Open-End Questions and Problems

05/15/17
CHEM 60 / Exam 2 / Name: / ANSWER KEY

Open-End Questions and Problems

Please read the following. To receive full credit for a question or a problem, in addition to the correct answer, you must show a neat, complete, and logical method of solution where each number is labeled with the appropriate unit and the final answer is rounded to the correct number of significant digits. The correct answer without any work shown will generally get zero credit! When an explanation is required, it should be brief, but accurate and complete.

There are 8 questions(on 5 pages) for a total of 56points. (Exam 1 part with multiple choice questions is worth 44 points. Both parts combined are 100 points total.

1.(7 points)

Which characteristic of an element was used to arrange
elements in the original Mendeleev's periodic table? / atomic weight
Which characteristic of an element is used to arrange
elements in the modernperiodic table? / atomic number
The number above the symbol of the element in each
cell of the periodic table is called / atomic number
The number below the symbol for each of the most
common elements is given without parentheses and it is / atomic weight
The number below the symbol for some of the elements is given in parentheses and without a decimal point and it is / mass number of one the element's isotopes

Elements that belong to the same group in the periodic table have similarities in chemical properties. What is the explanation of such similarities in modern chemistry?

Such similarities are due similar arrangement of the valence electrons in the atoms of the elements that belong to the same group in the periodic table.

Give two examples of similarities in chemical properties for the family of alkaline earth metals in the form of two generic chemical equations in which M stands for an atom of an alkali metal. (Hint: Reactions of those metals with water, with oxygen, or with any halogen.)

M(s) + 2 HOH(l) → M(OH)2(aq) + H2(g)M = Ca, Sr, or Ba

2 M(s) + O2(g) → 2 MO(s)2 M(s) + Cl2(g) → 2 MCl2(s)

Give two examples of similarities in chemical properties for the family of halogens in the form of two generic chemical equations in which X stands for an atom of a halogen. (Hint: Reactions of halogens with hydrogen and with any alkali metal.) X = F, Cl, Br, or I

H2(g) + X2 → 2 HX(aq)2 Na(s) + X2 → 2 MX(s)

Give the statement of the periodic law.

When elements are arranged in order of increasing atomic number, elements with similar chemical behavior occur at periodic (regularly recurring) intervals.

2.(7 points) Fill the blanks in the following table.

Neon / Nitrogen / Ammonia
The number of molecules in 1.00 mole of / none / 6.02×1023 / 6.02×1023
The number of atoms in 1.00 mole of / 6.02×1023 / 1.20×1024 / 2.41×1024
The number of elements in 1.00 mole of / one / one / two
The number of substances in 1.00 mole of / one / one / one
The number of compounds in 1.00 mole of / none / none / one
The mass in grams of 1.00 mole of / 20.18 / 28.02 / 17.03
The volume in liters of 1.00 mole of** / 24.5 / 24.4 / 23.3

** Avogadro's law (initially put forward by Avogadro in 1811 as a hypothesis; the Avogadro constant is named after Avogadro, but was not determined by him) states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules (not atoms unless the gas is one of the noble gases). Significant deviation for ammonia is due to relatively strong attractive forces between ammonia molecules.

3.(7 points) Give two examples of each. Note: (i) do not to use the same substance as an example more than once; (ii) paper, plastic, wood, a stick, gasoline, milk, food are not pure chemical substances and should not be used as examples.

Physical Change (a change that involves a specific pure substance)
Example 1:
ice melts
Example 2:
sugar (sucrose)dissolvesinwater / Chemical Change (a change that involves a specific pure substance)
Example 1:
potassiumreactswithwater toproducepotassiumhydroxideandhydrogengas
Example 2:
sulfur burnsinairtoproducesulfurdioxide
Physical Property (specific property of a specific pure substance)
Example 1:
density of aluminumis2.70cm3
Example 2:
sugar (sucrose)dissolvesinwater / Chemical Property (specific property of a specific pure substance)
Example 1:
chlorinereactswithirontoproduceiron(III)chloride
Example 2:
neondoesnotreactwithotherchemicalelementstoformany stablecompounds

4.(7 points) Refer to the molecular diagram given below to answer the questions that follow.

Write the balanced chemical equation for the reaction schematically depicted in the diagram.

2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(g)

Which substance is the limiting reactant? What is the percent yield of the reaction? Briefly explain.

Of the two reactants (the substances entering the reaction), NH3 and N2O, at the end of the reaction only NH3 is still present. N2O is completely gone. Therefore, N2O is the limiting reactant and the percent yield of the reaction is 100% (no calculations are needed).

5.(7 points)Potassium superoxide, KO2, reacts with carbon dioxide to produce potassium carbonate and oxygen. Complete the following table for the reaction of 25.0 g of potassium superoxide with 15.0 g of carbon dioxide.

Balanced Chemical Equation / 4 KO2(s) / + / 2 CO2(g) / → / 2 K2CO3(s) / + / 3 O2(g)
Mass before the reaction / 25.0 g / 15.0 g / 0 g / 0 g
Molar mass / 71.10 g/mol / 44.01 g/mol / 138.20 g/mol / 32.00 g/mol
Moles before the reaction / 0.352 mol / 0.341 mol / 0 mol / 0 mol
Moles consumed (–) or moles produced (+) / −0.352 mol / −0.176 mol / +0.176 mol / +0.264 mol
Moles after the reaction / 0 mol / 0.165 mol / 0.176 mol / 0.264 mol
Mass after the reaction / 0 g / 7.26 g / 24.3 g / 8.45 g

6.(7 points) Briefly answer the following questions.

(a)Compare the passage of electricity through a wire and through a solution.

The carriers of electricity in a wire (solid metal) are electrons and in a solution, ions.

(b)How can it be that aqueous solutions of all ionic compounds conduct electricity but aqueous solutions of covalent molecular compounds may or may not conduct electricity?

When an ionic solid dissolves in water, the ions of which it is built go into the solution. A solution containing ions conducts electricity. Covalent compounds soluble in water are molecular compounds; in general, their aqueous solutions contain no ions, just molecules. However, molecules of some covalent compounds, acids and ammonia, react with water molecules to produce ions in aqueous solutions. Aqueous solutions of those compounds conduct electricity.

7.(7 points)Consider aqueous solutions in two separate beakers before mixing. Schematically represent the content of one of the beakers as a solution magnesium chloride and the content of the other beaker as a solution of silver nitrate. Draw 6 formula units of one substance in the small beaker on the left and 6 formula units of the other substance in the small beaker on the right. This will represent equal numbers of moles of the two substances. Use the following symbols in your drawings: circles for magnesium ions, diamonds for chloride ions, squares for silver atoms, and triangles for nitrate ions. Draw a microscopic level representation showing the result of mixing of the two solutions and write the three equations to represent the reaction

Before Mixing / After Mixing
magnesium chloride
solution / silver nitrate
solution / /
/ magnesium ion /
/ chloride ion
/ silver ion
/ nitrate ion

FFE: MgCl2(aq) + 2 AgNO3(aq) → Mg(NO3)2(aq) + 2 AgCl(s)

CIE:Mg2+(aq) + 2Cl−(aq) + 2Ag+(aq) + 2NO3−(aq) → Mg2+(aq) + 2NO3−(aq) + 2AgCl(s)

NIE:Ag+(aq) + Cl−(aq) → AgCl(s)

8.(7 points) An aqueous solution containing 0.0200 mol of aluminum sulfate and an aqueous solution containing 0.0400 mol of barium nitrate are mixed.

(a)Write the equations to represent the reaction.

FFE:Al2(SO4)3(aq) + 3 Ba(NO3)2(aq) → 2 Al(NO3)3(aq) + 3 BaSO4(s)

CIE:2Al3+(aq) + 3SO42−(aq) + 3Ba2+(aq) + 6NO3−(aq) → 2Al3+(aq) +6NO3−(aq) + 3BaSO4(s)

NIE:Ba2+(aq) + SO42−(aq) → BaSO4(s)

(b)Fill the blanks in the table.

Aluminum ion / Sulfate ion / Barium ion / Nitrate ion / Precipitate
Chemical formula / Al3+ / SO42− / Ba2+ / NO3− / BaSO4
Moles before the reaction / 0.0400 / 0.0600 / 0.0400(L.R.) / 0.0800 / 0
Change (in moles) in the course of the reaction / 0 / −0.0400 / −0.0400 / 0 / +0.0400
Moles after the reaction / 0.0400 / 0.0200 / 0 / 0.0800 / 0.0400

(c)Calculate the mass of the precipitate in grams. Show work.

(0.0400 mol BaSO4) × (233.39 g BaSO4) / (1 mol BaSO4) = 9.34 g BaSO4