
Concept explainers
Interpretation:
The materials that show good analogy for anisotropy are to be selected.
Concept introduction:
Anisotropy is the phenomenon by which a material shows different values of a given property in different directions. It is a characteristic of those materials thathave their constituent particles in an ordered arrangement. All the crystalline solids are anisotropic in nature.
The phenomenon in which a material shows the same values of a given property in different directions is termed as isotropy. It is a characteristic of those materials thathave their constituent particles in unordered or random arrangement. All the liquids, gases, and amorphous solids are isotropic in nature.
A material can either be isotropic or anisotropic, but not both.

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Chapter 12 Solutions
Chemistry
- Determine the structures of the missing organic molecules in the following reaction: + H₂O +H H+ Y Z ☑ ☑ Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X, Y, and Z. You may draw the structures in any arrangement that you like, so long as they aren't touching. Molecule X shows up in multiple steps, but you only have to draw its structure once. Click and drag to start drawing a structure. AP +arrow_forwardPlease help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWLarrow_forwardDraw the skeletal ("line") structure of 2-hydroxy-4-methylpentanal. Click and drag to start drawing a structure. Xarrow_forward
- Determine whether the following molecule is a hemiacetal, acetal, or neither and select the appropriate box below. Also, highlight the hemiacetal or acetal carbon if there is one. hemiacetal acetal Oneither OHarrow_forwardWhat is the missing reactant R in this organic reaction? ་ ་ ་ ་ ་ ་ ་ ་ ་ ་ +R H3O+ • Draw the structure of R in the drawing area below. N • 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.arrow_forwardWrite the systematic name of each organic molecule: H structure H OH OH H OH name ☐ OHarrow_forward
- 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
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningLiving By Chemistry: First Edition TextbookChemistryISBN:9781559539418Author:Angelica StacyPublisher:MAC HIGHER

