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3 years ago

If the bond enthalpy for a C-H bond is 413 kJ, what will happen when the C-H bond is broken?

Physics

1 answer:

pentagon [3]3 years ago

8 0

Explanation:

Bond Enthalpy : It is defined as amount of energy required to break a the particular bond in there gaseous state. It is also known as bond energy. It units are kJ/mol.

Breaking of a bond is an Endothermic process (energy absorbed from the surroundings).

Formation of bond is an Exothermic process (energy is released to the surroundings).

If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.

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Calculate the impulse imparted when a 3,000-kg car hits a wall at 60 . m/s and comes to a stop.

Pani-rosa [81]

Impulse = change in momentum

The answer is 0.

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2 years ago

Why is it correct to say that the linear electron flow of the light reactions directly captures energy in nadph but only indirec

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Answer:

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2 years ago

A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located

const2013 [10]

Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

$v = \lambda f$

here we know that the wavelength of the wave is

$\lambda = 2.5 m$

$f = 10 Hz$

now speed of the wave is given as

$v = 10(2.5)$

$v = 25 m/s$

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$t = \frac{L}{v}$

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4 0

3 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

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3 years ago