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Salsk061 [2.6K]
4 years ago
14

The second ionization energy is the energy required to remove the _______ Electron after a __________ one has been removed.

Physics
1 answer:
Sindrei [870]4 years ago
6 0

The second ionization energy is the energy required to remove the <u>second </u>electron after a <u>valence</u> one has been removed.

<h3><u>Explanation:</u></h3>

For an element, the first ionization energy is defined as the amount of energy required to remove one electron from the outermost valence shell of a neutral atom. Removing one electron increases the number of protons, making it a 1+ ion.  

The nucleus (protons) has more bonding to the electrons with negative charge and thus more energy is required if another electron needs to be removed. This higher energy required to remove second electron from a 1+ ion (after the first one has been removed) is termed as the second ionization energy. Second ionization energy leads to formation of a 2+ ion. Similarly, third ionization energy is higher than second ionization energy.

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what is q if 28.6 g of water is heated from 22.0°c to 78.3°c? the specific heat of water is 4.184 j/g·°c.
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Answer:

6736 J

Explanation:

Knowing that Q = cmΔT:

Q = 4.184 J/(g·°C) · 28.6g · (78.3 °C - 22.0 °C) = 6736 J.

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The analysis of how people relate to each other is known as
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Anthropology 

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How much time does it take a person to walk 12km north at the velocity at a velocity of 6.5 km/hrs?
alukav5142 [94]

"6.5 km/hr" is not a velocity.  It's just a speed, so
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If he's walking in any other direction, it takes him longer than that.

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6 0
3 years ago
Read 2 more answers
What I ferromagnetism??
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Answer:

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You work for the city water department and need to pump 3400 liters/minute of water from a tank at ground level into a vented (i
umka21 [38]

Answer:

P_E=46.2778\ kW

v=1.804\ m.s^{-1}

Explanation:

Given:

  • flow rate of water, \dot{V}=3400\ L.min^{-1}=3.4\ m^3.min^{-1}

<em>∵Density of water is 1 kg per liter</em>

∴mass flow rate of water, \dot{m}=3400\ kg.min^{-1}

  • height of pumping, h=75\ m
  • efficiency of motor drive, \eta=0.9
  • diameter of pipe, D =0.2\ m

<u>Now the power required for pumping the water at given conditions:</u>

P=\dot{m}.g.h

P=\frac{3400}{60} \times 9.8\times 75

P=41650\ W

<u>Hence the electric power required:</u>

P_E \times \eta=P

P_E \times 0.9=41650

P_E=46.2778\ kW

<u>Flow velocity is given as:</u>

v=\dot{V}\div a

where: a = cross sectional area of flow through the pipe

v=\frac{3.4}{60}\div (\pi.\frac{0.2^2}{4} )

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3 0
3 years ago
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