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3 years ago

13

Which three of the following are characteristics of ionic bonding?

1 answer:

Lyrx [107]3 years ago

7 0

Gains at least one nutrient

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<u><em>If I'm wrong please tell me so I can correct my answer.</em></u>

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2 years ago

Read 2 more answers

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

3 0

3 years ago