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Question


In the PhET simulation window, click the Macro menu in the top left corner of the screen. This view

gives a view of the beaker at a macroscopic level (as your naked eye would see it). The Micro menu shows what happens to sugars and salts at the molecular level when they dissolve in water (note that you can use the arrows to switch to other type of solutes). Use both the Macro and Micro menus in the PhET simulation to help complete the following statements regarding solutions.

Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.


ANSWER:

Electrolytes and Nonelectrolytes

Electrolytes, typically known as salts, dissolve in water to form ions in solution. For some of these salts, the ion can be a polyatomic ion, which is a charged particle that consists of more than one element, e.g., NO3− .

Salts are not the only ionizable compounds. Acid and base compounds, where H+ serves as the cation, can dissociate in water. The molecular formulas for organic acids are commonly written with a −COOH ending or with the acidic hydrogen first to distinguish them from nonacidic isomers, e.g., C2H5COOH or HC3H5O2 for propanoic acid.

Some molecules dissolve in water without forming ions, thus they would be considered nonelectrolytes. These molecules are formed by covalent bonds and are able to dissolve due to interactions between their partial charges and the partial charges on water. In the case of water, this type of interaction is known as hydrogen bonding. Hydrogen bonding can occur between molecules in which hydrogen is covalently bonded to an electronegative element (oxygen, nitrogen, fluorine), and partially positive hydrogen will interact with an electronegative atom on another molecule.


Part B

Sort the following compounds based on whether or not they will behave as electrolytes that dissolve via ionization or as soluble nonelectrolytes that dissolve via hydrogen bonding.

Drag the appropriate formulas to their respective bins.

ANSWER:

Help Reset

partial chargesionsopposite

rm NaCl">NaCl

solutesugarnegativelypositively

rm CaCl_2">CaCl2

  • 1. Pure water contains only water molecules that interact strongly with each other due to their , which are graphically depicted as δ+ and δ−.
  • 2. Solutions are formed when a like a salt or sugar becomes homogeneously distributed in a solvent like water, and this distribution can be viewed in the Micro view.
  • 3. When salts dissolve, they separate into individual that strongly interact with the water molecules.
  • 4. Binary salts are made up of two elements at varying ratios, where one element is a charged cation, and the other is a charged anion.
  • 5. In the Micro view, each shake of the container releases 6 molecules of its respective solute, but 6 molecules of the salt actually produce more ions in solution than 6 molecules of the salt .
  • 6. Not all soluble molecules are salts, e.g., a covalent species like readily dissolves in water without forming ions.
  • 7. The reason they dissolve is because their partially charged atoms are able to associate with the partially charged atoms of water molecules, and these attractive forces occur as long as they are between atoms with charges.


partial chargesionsopposite

rm NaCl">NaCl

solutesugarnegativelypositively

rm CaCl_2">CaCl2Electrostatic Interactions

Solutes are constantly in motion when dissolved in a solvent. While in motion, similar charges on atoms will repel each other (+/+, +/δ+ , δ+ /δ+ , –/–, –/δ− , δ− /δ− ), whereas opposite charges will attract atoms to each other (+/–, +/δ− , –/δ+ , δ+ /δ− ). The oppositely charged interactions require the least amount of energy, thus molecules will predominantly orient in a manner that benefits electrostatic attraction.


Part C

Move each water molecule (each has a unique orientation) to the most ideal locations in the depicted scenario such that the strongest combination of electrostatic attractions are occurring.

Drag the appropriate water molecule orientations to their respective targets.

ANSWER:

Help Reset
  • PH3
  • C6H12O6
  • K2SO4
  • H2CO3
  • CaCl2
  • C12H22O11
  • Co(ClO)2
  • NaNO3
  • C3H8O3
  • Electrolytes that dissolve via ionization
  • Nonelectrolytes that dissolve via hydrogen bonding

Electrolytes, Conductivity, and Resistance

Conductivity is the ability for materials to allow the flow of electrons, ions, or both. This movement of charged species is generally known as current, and the prevention or inhibition of current is known as resistance. Even when there are ions in solution, resistance still exists, and there are limitations to the amount of current that can flow. This resistance causes the charged species to separate over some distance and creates a potential (also known as voltage). Increasing the concentration of ions reduces resistance. Conductive metals like silver, copper, and gold have negligible resistance to the flow of electrons; therefore resistors are always included in circuits to control the voltage.

V = IR

where V represents voltage (in volts, V ), I represents current (in amperes, A ), and R represents resistance (in ohms, Ω ).

Batteries are constructed to have an internal resistance that produces a known potential, which is why consumer batteries are rated at 1.5 V , 9 V , 12 V , etc. The following describes what happens when a circuit is shorted: a lower resistance pathway is introduced to a designed circuit, and electric current would flow through that new pathway instead since electrons and ions always travel the path of least resistance, and both the resistance and voltage would be considerably reduced (they can even approach values of zero).


Part D

In the PhET simulation, click the Macro menu in the top left corner of the screen. Notice the circuit available in the Conductivity tab on the right side of the screen. This circuit contains a negative electrode (green bar), positive electrode (red bar), battery, and light bulb. Drag this circuit to the beaker. Notice that you can change the length of the wire for both the electrodes by dragging them up or down (provide images with wire lengths changed). Sort the following the submersion scenarios according to whether they will result in a short circuit, a completed circuit that conducts electricity, or an incomplete circuit that does not conduct electricity.

Drag the appropriate items into the respective bins.

ANSWER:

Sample Exercise 4.1 Practice Exercise 1 with feedbackPart A - Relating Relative Numbers of Anions and Cations to Chemical FormulasIf you have an aqueous solution that contains 1.5 mol of HCl , how many moles of ions are in the solution?

ANSWER:

Help Reset
  • both electrodes in glucose solution
  • battery in pure water
  • battery in CaCl2 solution
  • negative electrode in pure water
  • both electrodes in CaCl2 solution
  • both electrodes in NaCl solution
  • negative electrode in NaCl solution
  • both electrodes in pure water
  • positive electrode in glucose solution
  • battery in NaCl solution
  • Does not complete the circuit
  • Completes the circuit
  • A short circuit exists

Correct HCl is a strong electrolyte that completely ionizes in water, thus 1.5 mol of hydrogen ions are formed and 1.5 mol of chloride ions are formed to give a total of 3.0 mol of ions when in aqueous solution.

Problem 4.19

Formic acid, HCOOH, is a weak electrolyte.

Part A

What solute particles are present in an aqueous solution of this compound?

Express your answer as a chemical formula. If there is more than one answer, enter each answer separated by a comma.

ANSWER:

1.0 1.5 2.0 2.5
3.0

Correct

Part B

Write the chemical equation for the ionization of HCOOH.

Express your answer as a balanced chemical equation. Identify all of the phases in your answer.

ANSWER:

HCOOH,H+,HCOO−

Correct

Net Ionic Equations

In many chemical reactions of two or more compounds, ions are present in the solution that do not participate in the reaction. These ions, known as spectator ions, can be removed from the chemical equation describing the reaction. A chemical equation that has the spectator ions removed is known as a net ionic equation.

Reactions between ionic compounds often result in the precipitation of an insoluble compound. Spectator ions and precipitation products of a chemical reaction can be predicted by consulting an ionic substance solubility table. Here is one such table:


HCOOH(aq)⇌H+(aq)+HCOO−(aq)

1 Most salts containing alkali metal or ammonium (NH4+) cations are soluble.
2 Most salts containing NO3− are soluble.
3 Most salts containing Cl− , Br− , and I− are soluble, except with metals such as Ag+ , Pb2+ , and Hg22+ .
4 SO42− salts with +1 cations and Mg2+ are soluble, but with most +2 cations these salts are insoluble.
5 Most compounds with OH− anions and group 1 or 2 cations are soluble. Notable exceptions are Be(OH)2 and Mg(OH)2 .

Entering chemical equations

When entering chemical equations, please follow the following formatting guidelines
  • Add phases to all chemical reactions unless asked otherwise. When adding phases, make sure not to place them as a subscript.
  • Do not stack charges and subscripts. To add a charge to an ion with a subscript, first enter the subscript. For example, to type SO42− start by typing SO4. Then use the arrow key to arrow over before adding a superscript with the charge.
If you make a formatting error, you will see an message such as "Check your placement of subscripts and superscripts." or "There is an error in your submission. Make sure you have formatted it properly." You will not lose any credit, but the answer will need to be edited and re-submitted. Part A

What is the net ionic equation of the reaction of FeCl2 with NaOH?

Express you answer as a chemical equation including phases.

ANSWER: Incorrect; correct answer displayed

Part B

What is the net ionic equation of the reaction of MgSO4 with Sr(NO3)2?

Express you answer as a chemical equation including phases.

ANSWER:

Fe2+(aq)+2OH−(aq)→Fe(OH)2(s)

Incorrect; correct answer displayed

Sample Exercise 4.3 Practice Exercise 1 with feedbackPart A - Predicting a Metathesis ReactionWill a precipitate form when solutions of Ba(NO3)2 and KOH are mixed?

ANSWER:

Sr2+(aq)+SO42−(aq)→SrSO4(s)

Correct Ba(NO3)2 and KOH mix to form Ba(OH)2 and KNO3 which are both soluble in water thus no precipitate will be formed.

Sample Exercise 4.4 Practice Exercise 1 with feedbackPart A - Writing a Net Ionic EquationWhat happens when you mix an aqueous solution of sodium nitrate with an aqueous solution of barium chloride?

ANSWER:

Yes
No

Correct

± Neutralizing an Acid Spill

A tanker truck carrying 5.04×103kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solution is 95.0%H2SO4 by mass and has a density of 1.84 g/mL.

Part A

Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 5.04×103kg of sulfuric acid solution?

Express your answer with the appropriate units.

ANSWER:

There is no reaction; all possible products are soluble. Only barium nitrate precipitates. Only sodium chloride precipitates. Both barium nitrate and sodium chloride precipitate.
Barium chloride is not soluble and it stays as a precipitate.

Correct

Problem 4.36

An aqueous solution of an unknown solute is tested with litmus paper and found to be acidic. The solution is weakly conducting compared with a solution of NaCl of the same concentration.

Part A

Which of the following substances could the unknown be: KOH, NH3, HNO3, KClO2, H3PO3, CH3COCH3 (acetone)?

Check all that apply.

ANSWER:

5170 kg

Incorrect; correct answer displayed

Oxidation-Reduction Reactions

Oxidation-reduction reactions (often called "redox" for short) are reactions that involve the transfer of electrons from one species to another. Oxidation states, or oxidation numbers, allow chemists to keep track of these electron transfers. In general, one element will lose electrons (oxidation), with the result that it will increase in oxidation number, and another element will gain electrons (reduction), thereby decreasing in oxidation number.

An oxidizing agent is an element or compound in a redox reaction that oxidizes another species and itself gets reduced and is therefore the electron acceptor in the reaction.

A reducing agent is an element or compound in a redox reaction that reduces another species and itself gets oxidized and is therefore the electron donor in the reaction.

As a summary, keep in mind the following:

  • Oxidation means an increase in oxidation state and a loss of electrons and involves a reducing agent.
  • Reduction means a decrease in oxidation state and a gain of electrons and involves an oxidizing agent.
Part A

Which element is oxidized in this reaction?

2CuO+C→2Cu+CO2

Enter the chemical symbol of the element.

ANSWER:  is oxidized

Part B

This question will be shown after you complete previous question(s).

Part C

Which element is reduced in this reaction?

2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH

Enter the chemical symbol of the element.

ANSWER:  is reduced

Part D

This question will be shown after you complete previous question(s).

Activity Series

One of the simplest types of oxidation-reduction reactions is the reaction of an aqueous cation with a free element. Whether or not a cation-element combination will produce a net change depends on the relative tendency of the various species to gain or lose electrons. An activity series ranks elements in order of their tendency to act as reducing agents. The elements at the top of the activity series are strong reducing agents and are easily oxidized. The elements at the bottom of the series are weak reducing agents. In other words, the reactions near the bottom tend to go backward rather than forward. The activity series can be used to predict reactions using the simple guideline: Any element higher on the activity series will reduce the cation of any element lower on the activity series. An abbreviated series is shown in the following:

KOH NH3 HNO3 KClO2 H3PO3 CH3COCH3
Oxidation reaction
Strongly
reducing
K K++e−
Ca Ca2++2e−
Mn Mn2++2e−
Cr Cr3++3e−
Fe Fe2++2e−
Sn Sn2++2e−
H2 2H++2e−
Cu Cu2++2e−
Weakly
reducing
Pt Pt2++2e−
Au Au3++3e−
Part A

Consider four beakers labeled A, B, C, and D, each containing an aqueous solution and a solid piece of metal. Identity the beakers in which a chemical reaction will occur and those in which no reaction will occur.

  1. Mn(s)+Ca(NO3)2(aq)
  2. KOH(aq)+Fe(s)
  3. Pt(NO3)2(aq)+Cu(s)
  4. Cr(s)+H2SO4(aq)
Drag each item to the correct bin.

ANSWER:

Part B

Consider the following data for five hypothetical elements: Q, W, X, Y, and Z. Rank the elements from most reactive to least reactive. Help Reset

  • A
  • B
  • C
  • D
  • Reaction
  • No reaction
Combination Observation of reaction
Q+W+ Reaction occurs
X+Z+ No reaction
W+Z+ Reaction occurs
Q++Y Reaction occurs
Place the element symbols from most to least reactive. To rank items as equivalent, overlap them.

ANSWER:

Sample Exercise 4.10 Practice Exercise 1 with feedbackPart A - Determining When an Oxidation-Reduction Reaction Can OccurWhich of these metals is the easiest to oxidize?

ANSWER:

Which of these metals is the easiest to oxidize?
Help Reset
  • Q
  • W
  • X
  • Y
  • Z
Least reactive Most reactive The correct ranking cannot be determined.
Problem 4.52Part A

Which of the following are redox reactions?

P4(s)+10HClO(aq)+6H2O(l)→4H3PO4(aq)+10HCl(aq)

Br2(l)+2K(s)→2KBr(s)

CH3CH2OH(l)+3O2(g)→3H2O(l)+2CO2(g)

ZnCl2(aq)+2NaOH(aq)→Zn(OH)2(s)+2NaCl(aq)

Check all that apply.

ANSWER:

Check all that apply.
sodium gold lithium iron
aluminum
Part B

For those reactions that are redox, indicate which elements are oxidized.

Express your answers as chemical symbols separated by a commas.

ANSWER:

P4(s) + 10HClO(aq) + 6H2O(l) → 4H3PO4(aq) + 10HCl(aq) Br2(l) + 2K(s) → 2KBr(s) CH3CH2OH(l) + 3O2(g) → 3H2O(l) + 2CO2(g)
ZnCl2(aq) + 2NaOH(aq) → Zn(OH)2(s) + 2NaCl(aq)
Part C

For those reactions that are redox, indicate which elements are reduced.

Express your answers as chemical symbols separated by a commas.

ANSWER:

Part D For those reactions that are not redox, indicate whether they are precipitation or neutralization reactions.

ANSWER:

For those reactions that are not redox, indicate whether they are precipitation or neutralization reactions.
± Ion Concentration

Ion concentration refers to the molar concentration of an ion in solution. It may be identical to, or greater or less than, the molar concentration of the compound containing the ion that was used to make the solution.

For soluble salts, the molarity of a particular ion is equal to the molarity of that compound times the subscript for that ion. For example, 1 M of AlCl3 is 1 M in Al3+ and 3 M in Cl− . 1 M of (NH4)2SO4 is 2 M in NH4+ and 1 M in SO42− .

Part A

What is the concentration of K+ in 0.15 M of K2S?

Express your answer to one decimal place and include the appropriate units.

ANSWER:

precipitation
neutralization
Part B If CaCl2 is dissolved in water, what can be said about the concentration of the Ca2+ ion?

ANSWER:

If rm CaCl_2 is dissolved in water, what can be said about the concentration of the rm Ca^{2+} ion?
Part C

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 750. mL of a solution that has a concentration of Na+ ions of 0.900 M ?

Express your answer to three significant figures and include the appropriate units.

ANSWER:

It has the same concentration as the Cl− ion. Its concentration is half that of the Cl− ion. Its concentration is twice that of the Cl− ion.
Its concentration is one-third that of the Cl− ion.
Sample Exercise 4.13 Practice Exercise 1 with feedbackPart A - Using Molarity to Calculate Grams of SoluteWhat is the concentration of ammonia in a solution made by dissolving 3.75 g of ammonia in 120.0 L of water?

ANSWER:

What is the concentration of ammonia in a solution made by dissolving 3.75 rm g of ammonia in 120.0 rm L of water?
mass of Na3PO4 = 
Problem 4.68

One cup of fresh orange juice contains 137 mg of ascorbic acid (vitamin C, C6H8O6).

Part A

Given that one cup = 261.8 mL calculate the molarity of vitamin C in organic juice.

Express your answer with the appropriate units.

ANSWER:

3.78×10−2 M 7.05 M 1.84×10−3 M 1.84 M
0.0313 M
Sample Exercise 4.15 Practice Exercise 1 with feedbackPart A - Using Mass Relations in a Neutralization ReactionHow many milligrams of sodium sulfide are needed to completely react with 25.00 mL of a 0.0100 M aqueous solution of cadmium nitrate, to form a precipitate of CdS(s) ?

ANSWER:

How many milligrams of sodium sulfide are needed to completely react with 25.00 rm mL of a 0.0100 rm M aqueous solution of cadmium nitrate, to form a precipitate of rm CdS (s)?
M = 
Sample Exercise 4.17 Practice Exercise 1 with feedbackPart A - Determining the Quantity of Solute by TitrationA mysterious white powder is found at a crime scene. A simple chemical analysis concludes that the powder is a mixture of sugar and morphine (C17H19NO3 ), a weak base similar to ammonia. The crime lab takes 10.00 mg of the mysterious white powder, dissolves it in 100.00 mL water, and titrates it to the equivalence point with 2.84 mL of a standard 0.0100 M  HCl solution. What is the percentage of morphine in the white powder?

ANSWER:

13.8 mg 19.5 mg 23.5 mg 32.1 mg
39.0 mg

Correct The number of moles of morphine that reacted with HCl was calculated by assuming a 1:1 stoichiometric ratio. Thus, the moles of morphine at the equivalence point was equal to the moles of HCl (0.0100M×0.00284L=2.84×10−5mol). The mass of morphine reacted was found by multiplying the moles reacted by the molar mass of morphine (0.008094 g). To determine the percentage of morphine in the white powder, the mass of morphine reacted was divided by the total mass of the white sample

0.008094g0.010g ×100 = 81.0%

Score Summary: Your score on this assignment is 40.2%.

You received 6.83 out of

Quiz Module 5 Page 1 of 11 Quiz Module 5 Due; 11:59pm on Sunday, March 19, 2017 You will recei e no credit for items you complete after the assignment is due. Grading Policy Give It Some Thought 6.1 Part A Our bodies are penetrated by X-rays but not by visible iight. Is this because X-rays travel faster than visible light? ANSWER: https://session.masteringchemistry.com/myct/assignmentPrintView?displayMode=studentVi. .. 4/4/2017
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Quiz Module 5 Page 2 of 11 Incorrect; correct answer displayed Problem 6.84 In August 2011, the Juno spacecraft was launched from Earth with the mission of orbiting Ju iter in 2016. The closest distance between Jupiter and Earth is 391 million miles. Part A If it takes 5.0 years for Juno to reach Jupiter, what is its average speed in mi/hr over this period? Express your answer using two significant figures. ANSWER: ® no O yes Part B Once Juno reaches Jupiter, what is the minimum amount of time it ta es for the transmitted signals to tra el from the spacecraft to Earth? Express your answer using three significant figures. ANSWER: V = I mi/hr Problem 6.3 The following diagrams represent two electromagnetic waves. Part A Which wave corresponds to the higher-energy ra iation? ANSWER: t = I min Problem 6.33 Molybdenum metal must absorb radiation with a minimum frequency of 1.09 10 ®before it can emit an electron from its surface ia the photoelectric effect. Part A What is the minimum energy needed to produce this effect? ANSWER: O Wave (b) has the higher energy. https://session.mastermgchemistry.com/myct/assignnientPrintView?displayMode=studentVi. .. 4/4/2017
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