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3 years ago

9

Which one of the following equations correctly represents the process relating to the ionization energy of X?

1 answer:

Naily [24]3 years ago

7 0

Answer:

The correct option is E. X(g) → X⁺(g) + e⁻

Explanation:

The electrons are attracted to the nucleus and it is necessary to provide energy to start them. Then the ionization energy is the energy necessary to pull an electron to a gaseous atom, isolated and in a fundamental state. The electrons in the last shell are the most weakly attracted to the nucleus, so these are the lost electrons.

In this way, the gaseous element is converted to a gaseous cation (positively charged ion).

Taking this definition into account, <u><em>the correct option is E. X(g) → X⁺(g) + e⁻</em></u>

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How much would could a woodchuck chuck if a woodchucker could chuck wood?

katovenus [111]

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3 years ago

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Bond [772]

Answer:

convert 1 milligram to 1 gram:

Explanation:

1 milligram is 0.001 gram.

so... 2 milligram is 0.002

3 milligrams is 0.003

and so on.

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3 years ago

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ira [324]

Answer:

The current would stop

Explanation:

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3 years ago

In a circuit, three pieces of wire labeled A, B and C are joined at a common point, D. If wire B carries 1.5 mA in a direction a

Alex_Xolod [135]

Answer:

correct option is a. 0.2 mA toward D

Explanation:

given data

B carries = 1.5 mA

C carries current = 1.3 mA

solution

we take positive direction of current going away from the point D

and negative direction of current coming towards point D

so we use here kirchoff's current law that is

iA + iB + iC = 0 ......................1

iA + 1.5 + (-1.3) = 0

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correct option is a. 0.2 mA toward D

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3 years ago

The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

$P_x=rcos\theta$

$P_y=rsin\theta$

where $\theta$ is angle made by horizontal bar with x axis

Velocity at y=150 mm

$150=300sin\theta$

thus $\theta =30^{\circ}$

position of $P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}$

$P_x=259.80 mm$

$P=259.80\hat{i}+150\hat{j}$

Velocity at this instant

$u_x=-rsin\theta =300\times sin30=-150 mm/s$

$u_y=rcos\theta =300\times cos30=259.80 mm/s$

4 0

3 years ago