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4 years ago

Does the lattice energy of an ionic solid increase or decrease as the charges or sizes of the ions increase?

Physics

1 answer:

alexandr1967 [171]4 years ago

7 0

Answer:

Explanation:

Lattice energy is the energy required to separate one mole of an ionic solid compound into its components gaseous cations and anions.

Due to increase in size of the ions, the lattice energy decreases while the lattice energy increases as the charge of the ions increases.

When the size increase, the distance between the nuclei also increase leading a decrease the force of attraction between the nuclei

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0

3 years ago

6.For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isot

Mama L [17]

52, mass number equals protons + neutrons

3 0

3 years ago

A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x =

frutty [35]

Answer:

θ = (7π / 3) rad

Explanation:

given,

displacement of simple harmonic motion along x-axis

equation is given as

x = 5 sin (π t + π/3 )

general equation of simple harmonic motion

x = A sin θ

θ is the phase angle

θ = π t + π/3

at t = 2 s

$\theta =\pi \times 2 +\dfrac{\pi}{3}$

$\theta =\dfrac{7\pi}{3}\ rad$

Phase of the motion at t =2 s is θ = (7π / 3) rad

7 0

3 years ago

You throw a baseball directly upward at time ????=0 at an initial speed of 14.9 m/s. What is the maximum height the ball reaches

xeze [42]

Answer:

Maximum height, h = 11.32 meters

Explanation:

It is given that,

The baseball is thrown directly upward at time, t = 0

Initial speed of the baseball, u = 14.9 m/s

Ignoring the resistance in this case and using a = g = 9.8 m/s²

We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

$v^2-u^2=2ah$

At maximum height, v = 0

and a = -g = -9.8 m/s²

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

$h=\dfrac{0-(14.9\ m/s)^2}{2\times -9.8\ m/s^2}$

h = 11.32 meters

Hence, the maximum height of the baseball is 11.32 meters.

8 0

3 years ago