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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

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A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an
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The radius of the curved road at the given condition is 54.1 m.

The given parameters:

  • <em>mass of the car, m = 1000 kg</em>
  • <em>speed of the car, v = 50 km/h = 13.89 m/s</em>
  • <em>banking angle, θ = 20⁰</em>

The normal force on the car due to banking curve is calculated as follows;

Fcos(\theta) = mg

The horizontal force on the car due to the banking curve is calculated as follows;

Fsin(\theta) = \frac{mv^2}{r}

<em>Divide </em><em>the second equation by the first;</em>

\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

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2 years ago
A ball is dropped off a cliff and falls for 24 seconds. How far did the ball fall from the top down
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A gasoline engine receives 200 J of energy from combustion and loses 150 J as heat to the exhaust. What is its efficiency? 75% 2
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Two resistors have resistances R(smaller) and R(larger), where R(smaller) &lt; R(larger). When the resistors are connected in se
ASHA 777 [7]

Answer:

R = 9.85 ohm , r = 0.85 ohm

Explanation:

Let the two resistances by r and R.

when they are connected in series:

V = 12 V

i = 1.12 A

The equivalent resistance when they are connected in series is

Rs = r + R

So, By using Ohm's law

V = i Rs

Rs = V / i = 12 / 1.12 = 10.7 ohm

R + r = 10.7 ohm    .... (1)

When they are connected in parallel:

V = 12 V

i = 9.39 A

The equivalent resistance when they are connected in parallel

R_{p}=\frac{R+r}{rR}

So, By using Ohm's law

V = i Rp

Rp = V / i = 12 / 9.39 = 1.28 ohm

\frac{R+r}{rR}=1.28    .... (2)

by substituting the value of R + r from equation (1) in equation (2), we get

r R = 8.36 ..... (3)

R-r = \sqrt{\left ( R+r \right )^{2}-4rR}

R-r = \sqrt{\left ( 10.7 \right )^{2}-4\times 8.36}=9 ..... (4)

By solvng equation (1) and (4), we get

R = 9.85 ohm , r = 0.85 ohm

8 0
3 years ago
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