1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
zmey [24]
3 years ago
9

Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the

element with the most negative electron affinity, E ea E_ea ?
2s2
2s2 2p2
2s2 2p5
2s2 2p6
Physics
2 answers:
kramer3 years ago
4 0

Answer:

2s2 2p6.

Explanation:

Electron affinity is a term used to measure of the attraction between the incoming electron and the nucleus - the stronger the attraction, the more energy is released. The factors which affect this attraction are: nuclear charge, distance, electronic configuration and screening.

Comparing the following electronic configuration of their outermost shell:

2s2

2s2 2p2

2s2 2p5

2s2 2p6

Generally, Eae increases across the period(more positive), where the noble gases are the most negative. This is because of their fully filled shells.

Therefore, 2s2 2p6 has fully filled orbitals, therefore it will have the highest negative electron affinity.

grandymaker [24]3 years ago
3 0

Answer:

he configuration with the highest electronic affinity is 2s2 2p5

Explanation:

Electronic affinity is the variation of energy when we add an electron to a neutral atom to form an ion

When an electron is added, it must occupy a space is the sub-level of the atom, giving more stability when it approaches the configuration of a complete shell with eight electrons (noble gas), so the affinity must increase when moving in a period Group VIII noble gases)

Let's examine the given settings

In this case, when adding an electron, 2s2 is very far from a complete level configuration, so its affinity must be small.

2s2 2p2 when adding an electro the one has a little more affinity, but is still a long way from a full shell, it would be missing 3 electrons

2s2 2sp5 this is the atom with the highest electronic affinity, since i = that when adding an electron the ion has the configuration of a noble gas. This is the most stable on the list

2s2 2p6 already has a full shell making it very difficult to insert an electron into this atom.

In summary, the configuration with the highest electronic affinity is 2s2 2p5

You might be interested in
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in
zhannawk [14.2K]

Answer:

a. F = 245 Newton.

b. Workdone = 392 Joules.

c. Power = 196 Watts

Explanation:

Given the following data;

Mass = 25kg

Distance = 1.6m

Time = 2secs

a. To find the force needed to lift the mass (in N );

Force = mass * acceleration

We know that acceleration due to gravity is equal to 9.8

F = 25*9.8

F = 245N

b. To find the work done by the student (in J);

Workdone = force * distance

Workdone = 245 * 1.6

Workdone = 392 Joules.

c. To find the power exerted by the student (in W);

Power = workdone/time

Power = 392/2

Power = 196 Watts.

5 0
2 years ago
The gravitional energy of an object is always measured realative to the ?
zaharov [31]
I would say that the answer would be MASS.
4 0
2 years ago
Read 2 more answers
Is 5g greater than 20.43 mg
worty [1.4K]

yes that's true 6g is larger

4 0
3 years ago
Read 2 more answers
From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s
Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

8 0
3 years ago
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a
frosja888 [35]

Answer:

a) Yes

b) 7 rad/s

c) 0.01034 J

Explanation:

a)

Yes the angular momentum of the block is conserved since the net torque on the block is zero.

b)

m = mass of the block = 0.0250 kg

w₀ = initial angular speed before puling the cord = 1.75 rad/s

r₀ = initial radius before puling the cord = 0.3 m

w = final angular speed after puling the cord = ?

r = final  radius after puling the cord = 0.15 m

Using conservation of angular momentum

m r₀² w₀ = m r² w

r₀² w₀ = r² w

(0.3)² (1.75) = (0.15)² w

w = 7 rad/s

c)

Change in kinetic energy is given as

ΔKE = (0.5) (m r² w² - m r₀² w₀²)

ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)

ΔKE = 0.01034 J

6 0
3 years ago
Other questions:
  • What is the difference between a direct current and an alternating current?
    13·2 answers
  • I NEED HELP ASAP!!!
    9·1 answer
  • Why are some scientists skeptical of the scientists results and conclusions
    6·1 answer
  • Jkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
    15·1 answer
  • a student sees a newspaper ad for an apartment that has 1330 square feet of floor space how many square meters are there?
    9·1 answer
  • Witch of the following is an example of a pull
    15·1 answer
  • What is the constraint forces
    7·1 answer
  • PLEASE HELP WITH THE 6 FOLLOWING SCIENCE QUESTIONS (the topic is circuit symbols and equations):
    5·1 answer
  • Write the examples of unit​
    11·1 answer
  • An unknown element shiny in color, malleable, conducts electricity, and is in group 2 on the periodic table is a non-metal.
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!