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2 years ago

Which of the following best summarizes the relationship between dehydration reactions and hydrolysis?

Physics

1 answer:

Softa [21]2 years ago

3 0

Answer:

Dehydration reactions assemble polymers, and hydrolysis reactions break down polymers.

Explanation:

dehydration reaction is a conversion that involves the loss of water from the reacting molecule or ion.

Hydrolysis is defined as any chemical reaction in which a molecule of water ruptures one or more chemical bonds.

Dehydration reactions link monomers together into polymers by releasing water, and hydrolysis breaks polymers into monomers using a water molecule. Monomers are just single unit molecules and polymers are chains of monomers.

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9.5 g bullet has a speed of 1.5 km/s what is the kinetic energy of the bullet

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2 years ago

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Convert the following to gram. a,250 kg b373 mg c,10 quanital,15 ton

rusak2 [61]

Explanation:

250 KG = 250000 Grams

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2 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;

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

<em />

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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2 years ago

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kogti [31]

Answer:

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Explanation:

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2 years ago