## Evaporation and Intermolecular Attractions Lab.

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### 資料紹介

1.Calculate the mean, standard deviation and percent standard deviation of your t-values.
Include the values in a table.

2.Two of the liquids, 2-propanol and acetone, had significantly different t-values. Explain the difference in t values of these substances, based on their intermolecular forces. Why would you have expected them to have similar t-values?
2-propanol and acetone expected to have similar t-value because, both have 3 carbons for main chain. However, there was significance difference in t-values of these two substances. The reason for this is because of the hydrogen bonding that 2-propanol has. Hydrogen bonding is very strong bonding so it makes lower done the t-value.

3.Which of the alcohols studies has the strongest intermolecular forces of attraction? The weakest intermolecular forces? Explain using the results of this experiment.
2-propanol has the strongest intermolecular forces of attraction because, there are hydrogen bonding in the center. And the acetone has the weakest intermolecular forces because it doesn’t have any hydrogen bonding. Hydrogen bonding is a strong bonding so it difficult to change temperature.

4.Plot a graph of the mean t values of the three alcohols versus their respective molecular weights. Plot molecular weight on the horizontal axis and t on the vertical axis.

### 資料の原本内容( この資料を購入すると、テキストデータがみえます。 )

*Processing;
Calculate the mean, standard deviation and percent standard deviation of your t-values.
Include the values in a table.
*Data Table of 4 liquids, that what we observed in a experiments.
Min (℃) Max (℃) T-value (℃) Mean (℃) Standard Deviation (℃) Standard Deviation (%) Methanol 9.0 21.1 12.1 13 1 8% 9.5 23.6 14.1 8.9 21.4 12.5 Ethanol 15.2 21.9 6.7 6.8 0.6 9% 14.8 21.1 6.3 14.6 22.1 7.5 2-propanol 15.1 20.5 5.4 7 1 18% 15.7 23.5 7.8 15.4 22.6 7.2 Acetone 8.2 22.3 14.1 14.2 0.2 2% 8.2 ..

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