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3 years ago

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The electrostatic attractions in crystalline salts, such as sodium chloride, are formed by __________ between positive and negat

ive ions. A) sharing electrons B) a sea of electrons C) dipole attractions D) transferring electrons

2 answers:

nikklg [1K]3 years ago

8 0

D transferring electrons because that causes electricity

mamaluj [8]3 years ago

7 0

<h3><u>Answer;</u></h3>

<em>D) transferring electrons</em>

The electrostatic attractions in crystalline salts, such as sodium chloride, are formed by<em><u> </u></em><em><u>transfer of electrons</u></em> between positive and negative ions.

<h3><u>Explanation;</u></h3>

<u><em>Crystalline salts such as sodium chloride are formed as a result of ionic bonds. </em></u>Ionic bonds are types of bonds that are formed between non metals and non metals as a result of transfer of electrons.
<em><u>Transfer of electrons occurs when one atom involved in ionic bond formation looses electrons and the other gains electrons.</u></em> The result is that both atoms form ions, where one is a positively charged ion (cation) while the other is a negatively charged ion (anion). T<em><u>he attraction between the cation and anion results to the formation of ionic bond which results to formation of crystalline structures such as sodium chloride.</u></em>

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An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to

elena55 [62]

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Initial speed of the electron, $u=3\times 10^5\ m/s$

Final speed of the electron, $v=7\times 10^5\ m/s$

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}$

$a=4\times 10^{12}\ m/s^2$

Force exerted on the electron is given by :

$F=m\times a$

$F=9.11\times 10^{-31}\times 4\times 10^{12}$

$F=3.64\times 10^{-18}\ N$

(b) Let W is the weight of the electron. It can be calculated as :

$W=mg$

$W=9.11\times 10^{-31}\times 9.8$

$W=8.92\times 10^{-30}\ N$

Comparison,

$\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}$

$\dfrac{F}{W}=4.08\times 10^{11}$

Hence, this is the required solution.

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4 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

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3 years ago

How much metabolic energy is required for a 68kg person to run at a speed of 15km/hr for 15min ?

irga5000 [103]

<span>What I have here is exactly the same problem, however, with the time changed to 19 mins:

metabolic energy = metabolic power*time = 1.150*19*60 = 1.311 kJ..corresponding to 1.311/4.186 = 313,2 Cal or kcal

If we reasonably assume a metabolic eff.cy of 20%, it means we need to assume food for 1500 Cal approx.

Just plug the value t=15min to the equation and you will surely get the correct answer.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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3 years ago

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The choices are:

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Answer:

The answer is letter e, Tension Force.

Explanation:

Force refers to the "push" and "pull" of an object, provided that the object has mass. This results to acceleration or a change in velocity. There are many types of forces such as <em>Normal Force, Gravity Force, Applied Force, Friction Force, Tension Force and Air Resistance Force.</em>

The situation above is an example of a "tension force." This is considered the force that is being applied to an object by strings or ropes. This is a type force that allows the body to be pulled and not pushed, since ropes are not capable of it. In the situation above, the tension force of the rope is acting on the bag and this allows the bag to be pulled.

Thus, this explains the answer.

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3 years ago

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lana66690 [7]

To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

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4 years ago