All categories

3 years ago

Why it is easier to remove a 2p electron from an oxygen atom than from a nitrogen atom?

Physics

1 answer:

DedPeter [7]3 years ago

7 0

Answer:

The electrons in oxygen are paired while in nitrogen, they are not.

Explanation:

To analyse this we start with writing out the ground state electronic configurations for both elements.

Oxygen: 1s²2s²2p4 meaning the p subshell has the following arrangement of electrons ↑↓ ↑ ↑

Nitrogen : 1s²2s²2p³ meaning the p subshell has the following arrangement of electrons ↑ ↑ ↑

Clearly the paired electron in oxygen will be experiencing repulsion from the electron it shares an orbital with causing it to be removed easily. The electrons in nitrogen are unpaired, each orbital is singly occupied

You might be interested in

Which would be a good analogy of wave motion

inessss [21]

I think (C )
----------------

3 0

3 years ago

Read 2 more answers

A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat

Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

$m g =\dfrac{mv^2}{r}$

$9.81 =\dfrac{v^2}{0.675}$

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

where, $I =\dfrac{2}{5}mr^2$

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

$0.7 u^2 + g H = 0.7 v^2 + g(2R)$

$0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)$

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial speed is 3.35 m/s

8 0

3 years ago

5. Graph A below plots a race car's speed for 5 seconds. The car's rate of acceleration is 6 m/s^2

Georgia [21]

Answer:

The answer is below

Explanation:

We are to check if the statement is true of false. If it is false, we correct the statement.

Solution:

Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.

The x coordinate represent the time and the y coordinate velocity.

5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:

Acceleration = $\frac{24-0}{4-0}=6\ m/s^2$

This is correct.

6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:

Acceleration = $\frac{12-12}{4-0}=0\ m/s^2$

This is false.

Therefore the acceleration of graph B is 0 m/s².

8 0

3 years ago

What is the difference between a general and a specific tolerance, and how can you tell the difference on a drawing?

Gnesinka [82]

<span>During the reading of a blueprint, a general tolerance pertains to all dimensions that are not independently identified. All tolerances are +\- .030. General tolerances are mostly found in the blueprint’s set of information. Let’s say for example, a compact tolerance is required, then a specific tolerance is considered for specific areas of the blueprint. These kinds of tolerances are commonly found along the affected area.</span>

6 0

3 years ago

Object A is 71 degrees and object B is 75 degrees how will thermal energy flow

Tasya [4]

Given :

Object A is 71 degrees and object B is 75 degrees .

To Find :

How will thermal energy flow.

Solution :

We know, by law of thermodynamics thermal energy will flow from higher temperature to lower temperature.

So, in the given question energy will flow from object B from object A.

Hence, this is the required solution.

3 0

3 years ago