PreparED Study Materials

CHM 1025: Introduction to Chemistry

School: University of Florida

Number of Notes and Study Guides Available: 30

Notes

Study Guides

Videos

Element Classification: Rep, Trans, Inner-Trans, Noble Gases
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We categorize elements into representative, transition, inner-transition, and noble gases based on their positions in the periodic table and general characteristics, without revealing specific element classifications.

Diprotic Acid Titration: Determining Molar Mass
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Discover how to find the molar mass of an unknown diprotic acid through a titration experiment with 0.1288 M NaOH. Learn step by step in this chemistry tutorial.

Understanding Boiling Points: Ethylene Glycol, Alcohols & Ethers Expla
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This video offers insights into the properties and boiling points of different alcohols and ethers, including antifreeze agents like ethylene glycol. We also discuss how molecular structures and interactions impact boiling points, shedding light on key differences among related compounds.

Vaporization vs. Condensation: Heat Absorption & Release
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Discover the contrasting processes of vaporization and condensation and their thermal nature. Understand why boiling is endothermic, absorbing heat, while condensation is exothermic, releasing heat, and how these processes impact ?H values.

Electrons in outer shell of: Potassium, Calcium, Aluminum
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This video explains the importance of the number of valence electrons in an element's chemical behavior and demonstrates how to determine the valence electrons for potassium, calcium, and aluminum based on their positions in the periodic table. Valence electrons dictate an element's reactivity with other elements, making this knowledge essential for understanding chemical reactions.

Electronegativity & Acid Strength: Decoding Acid Formulas
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This video discusses how the electronegativity of attached groups affects the strength of acids. Comparing pairs such as Chloroacetic vs Fluoroacetic and Acetic vs Nitroacetic acids, we highlight which is the stronger acid in each pair and why.

Textbook Solutions (0)

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