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3 years ago

The barrier to C-C bond rotation in iodoethane is 13 kJ/mol. What energy can you assign to an H-I eclipsing interaction?

Physics

1 answer:

n200080 [17]3 years ago

7 0

Answer:

5J/mol

Explanation:

H-H eclipsed = Torsional strain = 4.0 kJ/mol

H-CH3 eclipsed = Mostly torsional strain = 6.0 kJ/mol

CH3-CH3 eclipsed =Torsional and steric strain = 11.0 kJ/mol

CH3-CH3 gauche = Steric strain = 3.8 kJ/mol

You know each H-H eclipsed is 4 kJ/mol (8 kJ/mol total) and the reported barrier is 13 kJ/mol. Therefore the H-I eclipsed interaction must be 5 kJ/mol, (13-8 = 5).

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A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago

"Which of the following best describes the circuit shown below?

Anettt [7]

The figure shown above is series combination as the two resistors (bulb) are there which are then connected to the battery
so i conclude from the above options given the option is B
hope it helps

3 0

3 years ago

Read 2 more answers

A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find

vladimir1956 [14]

Answer:

i) The period of the particle is 0.3 seconds

ii) The angular velocity is approximately 20.94 rad/s

iii) The linear velocity is approximately 1.047 m/s

iv) The centripetal acceleration is approximately 6.98 m/s²

Explanation:

The given parameters are;

The number of revolution of the particle, n = 800 revolution

The time it takes the particle to make 800 revolutions = 4 minutes

The dimension of the circle = 5 cm = 0.05 m

Given that the dimension of the circle is the radius of the circle, we have;

i) The period of the particle, T = The time to complete one revolution

T = 1/(The number of revolutions per second)

∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s

The period, T = 3/10 seconds = 0.3 seconds

ii) The angular velocity, ω = Angle covered/(Time)

800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes

∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3

The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s

iii) The linear velocity, v = r × ω

∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s

iv) The centripetal acceleration, $a_c$ = v²/r

∴ The centripetal acceleration, $a_c$ = (π/3)²/(0.05) = 20·π/9

The centripetal acceleration, $a_c$ = 20·π/9 m/s² ≈ 6.98 m/s²

4 0

3 years ago

Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone o

Sholpan [36]

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .

apparent frequency = $f_o\times\frac{V+v_o}{ V-v_s}$

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency = $f\times \frac{340+6}{340-10}$

= $f\times \frac{346}{330}$

So m = 346 , n = 330 .

8 0

3 years ago

how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy

Blababa [14]

This question is probably referring to heat energy transferring from the car to its surroundings.

4 0

4 years ago