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Sholpan [36]
3 years ago
9

A solid solute is added to a liquid solvent. The solution is stirred and heated. How does stirring and heating affect the soluti

on?
A) Stirring increases the rate of dissolving, but heating decreases the rate.

B) Both stirring and heating increase the rate of dissolving.

C) Both stirring and heating decrease the rate of dissolving.

D) Stirring decreases the rate of dissolving, but heating increases the rate.

Physics
2 answers:
WITCHER [35]3 years ago
8 0

Answer:

B) Both stirring and heating increase the rate of dissolving.

Explanation:

When we dissolve a solute in a solvent, it is separated into its constituting ions.

For example, if we dissolve salt in water, The NaCl is disassociated into its ions Na^{+} and Cl^{-} because of the interaction of H^{+} and OH^{-} ions present in water.

if we can increase the amount of salt interacting with water,  we can increase the rate of dissolving.

The picture below will provide a better understanding about the area of reaction.

In the figure a, salt is accumulated at the bottom, in this case only the salt molecules at the upper surface will be able to react with water.

In the figure b, salt grains are moving around the container. This increases the number of molecules interacting with water. This leads to quicker dissolving.

We can make the salt grains move by either heating or by stirring.   So both stirring and heating will increase the rate of dissolving

bija089 [108]3 years ago
5 0

because of conduction the metal spoon is a conductor so the heat is getting traveled through the spoon and into your hand

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Answer:

its speed will be less than V

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1/2mV² + 1/2Iω² = mgd

1/2mV² = mgd - 1/2Iω²

V² = [2(mgd - 1/2Iω²)/m]

V = √[2(mgd - 1/2Iω²)/m]

When the four balls are moved inward closer to the drum, their rotational inertia increases and also its angular speed which thus causes an increase in rotational kinetic energy. But, since the box still falls the same distance of d, its final kinetic energy plus rotational kinetic energy of the drum plus balls still equals its initial potential energy

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With its new speed is now V' at the end of d,

1/2mV'² + 1/2I'ω'² = mgd

1/2mV'² = mgd - 1/2I'ω'²

V² = [2(mgd - 1/2I'ω'²)/m]

V' = √[2(mgd - 1/2I'ω'²)/m]

Since I' and ω' increase, the rotational kinetic energy of the drum and balls (1/2I'ω'²) increases. Thus, the difference (mgd - 1/2I'ω'²) < (mgd - 1/2Iω²) which implies that the kinetic energy of the box decreases. Hence, since its kinetic energy decreases, its speed V' also decreases.

So,  V' < V

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