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Anon25 [30]
3 years ago
5

Lab: Calorimetry and Specific Heat

Chemistry
1 answer:
gayaneshka [121]3 years ago
5 0

Answer:

Title for the lab--->Calorimetry and Specific Heat

Name:

Teacher:

Date:

Purpose: Explore how the specific heat of a substance can be determined using a “coffee cup” calorimeter.

Question: How can you determine the specific heat of a metal using a calorimeter?

Hypothesis: If a metal has a low specific heat, then the metal could make an excellent material for cookware because it takes less heat to raise the temperature of the metal to cook food.

MATERIALS:

- Two polystyrene coffee cups

- Lid for polystyrene cup

- Thermometer      

- 250 mL beaker        

- 400 mL beaker    

- Ring stand    

- Two buret clamps    

- Hot plate  

- Test-tube holder    

- 50 mL graduated cylinder  

- Water    

- aluminum, copper, iron, and lead powder    

- Analytical balance

PROCEDURE:

1) Set up a 600 mL beaker and add 400 mL of dis"lled water inside and place it on a hot plate un"l it boils. Also set up 2 dry Styrofoam cups with a lid aside and add 100.0 mL tap water (calorimeter water) inside.

2) In the boiling water, place the unknown metal sample inside for about 30 min. Before though, weigh it accurately and record the mass on your data sheet.

3) After 30 min, take the temperature of boiling water to the nearest 0.1 °C (Ini"al temp of metal), then take the metal out of the boiling water and replace it into Styrofoam cups filled with room temperature tap water. Ini"al temperature of water is the room temperature of the tap water.

4) Take every 15 seconds for about 2 minutes the temperature of water inside the calorimeter with a thermometer inserted through the hole in the lid. Gently swirl the cup to mix the contents and record the temperature. Use the highest temperature.

5) Calculate the specific heat of the metal sample.

then the table will be in the picture i posted on here  

next is your ---> Analysis /Conclusion

Analysis:

Analyzed from the data above, we know the temperature that is measured depends on the amount of mole in the solution. If the amount of mole in the solution is large, the temperature will be as well. After a few moments, the temperature rises and falls, and in some instances, it remains steady.

Conclusion:

In conclusion, I do believe my hypothesis was supported through my experiments. Through said experiments, I identified the unknown metal as we calculated the changes in the heat, and found that the metals that should be put to use are copper and iron metals. Copper and iron have a low specific heat, which makes these ideal metals when creating cookware. When a metal has low specific heat, it needs less energy to produce heat and it cools down with less energy as well. For these reasons, my hypothesis is supported.

Explanation:

hope this helps :)

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Veronika [31]
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that is 3M = moles of solute/ 4 Kg
multiply  both side by  4

moles of the  solute  is therefore = 12  moles

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1 mole =6.02 x10^23  molecules

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4 years ago
Construct a three-step synthesis of trans-2-pentene from acetylene by dragging the appropriate formulas into the bins. Note that
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Answer:

The three-step synthesis of trans-2-pentene from acetylene is as follows.

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Explanation:

Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.

The chemical reaction of each step of chemical reactions is as follows.

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Answer:

Explanation:

A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.

       This is an arrhenius Base

According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :

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Answer:

1,15mL = V₂

Explanation:

Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:

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<em>Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.</em>

The V₁ is 1.23mL

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Replacing:

1.23mL*285.15K = V₂*305.15K

<h3>1,15mL = V₂</h3>

<em />

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