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3 years ago

Lab: Energy Transfer

Chemistry

1 answer:

tester [92]3 years ago

4 0

2. What procedure did you use to complete the lab?

Outline the steps of the procedure in full sentences.

Section II: Observations and Conclusions

3. What charts, tables, or drawings would clearly show what you have learned in this lab?

Each chart, table, or drawing should have the following items:

a. An appropriate title

b. Appropriate labels

4. If you could repeat the lab and make it better, what would you do differently and why?

There are always ways that labs can be improved. Now that you are a veteran of this lab and have experience with the procedure, offer some advice to the next scientist about what you suggest and why. Your answer should be at least two to three sentences in length.

Writing the Lab Report

Now you will use your answers from the four questions above to write your lab report. Follow the directions below.

Section I: Overview of Lab

Use your answers from questions 1 and 2 (above) as the basis for the first section of your lab report. This section provides your reader with background information about why you conducted this lab and how it was completed. It should be one to two paragraphs in length.

Section II: Observations and Conclusions

Use your answers from questions 3 and 4 (above) as the basis for the second section of your lab report. This section provides your reader with charts, tables, or drawings from the lab. You also need to incorporate your answers to the follow-up questions (from the Student Guide) in your conclusions.

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Answer: A it goes hot to cold not cold to hot.

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The data table below shows four different chemical reactions, the temperature change of each, and the duration of the change in

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Reaction 4

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You look at the label on a container of shortening and see the words "hydrogenated vegetable oil." this means that during proces

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Increase in melting point;
Trans- arrangements of side chains around double bonds that remains in the hydrogenated fat.

Explanation:

Vegetable oil contain a larger ratio of double bonds among all its carbon-carbon bonds than animal fat such as butter does. Unlike carbon-carbon single bonds, structures connected to carbon-carbon double bonds are unable to rotate around the bonding axis. As a result, molecules rich in double bonds aren't as malleable or stack as tightly as those with a smaller number of double bonds do. The spacy molecular configuration hinders the formation of intermolecular forces, such that in nature in comparison with animal fats, vegetable oils tend to demonstrate lower melting points.

Hydrogenating vegetable oils reduce the number of double bonds per molecule while attaching extra hydrogen atoms to carbon atoms that used to form double bonds. This process would increase the strength of intermolecular interaction, hence raising the melting point.

The hydrogenation process does not necessary convert all double bonds to single bonds; some double bonds remains in the molecule, preventing the rotation of structures on their sides. Double bonds in naturally-occuring fatty acids tend to be of the cis- configuration, with hydrogen atoms connected to the same side of the carbon-carbon double bond. The high temperature involved in the hydrogenation process (around 90 degrees Celsius) can trigger the flipping of atoms connected to these double bonds to produce trans- fatty acids with hydrogen atoms bonded to opposite sides of the double bond.

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CaCO3(s) ∆→CaO(s) + CO2(g).

sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

Ca: 40.078;
C: 12.011;
O: 15.999.

What's the theoretical yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

$M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of CaCO₃ available:

$\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}$ .

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

$n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}$ .

Molar mass of CO₂:

$M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of the 0.400655 moles of $\text{CO}_2$ expected for the 40.1 grams of CaCO₃:

$m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}$ .

What's the percentage yield of this reaction?

$\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%$ .

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Answer:

Explanation:

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