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Anvisha [2.4K]
3 years ago
12

Which of the following is a typical property of an ionic compound

Physics
2 answers:
Yanka [14]3 years ago
6 0

The answer is C.

Ionic compounds are those that bring together anions and cations bonded together by ionic bonds. The electrostatic forces of the different charges are significant in the bonds that make them strong hence require high energy to break them (high melting point).  Due to the regular structure of ionic compound that tend to form lattices in solid form, when struck, they shatter along the lines of weakness of the lattice.  

DiKsa [7]3 years ago
5 0

<span>Ionic solids are rigid because the multiple interactions between oppositely charged ions hold the charged particles in relatively fixed positions.</span>

Electrical conductivity depends on ions being able to move freely. The conduction of heat depends partly on the movement of electrons in a substance. Electric charge is "locked" in the lattice positions of the ions in the solids. This means there is poor mobility of charges and poor conductivity of electricity and heat.








<span>Ionic solids are brittle and hard because the electrostatic attractions in the solid again hold the ions in definite positions. The electrostatic attractions must be overcome to move the ions. When the ions in the solid are shifted by some very strong force the positions of ions shift so that like charged ions are close together. This results in strong repulsions and the like charged ions move apart. The solid shatters and does not simply deform like a metal. Try crushing a few grains of salt in the bowl of a spoon with another spoon. The particles do not deform, they shatter.Ionic solids melt when the ions are heated and have enough energy to slide past one another. They are mobile and can act to carry electrical charge through the liquid. This explains why a molten ionic substance conducts electricity, but a solid ionic material doesn't. The ions move through the liquid to carry charge from one place to another.The dissolving process is like a tug of war. Soluble ionic substances are ripped apart by the solvent when the solid is dissolved. The solvent pulls the ions out of the solid and breaks the forces holding the crystal together. The solvent isolates the ions in an envelope of solvent particles. The ions are free to move and carry electric charge through the solution.Nonsoluble ionic solids are held together so tightly that the ions cannot be pulled out of the lattice by the solvent. The attractive forces in the solid are stronger than the attractions between the solvent and the ions. The solvent can't pull the ions out of the crystal.</span>

<span><span>Ionic compound formula<span>Ionic compound name</span>Melting point in degrees celsius<span>Solubility grams per 100 g watersuperscript is temperature</span></span><span>LiCllithium chloride613<span>45 g cold water, 128 g100</span></span><span>NaClsodium chloride801<span>209 g cold water, 284 g 100</span></span><span>KClpotassium chloride776<span>35 g cold water, 57 g 100</span></span><span>RbClrubidium chloride715<span>77 g cold water, 139 g100</span></span><span><span>MgF2</span>magnesium fluoride1396<span>0.0076 g cold water, insoluble 100</span></span><span><span>MgCl2</span>magnesium chloride708<span>54 g cold water, 72 g100</span></span><span><span>MgBr2</span>magnesium bromide695-700<span>101 g cold water, 126 g 100</span></span><span><span>MgI2</span>magnesium iodidegreater than 700<span>100 g <span>0 </span>cold water, 165 g 100</span></span></span>

 

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Explanation:

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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
Un prisma de cemento pesa 2500 N y ejerce una presión de 125 Pa, ¿cuál es el valor del área en la cual se apoya?
Eduardwww [97]

Answer:

Area = 20 m²

Explanation:

Given the following data;

Force = 2500 N

Pressure = 125 Pa

To find the area on which it rest;

Mathematically, pressure is given by the formula;

Pressure = \frac {Force}{area}

Making area the subject of formula, we have;

Area = \frac {Force}{pressure}

Substituting into the formula, we have;

Area = \frac {2500}{125}

Area = 20 m²

3 0
3 years ago
The force between a pair of .005 charges is 750 N. What is the distance between them?
Ganezh [65]

Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?

Answer:

17.32 m

Explanation:

From coulomb's Law,

F = kqq'/r²........................... Equation 1

Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.

make r the subject of the equation above

r = √(kqq'/F)..................... Equation 2

From the question,

Given: q = q' = 0.005 C, F = 750 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute these values into equation 2

r = √(9.0×10⁹×0.005×0.005/750)

r = √(300)

r = 17.32 m.

Hence the distance between the pair of charges = 17.32 m

6 0
3 years ago
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