is by friction. Charges are transferred from one body to another when an object is rubbed against another. This is charging by friction.

A sweater has negative charges hence when the balloon is rubbed against the sweater, fibers of the sweater lose electrons because electrons are leave it.

Learn more: brainly.com/question/830809

4 0

2 years ago

Formula for the distance (d) is given by d = rate*time. For example if you are traveling at 60 mph for 3 hours the distance trav

babunello [35]

Explanation:

Distance covered by the particle is given by:

Distance (d) = rate (v) × time (t)

Speed of Mary, v₁ = 50 mph

Speed of Jim, v₂ = 60 mph

It is assumed that, Mary and Jim leave at the same time. After one hour, Jim is 10 miles ahead.

Distance travelled by Jim, d₁ = (60t + 10)

Distance travelled by Mary, d₂ = 50t

The distance between Mary and Jim is greater than or equal to 100 miles.

$60t+10-50t\ge100$

$10t\ge90$

$t\ge9\ h$

So, Jim takes is 9 hours more than Mary to cover same distance. Hence, this is the required solution.

7 0

3 years ago

A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given

hichkok12 [17]

Answer:

???

Explanation:

3 0

4 years ago

A coin is dropped in a 15.0 m deep well.

labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0

3 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$