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WINSTONCH [101]
3 years ago
15

The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat

of sublimation for Cs is +76 kJ/mol, bond dissociation energy for 12Cl2 is +121 kJ/mol, Ei1 for Cs is +376 kJ/mol, and Eea for Cl(g) is −349 kJ/mol. What is the magnitude of the lattice energy for CsCl?
Physics
1 answer:
german3 years ago
5 0

Answer :  The magnitude of the lattice energy for CsCl is, 667 KJ/mole

Explanation :

The steps involved in the born-Haber cycle for the formation of CsCl :

(1) Conversion of solid calcium into gaseous cesium atoms.

Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)

\Delta H_s = sublimation energy of calcium

(2) Conversion of gaseous cesium atoms into gaseous cesium ions.

Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)

\Delta H_I = ionization energy of calcium

(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.

Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)

\Delta H_D = dissociation energy of chlorine

(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)

\Delta H_E = electron affinity energy of chlorine

(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.

Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)

\Delta H_L = lattice energy of calcium chloride

To calculate the overall energy from the born-Haber cycle, the equation used will be:

\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L

Now put all the given values in this equation, we get:

-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L

\Delta H_L=-667KJ/mole

The negative sign indicates that for exothermic reaction, the lattice energy will be negative.

Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole

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Answer:

<h3>473.8 m/s; 473.8 m/s</h3>

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

6 0
3 years ago
A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a
stepan [7]

Answer:

a) α = 1.875 \frac{rad}{s^{2} }

b) t = 8 s

Explanation:

Given:

ω1 = 0 \frac{rad}{s}

ω2 = 15 \frac{rad}{s}

theta (angular displacement) = 60 rad

*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

15^{2} = 0^{2} + 2(α)(60)

225 = 120α

α = 1.875 \frac{rad}{s^{2} }

b)

α = (ω2-ω1)/t

t = (ω2-ω1)/α = (15-0)/1.875 = 8

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4 0
3 years ago
The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree
lina2011 [118]
A) the periodic time is given by the equation;
 T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
   = 6.28 × √0.0673
    = 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
 thus; frequency = 0.614 Hz

b)  The vertical distance, the height is given by
 h= 0.66 cos 12
 h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
 mgh = 1/2mv²
  v² = 2gΔh = 2×9.81 × 0.01 
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v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
   = 1/2 × 0.310 × 0.443²
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4 0
3 years ago
Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated
Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>
  • Gravitation is the force of attraction between any two bodies.
  • Every body in the universe attracts every other body with a force.
  • This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
  • Mathematically we can expressed it as,

                        F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2

<h3>How to solve the problem?</h3>
  • Here, we have given with the data's,

                      M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m

  • Thus, the force of attraction between these two bodies will be,

               F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0
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Why is the force of gravity on your body weaker on the Moon than on the Earth?
andrezito [222]

Answer:

The moon’s gravity is weaker than the earth gravity due to the smaller in size as compared to the earth. As there is no atmosphere present on the moon gravity, so there are fewer chances of withstanding temperature.

Please  mark as brainliest

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