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Sholpan [36]
4 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]4 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]4 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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steposvetlana [31]
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Explanation:

<em>Hey</em><em>,</em><em> </em><em>there</em><em>!</em>

<em>Here</em><em>,</em><em> </em><em>In</em><em> </em><em>question</em><em> </em><em>given</em><em> </em><em>that</em><em>, </em>

<em>potential</em><em> </em><em>difference</em><em> </em><em>(</em><em>V</em><em>)</em><em>=</em><em> </em><em>8</em><em>V</em>

<em>resistance</em><em> </em><em>(</em><em>R</em><em>)</em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>ohm</em>

<em>Now</em><em>,</em>

<em>According</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Ohm's</em><em> </em><em>law</em><em>,</em>

<em>V</em><em>=</em><em> </em><em>R</em><em>×</em><em>I</em><em> </em><em> </em><em> </em><em> </em><em>{</em><em> </em><em>where</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>current</em><em>}</em>

<em>or</em><em>,</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>V</em><em>/</em><em>R</em>

<em>or</em><em>,</em><em> </em><em>I</em><em> </em><em>=</em><em> </em><em>8</em><em>/</em><em>1</em><em>0</em>

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