
Chemistry: Principles and Reactions
8th Edition
ISBN: 9781305079373
Author: William L. Masterton, Cecile N. Hurley
Publisher: Cengage Learning
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Chapter 5, Problem 73QAP
Interpretation Introduction
Interpretation:
The density and molar mass of an unknown gas are required to be determined.
Concept introduction:
Graham’s law may be formulated as:
where M denotes the molar volume of the gas.
That means the rate of effusion of a gas and the square root of the mass of the particles of the gas is inversely proportional.
Under the same conditions of temperature and pressure, the molar volume of all gases is the same and hence, the Graham’s law may be written as:
where d denotes the density of the gas.
The density of a gas is directly proportional to the molar mass of the gas (since molar volume is constant for all the gases).
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Chapter 5 Solutions
Chemistry: Principles and Reactions
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- Determine whether each of the following molecules is a hemiacetal, acetal, or neither and select the appropriate box in the table. CH3O OH OH OH hemiacetal acetal neither hemiacetal acetal neither Xarrow_forwardWhat is the missing reactant R in this organic reaction? N N དལ་ད་་ + R • Draw the structure of R in the drawing area below. • Be sure to use wedge and dash bonds if it's necessary to draw one particular enantiomer. Click and drag to start drawing a structure. ㄖˋarrow_forwardDraw the condensed structure of 4-hydroxy-3-methylbutanal. Click anywhere to draw the first atom of your structure.arrow_forward
- Using the bond energy values, calculate the energy that must be supplied or is released upon the polymerization of 755 monomers. If energy must be supplied, provide a positive number; if energy is released, provide a negative number. Hint: Avogadro’s number is 6.02 × 1023.arrow_forward-AG|F=2E|V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: Acidic solution -0.93 +0.38 -0.51 -0.06 H3PO4 →H4P206 H3PO3 H3PO2 → P→ PH3 -0.28 -0.50 → -0.50 Basic solution 3-1.12 -1.57 -2.05 -0.89 PO HPO →→H2PO2 P PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P2O6 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH, 0.0 -0.5- 2 3 9 3 -1.5 -2.0 Pa H,PO H,PO H,PO -3 -1 0 2 4 Oxidation state, N 2 b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) c) Elemental phosphorus tends to disproportionate under basic conditions. Use data in…arrow_forwardThese two reactions appear to start with the same starting materials but result in different products. How do the chemicals know which product to form? Are both products formed, or is there some information missing that will direct them a particular way?arrow_forward
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