CHEM 1151 GORDON: CHEM 1151 | StudySoup

PreparED Study Materials

CHEM 1151: CHEM 1151

School: Gordon College

Number of Notes and Study Guides Available: 1

Notes

Videos

Producing Ammonium Sulfate: Calculating the Required Ammonia
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Discover the process to determine the quantity of ammonia (NH?) required to produce a specific amount of ammonium sulfate ((NH?)?SO?). Through a step-by-step explanation, learn the application of the balanced chemical equation and molar mass conversions. Transform theoretical chemistry into practical knowledge with this insightful guide.

Comparing Viscosity: Why C5H11OH is 12x Thicker than C6H14 at 20°C
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Explore the intriguing contrast between Pentanol and Hexane's viscosity at 20 degrees Celsius. Uncover how molecular interactions influence a liquid's 'thickness' and discover why similar molecular weights can lead to vastly different substance properties

Why is Chloromethane Polar but Methane Nonpolar? Decoding Molecule Sha
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Explore the polar nature of Chloromethane and the nonpolar characteristics of Methane. Understand the impact of electronegativity differences and bond types. Deciphering molecular polarity through the lens of tetrahedral structures.

Quantifying Atoms Molecules & Moles: A Comprehensive Chemistry Guide
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Grasp the fundamental concept of moles in chemistry by equating it to the universally understood notion of a dozen. This video breaks down the usage of Avogadro's constant demonstrating calculations from moles to particles and vice versa. Using real-world examples like Carbon atoms Sulfur Dioxide molecules and Iron atoms viewers gain a clear understanding of moles and particle conversions

The original sulfur quantity (tons) for 26M tons SO?
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Explore the environmental impact of sulfur dioxide production, revealing how 26 million tons of this compound conceal 13 million tons of sulfur. It delves into the chemistry of this transformation, converting atomic and molecular masses, providing valuable insights into emissions from activities like burning coal and auto exhaust."

Diluting a 5.5 M KCl Solution to 0.100 M
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Learn the steps to make a 2.5 L of 0.100 M KCl solution from a 5.5 M stock solution in this easy-to-follow tutorial.

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