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Vaselesa [24]
3 years ago
14

Glenn is heating water in a teakettle. The hotter water at the bottom of the kettle rises. The cooler water near the top sinks a

nd warms. Which kind of transfer of heat is this?
conduction
convection
radiation
transmission
Chemistry
2 answers:
alexandr402 [8]3 years ago
5 0
B. convection

Hope this helps!
san4es73 [151]3 years ago
3 0

Answer: Option (b) is the correct answer.

Explanation:

A process in which a fluid is heated then cold material (more dense) sinks at the bottom whereas hotter material (less dense) rises at the top is known as convection.

In convection, heat transfers from burner to the vessel and further it reaches to the fluid present into the vessel.

Therefore, we can conclude that the given situation represents convection as the kind of transfer of heat.

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49.9 ml of a 0.00292 m stock solution of a certain dye is ddiluted to 1.00 L. the diluted solution has an absorbance of 0.600. w
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Complete Question

49.9 ml of a 0.00292 m stock solution of a certain dye is diluted to 1.00 L. the diluted solution has an absorbance of 0.600. what is the molar absorptivity coefficient of the dye

Answer:

The  value is  \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

Explanation:

From the question we are told that

   The volume of the stock solution is  V_1   =  49.9 mL  =  0.0499 \  L  

   The concentration of the stock solution is  C_1  =  0.00292 \  M

   The volume of the diluted solution is  V_2 =  1.00 \  L

   The absorbance is  A =  0.600

Generally the from the titration equation we have that

         C_1 * V_1 =  C_2 * V_2

=>      0.00292  * 0.0499 =  C_2 * 1

=>     C_2 = 0.0001457 \  M

Generally from  Beer's law we have that

      A  =  \epsilon  * l  *  C_2

=>   \epsilon  =  \frac{A}{ l  *  C_2 }

Here  l is the length who value is  1 cm because the unit of  molar  absorptivity coefficient of the dye is M^{-1} *  cm^{-1}

So

            \epsilon  =  \frac{0.600}{ 1   * 0.0001457   }  

=>       \epsilon  =  4118.1 \  M^{-1} cm^{-1}  

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