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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

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A scientist heats a flexible container full of neon gas. What will most likely happen to the container as the gas absorbs heat?

Volgvan

It will most likely get hotter and hotter and eventually exsplod from all the presure.

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3 years ago

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Consider an embedded system which uses a battery with a 17.15-Amp-Hour capacity. What is the maximum average current draw (in mi

Marianna [84]

Answer:

85.12 μAmp

Explanation:

The battery power output = 17.15 Amp-hr

If the battery is to last 23 years, we have to calculate how many hours there are in 23 years

in one year there are 24 hours x 365 day = 8760 hrs

in 23 years there are 23 x 8760 = 201480 hours

maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = 85.12 μA

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3 years ago

How much work is done on a small car if a 3150 N force is exerted to move it 75.5 m to the side of the road

irinina [24]

Answer:

Explanation:

Work = Force times displacement. Therefore,

W = 3150(75.5) so

W = 238000 N*m

6 0

2 years ago

A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power.

tatyana61 [14]

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

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3 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

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3 years ago