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emmainna [20.7K]
3 years ago
10

In the essay box, compose a scientific report which includes your observations, data, and conclusions. The following questions s

hould be answered in your conclusion.
Why does the first method for determining volume work only for a regularly shaped object?

Will the second method for determining volume work for any object or just an irregularly shaped one? Why?

Is one method of measurement more accurate than the other? Why or why not?

Would the displacement method of measurement work for a cube of sugar? What about a cork? Why or why not?

What did you find out from this investigation? Be thoughtful in your answer.


I need help writing the scientific report, please help:)
Chemistry
1 answer:
timama [110]3 years ago
4 0

Answer:

1. It depends what type of method you are using. if it is Height x Width x Length then it will not work for an irregular shape because it has extra pieces that would not be included.

2. The second method would work for both regular and irregular shapes because you would have to know or find out the volume of the regular shape to get the volume for the irregular shape.

3. It also depends on what you are doing, if you are doing a regular shape then use the first method, if it's an irregular shape then use the second method, if you do the maths correctly both should give you an accurate answer for what you want to achieve.

4. No, because the sugar would dissolve.

5. No, on this case the displacement method would not work because of the weight difference

Explanation:

All the answers for you!

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The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

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<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

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E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

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