assignment fill in the blank exercise 10.02

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10.2: Exercises

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Lab 10 Exercise  \(\PageIndex{1}\)

Identify the pats of a "neuron".

Lab 10 Exercise  \(\PageIndex{2}\)

Label the following:, lab 10 exercise  \(\pageindex{3}\), match these items or actions to their locations on a neuron: graded potentials, action.

Lab 10 Exercise  \(\PageIndex{4}\)

Label the following glia & related items: .

Lab 10 Exercise  \(\PageIndex{5}\)

Label the histological sections.

Lab 10 Exercise  \(\PageIndex{6}\)

Obtain slides of each of the following tissues, observe them, draw and label the significant features. SLIDES: Spinal cord, cerebral cortex, cerebellum, dorsal root ganglion, peripheral nerve, neuromuscular junction:

1.     Obtain a slide of nervous tissue from the slide box. Use any nervous tissue except peripheral nerve, there are no nerve cell bodies in a peripheral nerve section.

2.     View the slide on the second-highest objective. Search carefully until you find a clear, representative neuron in your field of view.

3.     In the circle below, draw the neuron you found. Only draw the single neuron. Do not draw any of the other material. Draw your structures proportionately to their size in your microscope’s field of view.

4.     Label any neural parts you can clearly recognize.

3.2 Determining Empirical and Molecular Formulas

Learning objectives.

By the end of this section, you will be able to:

• Compute the percent composition of a compound
• Determine the empirical formula of a compound
• Determine the molecular formula of a compound

The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.

Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition , defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

Example 3.9

Calculation of percent composition.

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

Check Your Learning

12.1% C, 16.1% O, 71.79% Cl

Determining Percent Composition from Molecular or Empirical Formulas

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH 3 ), ammonium nitrate (NH 4 NO 3 ), and urea (CH 4 N 2 O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH 3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.10 . As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound's elements.

Example 3.10

Determining percent composition from a molecular formula.

Note that these percentages sum to equal 100.00% when appropriately rounded.

Determination of Empirical Formulas

As previously mentioned, the most common approach to determining a compound’s chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative numbers , not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding numbers of atoms (in moles) are:

Thus, this compound may be represented by the formula C 0.142 H 0.284 . Per convention, formulas contain whole-number subscripts, which can be achieved by dividing each subscript by the smaller subscript:

(Recall that subscripts of “1” are not written but rather assumed if no other number is present.)

The empirical formula for this compound is thus CH 2 . This may or may not be the compound’s molecular formula as well; however, additional information is needed to make that determination (as discussed later in this section).

Consider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O. Following the same approach yields a tentative empirical formula of:

In this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding Cl 2 O 7 as the final empirical formula.

In summary, empirical formulas are derived from experimentally measured element masses by:

• Deriving the number of moles of each element from its mass
• Dividing each element’s molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula
• Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained

Figure 3.11 outlines this procedure in flow chart fashion for a substance containing elements A and X.

Example 3.11

Determining a compound’s empirical formula from the masses of its elements.

Next, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe 1 O 1.5 ). Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:

The empirical formula is Fe 2 O 3 .

Link to Learning

For additional worked examples illustrating the derivation of empirical formulas, watch the brief video clip.

Deriving Empirical Formulas from Percent Composition

Finally, with regard to deriving empirical formulas, consider instances in which a compound’s percent composition is available rather than the absolute masses of the compound’s constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion.

Example 3.12

Determining an empirical formula from percent composition.

The molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element’s mass by its molar mass:

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:

Since the resulting ratio is one carbon to two oxygen atoms, the empirical formula is CO 2 .

Derivation of Molecular Formulas

Recall that empirical formulas are symbols representing the relative numbers of a compound’s elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in the previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.

Molecular formulas are derived by comparing the compound’s molecular or molar mass to its empirical formula mass . As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If the molecular (or molar) mass of the substance is known, it may be divided by the empirical formula mass to yield the number of empirical formula units per molecule ( n ):

The molecular formula is then obtained by multiplying each subscript in the empirical formula by n , as shown by the generic empirical formula A x B y :

For example, consider a covalent compound whose empirical formula is determined to be CH 2 O. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one O atom). If the compound’s molecular mass is determined to be 180 amu, this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:

Molecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:

Note that this same approach may be used when the molar mass (g/mol) instead of the molecular mass (amu) is used. In this case, one mole of empirical formula units and molecules is considered, as opposed to single units and molecules.

Example 3.13

Determination of the molecular formula for nicotine.

Next, calculate the molar ratios of these elements relative to the least abundant element, N.

The C-to-N and H-to-N molar ratios are adequately close to whole numbers, and so the empirical formula is C 5 H 7 N. The empirical formula mass for this compound is therefore 81.13 amu/formula unit, or 81.13 g/mol formula unit.

Calculate the molar mass for nicotine from the given mass and molar amount of compound:

Comparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:

Finally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:

C 8 H 10 N 4 O 2

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• Authors: Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson, PhD
• Publisher/website: OpenStax
• Book title: Chemistry 2e
• Publication date: Feb 14, 2019
• Location: Houston, Texas
• Book URL: https://openstax.org/books/chemistry-2e/pages/1-introduction
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10.1: General Characteristics of Viruses

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• Gary Kaiser
• Community College of Baltimore Country (Cantonsville)

Learning Objectives

• State 2 living and 2 nonliving characteristics of viruses.
• List 3 criteria used to define a virus.
• Discuss why bacteria can be cultivated on synthetic media such as nutrient broth whereas viruses cannot.
• Define bacteriophage.

Viruses are infectious agents with both living and nonliving characteristics. They can infect animals, plants, and even other microorganisms. Viruses that infect only bacteria are called bacteriophages and those that infect only fungi are termed mycophages . There are even some viruses called virophages that infect other viruses.

Recently, viruses have been declared as living entities based on the large number of protein folds encoded by viral genomes that are shared with the genomes of cells. This indicates that viruses likely arose from multiple ancient cells.

The vast majority of viruses contain only one type of nucleic acid: DNA or RNA, but not both. Virus are totally dependent on a host cell for replication (i.e., they are strict intracellular parasites.) Furthermore, v iral components must assemble into complete viruses (virions) to go from one host cell to another. Since viruses lack metabolic machinery of their own and are totally dependent on their host cell for replication, they cannot be grown in synthetic culture media. Animal viruses are normally grown in animals, embryonated eggs, or in cell cultures where in animal host cells are grown in a synthetic medium and the viruses are then grown in these cells.

• Viruses are infectious agents with both living and nonliving characteristics.
• Living characteristics of viruses include the ability to reproduce – but only in living host cells – and the ability to mutate.
• Nonliving characteristics include the fact that they are not cells, have no cytoplasm or cellular organelles, and carry out no metabolism on their own and therefore must replicate using the host cell's metabolic machinery.
• Viruses can infect animals, plants, and even other microorganisms.
• Since viruses lack metabolic machinery of their own and are totally dependent on their host cell for replication, they cannot be grown in synthetic culture media.

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10: The Mole

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• 10.1: Avogadro's Number
• 10.2: Conversions Between Moles and Atoms
• 10.3: Molar Mass
• 10.4: Conversions Between Moles and Mass
• 10.5: Conversions Between Mass and Number of Particles
• 10.6: Avogadro's Hypothesis and Molar Volume
• 10.7: Conversions Between Moles and Gas Volume
• 10.8: Gas Density As you know, density is defined as the mass per unit volume of a substance. Since gases all occupy the same volume on a per mole basis, the density of a particular gas is dependent on its molar mass. A gas with a small molar mass will have a lower density than a gas with a large molar mass. Gas densities are typically reported in g/L . Gas density can be calculated from molar mass and molar volume.
• 10.8: Mole Road Map
• 10.10: Percent Composition
• 10.11: Percent of Water in a Hydrate
• 10.12: Determining Empirical Formulas
• 10.13: Determining Molecular Formulas

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2.10: Exercises

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