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3 years ago

Which statement best describes ionic bonding?

Physics

2 answers:

qwelly [4]3 years ago

8 0

Answer:

Explanation:

Ionic bonds work when a metal gives an electron to a nonmetal and this difference in charge is how they bond.

Aliun [14]3 years ago

3 0

Answer:

Option A

Explanation:

=> A metal atom transfers or loses electron(s) to becomes an anion.

=> A non-metal atoms gains electron(s) to become cation.

=> The bond between anion and cation is an Ionic bond.

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A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20m\s. The bullet is brought to rest in a

prohojiy [21]

Answer:

1) F = 24 N

2) Distance = 1 m

Explanation:

We are given;

Mass; m = 120 g = 0.12 kg

Initial velocity; u = 20 m/s

Final velocity; v = 0 m/s since it came to rest.

Time; t = 0.1 s

We can calculate acceleration from Newton's first equation of motion;

a = (v - u)/t

a = (0 - 20)/0.1

a = -200 m/s²

1) magnitude of the resistance will be;

F = ma

F = 0.12 × (-200)

F = -24 N

Since, we are dealing with the magnitude, we will take the absolute value. Thus, F = 24 N

2) To find the distance moved by the bullet, we know that;

Distance = Average speed × time

Thus;

Distance = ((v + u)/2) × t

Distance = ((0 + 20)/2) × 0.1

Distance = 1 m

4 0

2 years ago

If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency

Pepsi [2]

Answer:

a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W

Explanation:

This exercise is about Faraday's law

fem = $- \frac{ d \Phi_B}{dt}$

where the magnetic flux is

Ф = B x A

the bold are vectors

A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

v = w r

w = 2π f

v = 2π f r

we substitute

fem = - 2π B r (2π f r)

fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

fem = - 4π² 428 10⁻⁶ 4π 0.10²

fem = - 16π³ 428 10⁻⁶ 0.010

fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

V = I R

I = V / R

I = -2.1514 10⁻⁴ / 0.00336

I = - 64.0 10⁻³ A

Electric power is

P = V I

P = 2.1514 10⁻⁴ 64.0 10⁻³

P = 1.38 10⁻⁶ W

6 0

3 years ago

If you were looking for a metalloid on the periodic table,the best place to look would be?

vovikov84 [41]

Answer:

Long the step line

Explanation:

8 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago

A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.

Vesnalui [34]

Answer:

Normal, Gravity, Friction, and Air Resistance.

Explanation:

When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:

Normal force: It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.

Gravity: Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.

Friction: Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.

Air resistance: air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.

Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."

4 0

3 years ago