Which configuration is common to the outer shells of the elements neon, argon, and krypton?sp8
s2dp6
s2p6
s2p4

Answers

Answer 1

Answer: Answer: S2P6

The electronic configuraiton of given elements are as follows:

Ne =[He]2s2 2p6
Ar = [Ne]3s2 3p6
Kr = [Ar]4s2 3d10 4p6

As it can be seen that for all three elements, their outermost orbital are completely filled, that is it has both s orbital, p orbital and d orbital fulfilled. Noble or inert gas atoms like Neon, Argon, Krypton have fulfilled valence shell. Fulfilled outermost orbital is the most stable electronic state, hence all elements tends to achieve such stability. These noble gas elements are called inert gas because of their fulfilled outermost shell. This means they don't react easily or take part in eletron donating, receiving or sharing. This is because, for all other elements except inert gas atoms, their valence shell is incomplete and they tend to react by other atoms so as to complete their outermost shell , which we call as duplet (in case of Helium like) or Octet state. Such elements either donate some electrons or receive some to acheive such stable state..

Related Questions

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find the initial speed of the rock.a. 3m/sb. 30.3 m/sc. None of the above
A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude
A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground first?
Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?
A 55-liter tank is full and contains 40kg of fuel. Find using Sl units: • Density p. • Specific Weight y • Specific Gravity Answer tolerance = 1%. Be sure to include units. The sign of the answers will not be graded, use a positive value for your answer. Your answers: p= (Enter a positive value) y = (Enter a positive value) SG = (Enter a positive value)

A satellite travels around the Earth in a circular orbit. What is true about the forces acting in this situation? A. The resultant force is the same direction as the satellite’s acceleration. B. The gravitational force acting on the satellite is negligible. C. There is no resultant force on the satellite relative to the Earth. D. The satellite does not exert any force on the Earth.

Answers

Answer:

A. The resultant force in the same direction as the satellite’s acceleration.

Explanation:

Launching a satellite in the space and then placing it in orbit around the Earth is a complicated process but at the very basic level it works on simple principles. Gravitational force pulls the satellite towards Earth whereas it acceleration pushes it in straight line.

The resultant force of gravity and acceleration makes the satellite remain in orbit around the Earth. It is condition of free fall where the gravity is making the satellite fall towards Earth but the acceleration doesn't allow it and keeps it in orbit.

Final answer:

In a circular orbit around the Earth, the resultant force acting on a satellite is in the same direction as its acceleration.

Explanation:

In a satellite orbiting the Earth in a circular orbit, there are several forces at play. The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in its orbit. The centripetal force acts towards the center of the circular orbit, while the satellite's acceleration is directed towards the center as well. Therefore, option A is correct: the resultant force is in the same direction as the satellite's acceleration.

The gravitational force acting on the satellite is not negligible; in fact, it is crucial in providing the necessary centripetal force. Therefore, option B is incorrect.

Option C is incorrect as well. There is a resultant force acting on the satellite relative to the Earth, which is responsible for keeping the satellite in its circular orbit.

Lastly, option D is also incorrect. According to Newton's third law of motion, the satellite exerts an equal and opposite force on the Earth, keeping the Earth and the satellite in orbit around their common center of mass.

Learn more about Satellites in Circular Orbit here:

brainly.com/question/36010275

#SPJ11

A 1500 kg car moving at 25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of 2.0 × 106 N/m. What is the maximum distance the spring compresses?

Answers

Answer:

x = 0.68 meters

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Speed of the car, v = 25 m/s

Spring constant of the spring, $k=2* 10^6\ N/m$

When the car hits the uncompressed horizontal ideal spring the kinetic energy of the car is converted to the potential energy of the spring. Let x is the maximum distance compressed by the spring such that,

$(1)/(2)mv^2=(1)/(2)kx^2$

$x=\sqrt{(mv^2)/(k)}$

$x=\sqrt{(1500* (25)^2)/(2* 10^6)}$

x = 0.68 meters

So, the spring is compressed by a distance of 0.68 meters. Hence, this is the required solution.

Final answer:

The maximum distance the spring compresses when a 1500 kg car moving at 25 m/s hits it, given a spring constant of 2.0 × 10⁶N/m, is approximately 0.53 meters or 53 centimeters.

Explanation:

In this specific problem, we can apply the conservation of energy principle, where the initial kinetic energy of the car is converted into potential energy stored in the spring when the car comes to a stop. The formula for kinetic energy is K = 1/2 × m× v² and for potential energy stored in a spring is U = 1/2×k × x², where m = mass of the car, v = velocity of the car, k = spring constant, and x = maximum distance the spring is compressed.

By setting the kinetic energy equal to potential energy (since no energy is lost), we get 1/2 × m×v² = 1/2×k×x². Solving this equation for x (maximum compression of the spring), we obtain x = sqrt((m×v²)/k). Substituting the given values, x = sqrt((1500 kg× (25 m/s)²) / (2.0 × 10⁶ N/m)), which yields approximately 0.53 meters or 53 centimeters. Therefore, the maximum distance the spring compresses is 53 cm.

Learn more about Conservation of Energy here:

brainly.com/question/35373077

#SPJ3

A diverging lens has a focal length of 23.9 cm. An object 2.1 cm in height is placed 100 cm in front of the lens. Locate the position of the image. Answer in units of cm. 007 (part 2 of 3) 10.0 points What is the magnification? 008 (part 3 of 3) 10.0 points Find the height of the image. Answer in units of cm.

Answers

Answer:

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance = 100 cm

v = Image distance

f = Focal length = -23.9 cm (concave lens)

$h_u$ = Object height = 2.1 cm

Lens Equation

$(1)/(f)=(1)/(u)+(1)/(v)\n\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\n\Rightarrow (1)/(v)=(1)/(-23.9)-(1)/(100)\n\Rightarrow (1)/(v)=(-1239)/(23900) \n\Rightarrow v=(-23900)/(1239)=-19.29\ cm$

Image is virtual and formed on the same side as the object, 19.29 cm from the lens.

Magnification

$m=-(v)/(u)\n\Rightarrow m=-(-19.29)/(100)\n\Rightarrow m=0.1929$

$m=(h_v)/(h_u)\n\Rightarrow 0.1929=(h_v)/(2.1)\n\Rightarrow h_v=0.1929* 2.1=0.40509\ cm$

The height of the image is 0.40509 cm

Image is upright as the magnification is positive and smaller than the object.

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?

Answers

Answer:

The change in momentum is $\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A = <5600, 7200,0>$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = <-4000 , -780 , 0> N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)$

$\Delta p = <-4000 , -780 , 0> * (0.3)$

$\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
OD. 23,219 ft

Answers

I believed the answer is d

Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?