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3 years ago

10

You can increase the rate solute dissolves in solvent by______

2 answers:

V125BC [204]3 years ago

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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21. A person standing 49.5m from the foot of a cliff claps his hands and hears an echo 0.3

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Answer:

330 m/s

Explanation:

The sound wave has to travel TO the cliff AND back = 2 * 49.5 = 99 m

magnitude of velocity = distance / time = 99m / .3 s = 330 m/s

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2 years ago

Describe how radio waves are different from sound waves.

Lilit [14]

Answer:

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Explanation:

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Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

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3 0

3 years ago

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60

miskamm [114]

Answer:

Answer:196 Joules

Explanation:

Hello

Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

$F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules$

Answer:196 Joules

I hope it helps

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3 years ago

The figure below shows a cylinder filled with an ideal gas, which has a moveable piston resting on it. The cylinder's volume is

Anton [14]

I uploaded the answer to $^{}$ a file hosting. Here's link:

bit. $^{}$ ly/3gVQKw3

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3 years ago