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3 years ago

You can increase the rate solute dissolves in solvent by______

2 answers:

5 0

Crushing particles of solute. The more surface area of the solute exposed to the solvent the faster the solute will dissolve and react with the solvent

geniusboy [140]3 years ago

5 0

Answer: Option (b) is the correct answer.

Explanation:

More is the number of solute particles present more will be the interaction take place between both the molecules of solute and solvent.

As a result, more will be the rate of reaction which means that when solute is added into the solvent after crushing it then it will readily dissolve into the solvent.

When we lower the temperature then there will occur decrease in kinetic energy of the particles. So, less number of collisions will take place between solute and solvent particles.

Hence, less will be the rate of reaction.

Thus, we can conclude that you can increase the rate solute dissolves in solvent by crushing particles of solute.

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3 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

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3 years ago

A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?

givi [52]

I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.

If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.

For the wave you described . . .

   Frequency = (speed) / (wavelength)

   = (6 m/s) / (12 m)

   = 0.5 / sec

   =    0.5 Hz .

(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)

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Answer:

Explanation:

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7 0

3 years ago

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