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3 years ago

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Why are ionic solids poor conductors of electricity? The lattice is brittle. The ions are neutral. The ions are of equal but opp

osite charge. The ions are in fixed positions.

1 answer:

sveta [45]3 years ago

6 0

The ions are in fixed positions.

Explanation:

Ionic solids are poor conductors of electricity because their ions are fixed in position. Their ions are not free to move about. They are fixed in crystal lattices.

For the conduction of electricity, compounds must possess free mobile electrons and moving ions in solution.
Ionic compounds are formed by the electrostatic attraction between a metallic and non-metallic ion.
They actually contain ions but their ions are locked up.
They are not free to move about.
Electrical conduction involves ion mobility.
In molten and aqueous forms, they are able to conduct electricity because their ions are then mobile.

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Ionic compound brainly.com/question/6071838

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A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1

balandron [24]

Answer:

The answer is " $143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}$ "

Explanation:

For point a:

Energy balance equation:

$\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\$

$W=0\\\\Q=0\\\\m_e=0$

From the above equation:

$\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\$

because the rate of air entering the tank that is $h_i$ constant.

$\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\$

Since the tank was initially empty and the inlet is constant hence, $m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\$

Interpolate the enthalpy between $T = 300 \ K \ and\ T=295\ K$ . The surrounding air

temperature:

$T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}$

Substituting the value from ideal gas:

$\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}$

Follow the ideal gas table.

The $u_2= 298.33\ \frac{kJ}{kg}$ and between temperature $T =410 \ K \ and\ T=240\ K.$

Interpolate

$\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}$

Substitute values from the table.

$\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\$

For point b:

Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. $\bar{R} \ is\ \frac{R}{M}$ (M is the molar mass of the gas that is $28.97 \ \frac{kg}{mol}$ and R is gas constant), and T is the temperature.

$n=\frac{pV}{TR}\\\\$

$=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\$

For point c:

Entropy is given by the following formula:

$\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}$

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3 years ago

21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe

DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

- The constant pushing force forward, of magnitude F = 40 N

- The frictional force, acting backward, $F_f$

Since the two forces are in opposite direction, the equation of motion is

$F-F_f = ma$

where

m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

a = 0

And therefore,

$F - F_f = 0\\F_f = F$

which means that the magnitude of the force of friction is also equal to 40 N.

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3 years ago

If a 50 W light bulb and a 90 W light bulb operate from 120 V, which bulb has a greater current in it?

PIT_PIT [208]

The 50 W bulb would have a greater current than the 90 W bulb

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A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i

snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle $=45^{\circ}$

Acceleration due to gravity on earth is $9.8 m/s^2$

Acceleration due to gravity on moon is $\frac{9.8}{6}=1.63 m/s^2$

Range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R_{earth}=\frac{u^2\sin 2\theta }{g}=5$ ----1

$R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}$ -----2

Divide 1 & 2

$\frac{5}{R_{moon}}=\frac{1}{6}$

$R_{moon}=30 m$

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3 years ago

two forces act at an angle of 135 degree.the forces are F1 = 50N and F2 = 25N respectively.graphically construct a parallelogram

olya-2409 [2.1K]

Answer:gay

Explanation:

gay

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3 years ago