SCIN131-Introduction to Chemistry-APUS-Final Assignment

Part 1 of 5 - 0.0/ 10.0 Points Please show as much work as possible for full credit. Question 1 of 16 0.0/ 5.0 Points What is the electron configuration for S? Electron configuration of S is 1s2 2s2 2p6 3s2 3p4 or [Ne] 3s2 3p4 Feedback: Electron Configurations are in Chapter 9 Question 2 of 16 0.0/ 5.0 Points State the octet rule. The octet rule states that atoms will react with each other in a way that will give them eight electrons on their valence shell ( outermost electron shell) It makes them stable.  Usually octet rule refers to electrons in s and p orbitals Feedback: This is discussed in chapter 10

Part 2 of 5 - 0.0/ 20.0 Points Question 3 of 16 0.0/ 5.0 Points An experiment shows that a 153-mL gas sample has a mass of 0.171 g at a pressure of 761 mm Hg and a temperature of 32oC. What is the molar mass of the gas? R=0.0821 L atm / mol K Using Ideal gas law PV = nRT 761 mm Hg = 1 atm 153 ml = 0.153 L 32 C = 305 K 1 * 0.153 = n * 0.0821 * 305 n = 0.153/ 25.04   n= 0.006 moles Number of moles = mass/Molar mass 0.006 = 0.171 / M Molar mass = 28.5 Feedback: Gas laws are in Ch 11 Question 4 of 16 0.0/ 5.0 Points Water can be formed according to the equation: 2H2 (g) + O2 (g) → 2H2O (g) If 8.0 L of hydrogen is reacted at STP, exactly how many liters of oxygen at STP would be needed to allow complete reaction? According to Avogadro law, equal volumes of different gasses contains equal number of moles   So mole ratio= volume ratio VH2:VO2=2:1  If Volume of H2=8.0L  Then Volume of O2=8.0/2 L Vol of O2 =4.0L Feedback: Gas Laws are found in Chapter 11 Question 5 of 16 0.0/ 5.0 Points Identify the Intermolecular forces from STRONGEST to WEAKEST (strongest on the top) and place the following compounds in the appropriate row by identifying which Intermolecular forces they have. Then rank the compounds from lowest boiling point to highest. CO2, Ne, H2O, CH4, HF Feedback: Intermolecular forces and an example of ranking boiling points can be found in Chapter 12. Question 6 of 16 0.0/ 5.0 Points Define Pressure. Pressure can be defined as force acting per unit area . Pressure =Force / Area  If area is small and force acting on it is high then that results in a higher pressure. If area is large and force acting on it is small, then that results in a lower pressure.

Part 3 of 5 - 3.0/ 20.0 Points Question 7 of 16 0.0/ 5.0 Points Calculate the mass of glucose (C6H12O6) in a 105-mL sample of a 1.61 M glucose solution. Number of moles = Molarity * Volume in litres n = 1.61 * 0.105 = 0.17 moles   1 mole Glucose weighs 180 grams. As molar mass of  glucose is 180 0.17 moles weighs 0.17*180 = 30.6 grams Question 8 of 16 3.0/ 5.0 Points Predict the products of the following Acid reactions: A)H3PO4(aq) + NaOH (aq) → Na3PO4 +  H20   ; B)  H2SO4(aq) + BaO(s) →  BaSO4  +  H20 are the equations balanced? no Answer Key: H2O|Na3PO4, H2O|Na3PO4, H2O|BaSO4, H2O|BaSO4, no Feedback: Acid reactions are found in Chapter 14 Question 9 of 16 0.0/ 5.0 Points Define “indicator” using the knowledge from your lab. An indicator is a substance that undergoes a distinct observable change when conditions in its solution change Example : A pH indicator changes color over a narrow range of pH values in solution  Question 10 of 16 0.0/ 5.0 Points Consider that there is 26.38 g NaCl in a solution that has a total volume of 1,780 mL with a density of 0.8495 g/mL. A) What is the molarity of NaCl? B)What is the mass/mass %? C)What is the molality? Given that there are 26.38 g of NaCl with total volume of 1780 mL and density of 0.8495 g/ml n= m/M = 26.38/ 58.5 = 0.451 mol NaCl A) Molarity = n/V = 0.451/ 1.78 = 0.253 M  B) d = m/v  So m = d×v  m = 1780× 0.8495 = 1512.11 g  Thus mass/mass% = 26.38/ 1512.11 = 0.0174 = 1.74 % C) 1512.11 g - 26.38 g = 1485.73 g water = 1.4857 kg water  Molality = 0.451 / 1.4857 = 0.3035 mol/kg

Part 4 of 5 - 0.0/ 20.0 Points Question 11 of 16 0.0/ 5.0 Points Provide an example of a common ether and its primary use. Example of an ether is Diethyl ether  It is a common ether and used as an early anesthetic It is also used as starting fluid for diesel engines ans as a refrigerant and in the manufacture of smokeless gunpowder and also used in perfumery Question 12 of 16 0.0/ 5.0 Points Explain why wet hair placed in curlers that is allowed to dry tends to retain the shape of the curl. Hair is made up of keratin proteins, natural oils, and hydrogen bonds. These ingredients work together to give hair its natural shape, color, etc. When we use a curling iron, the heat it produces breaks down hair’s hydrogen bonds, stripping away its natural oils and proteins. The heat changes hair’s texture, allowing to mold it to shape the curler have. The shape of a curler helps to mold hair into the shape of curler. As we hold hair in a particular shape for a few seconds, the bonds begin to reform into the new shape. The bonds will tend to hold hair in its new shape until they’re broken down again. Question 13 of 16 0.0/ 5.0 Points What is the difference between a simple sugar and a complex carbohydrate? Simple Sugars are made by plants, and complex carbohydrates are made by animals.  Complex carbohydrates are polymers of simple sugars.  Complex carbohydrates are liquid at room temperature, and sugars are solid. Question 14 of 16 0.0/ 5.0 Points Describe the goal of the human genome project. Goals of human genome project are to  Create genetic and physical maps of all chromosomes  Determine the sequence of 3 billion base pairs of DNA in the genome Identify the entire set of genes in the genome Analyze genetic variations among human (SNPs) Map and sequence the genome of model organisms Examine ethical, social and legal issues

Part 5 of 5 - 0.0/ 30.0 Points Question 15 of 16 0.0/ 15.0 Points  CO(g) + NO(g) → CO2(g) + N2(g) (a) How many grams of N2(g) will be formed from 5.19 x 1022 molecules of CO(g) at STP? (b) Name each chemical. When we balance the reaction, it becomes: 2 CO + 2 NO  2 CO2 + N2 Using Avagadro's Number, converting molecules of CO to mol of CO: 5.19 x 10^22 molecules divided by Avogadro's number (6.022 x 10^23 molecules per mol) yields 0.0862 mol CO According to the balanced reaction, for every 2 mol of CO consumed, 1 mol of N2 is produced.  Assuming NO(g) is in excess and all CO is consumed to produce the maximum theoretical yield, there should be 0.0431 mol of N2 produced. Converting mol of N2 to grams of N2 by multiplying 0.0431 mol of N2 by the molecular weight of N2 (which is 28.02 g/mol) to convert units into grams, yielding the answer to be 1.21 grams of N2 produced. b)   CO is Carbon Monoxide         NO is Nitrogen Monoxide    CO2 is Carbon Dioxide    N2 is Elemental/Natural Nitrogen Question 16 of 16 0.0/ 15.0 Points A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine:  C6H8O6(aq) + I2(aq) →    C6H6O6(aq) + HI(aq) (a) A 25.0 mL sample of juice requires 11.8 mL of a 0.0164 M  I2 solution for reaction. how many moles of ascorbic acid are in the sample? (b) What is the molarity of the acid? (c) If a person wanted to consume the FDA recommended 60 mg of Vitamin C, how many ounces of juice would that person need to consume? (4 qts = 128 fluid oounces; 1L = 1.057 qt) A) (11.8mL) ×(0.0164 M) = 1.9352 × 10^-4 moles of I2 for reaction to occur. The equation is balanced ( 2HI is ignored ) the I2 reacts with ascorbic acid at a 1:1 molar ratio. Therefore, if there is 1.9352 ×10-4 moles of I2 there are same number of moles of ascorbic acid B) If we have 1.9352 × 10^-4 moles of acid, dividing that by total volume of solution 25 mL gives molarity M = (1.9352 × 10^-4 )/ 25 ×10^-3 = 7.7 ×10^-3 M C) 60mg = 0.60 grams Moles of vitamin C = 0.60/176 = 3.4 × 10^-4 moles By scaling up and down 25 mL juice over 1.9352 × 10^-4 = 3.4 ×10^-4 moles vitamin c needed over X ml juice we get 44.2 ml = 1.49 ounces of juice

Part 1 of 5 - 10.0/ 10.0 Points Please show as much work as possible for full credit. Question 1 of 16 5.0/ 5.0 Points State the octet rule. The octet rule states that elements gain or lose electrons to attain an electron configuration of the nearest noble gas.  Feedback: This is discussed in chapter 10 Question 2 of 16 5.0/ 5.0 Points What is the electron configuration for S? Electronic configuration of Sulfur is [Ne] 3s23p4 Feedback: Electron Configurations are in Chapter 9

Part 2 of 5 - 20.0/ 20.0 Points Question 3 of 16 5.0/ 5.0 Points Describe (using what you know about gases) why a can of compressed air gets cold after you use it to clean your keyboard. The volume of gas in the can does not change much, but the pressure drops significantly. Correspondingly, the temperature drops, and the can gets cold. We can use the ideal gas law, approximately: PV = nRT. With V roughly constant, temperature varies with pressure. Question 4 of 16 5.0/ 5.0 Points An experiment shows that a 153-mL gas sample has a mass of 0.171 g at a pressure of 761 mm Hg and a temperature of 32oC. What is the molar mass of the gas? R=0.0821 L atm / mol K 1. V = 153/ 1000 = 0.153 L  P = 761/760 = 1 atm  R = .0821  T = 32 +273.15 = 305.15 K  n = ?  PV= nRT  1 x 0.153 = n x .0821 x 305.15  n = 0.00611 moles  Molar mass = g /mol = 0.171g/0.00611 = 27.98 g/mol  Feedback: Gas laws are in Ch 11 Question 5 of 16 5.0/ 5.0 Points A 375 mL sample of gas is initially at a pressure of 721 torr and a temperature of 32°C. If this gas is compressed to a volume of 286 mL and the pressure increases to 901 torr, what will be the new temperature of the gas (reported to three significant figures in °C)?  P1V1  =   P2V2                        T1         T2                                                             T2 =  P2V2T1                                                                   P1V1 Converting Celsius into Kelvin  Kelvin = 273+oC K=273+32 K=305 •   (721 torr x 375mL)  =  (901 torr  x  286mL) •            305K                              T2 •   T2 = (901 torr x 286mL x 305K) •                 (721 torr  x 375mL) •   T2 = 290.68K •   290.68-273= 17.68 •   T2 = 17.68oC Feedback: Gas Laws are found in Chapter 11 Question 6 of 16 5.0/ 5.0 Points Define Liquid. “Liquid is a physical state of matter characterized by…” A liquid is a physical state of matter characterized by kinetic energy and potential energy of about the same magnitude.

Part 3 of 5 - 19.0/ 20.0 Points Question 7 of 16 5.0/ 5.0 Points Define “indicator” using the knowledge from your lab. Chemical indicator is any substance that gives a visible sign, usually by a colour change, of the presence or absence of a threshold concentration of a chemical species, such as an acid or an alkali in a solution. An example is the substance called methyl yellow, which imparts a yellow colour to an alkaline solution. Question 8 of 16 5.0/ 5.0 Points If a 10.00mL sample of unknown concentration of HF is titrated with 28.94mL of 0.1040M NaOH, what is the concentration of HF? Using the formula M1*V1=M2*V2 Here, we are looking for M1. 10.00ml * x = 28.94mL * 0.1040M 10.00ml * x = 3.009M x =0.3009M Concentration of HF is 0.3009 M Feedback: Found in Chapter 14 and relates to the lab Question 9 of 16 5.0/ 5.0 Points Consider that there is 26.38 g NaCl in a solution that has a total volume of 1,780 mL with a density of 0.8495 g/mL. A) What is the molarity of NaCl? B)What is the mass/mass %? C)What is the molality? 26.38 g NaCl: n = m/M = 26.38 / 58.5 = 0.451 mol NaCl. A) Molarity = n / v = 0.451 / 1.78 = 0.253 mol/L. B) d = m / v   m = 1780 x 0.8495 = 1512.11 g.  26.38 / 1512.11 = 0.0174 = 1.74%. C) 1512.11 g - 26.38 g = 1485.73 g water = 1.48573 kg water.  Molality = 0.451 / 1.48573 = 0.30355 mol/kg. Question 10 of 16 4.0/ 5.0 Points Predict the products of the following Acid reactions: A)H3PO4(aq) + NaOH (aq) → Na3PO4 (aq) +  H2O   ; B)  H2SO4(aq) + BaO(s) →  BaSO4 (s)  +  H2O are the equations balanced? no Answer Key: H2O|Na3PO4, H2O|Na3PO4, H2O|BaSO4, H2O|BaSO4, no Feedback: Acid reactions are found in Chapter 14 Comment: See attached document for the correct answer.     Acid question APUS.docx     (50 KB)

Part 4 of 5 - 20.0/ 20.0 Points Question 11 of 16 5.0/ 5.0 Points Provide an example of a common amine and where you might find it. Amines are organic compounds and functional groups that contain a basic nitrogen atom with a lone pair. Amines are derivatives of ammonia, wherein one or more hydrogen atoms have been replaced by a substituent such as an alkyl or aryl group Many drugs are designed to mimic or to interfere with the action of natural amine neurotransmitters, exemplified by the amine drugs: Chlorpheniramine is an antihistamine that helps to relieve allergic disorders due to cold, hay fever, itchy skin, insect bites and stings. Question 12 of 16 5.0/ 5.0 Points Provide an example of a common ether and its primary use. Ethers are a class of organic compounds that contain an ether group—an oxygen atom connected to two alkyl or aryl groups—of general formula R–O–R'. These ethers can again be classified into two varieties, if the alkyl groups are the same on both sides of the oxygen atom then it is a simple or symmetrical ether. Whereas if they are different the ethers are called mixed or unsymmetrical ethers A typical example is the solvent and anesthetic diethyl ether, commonly referred to simply as "ether" (CH3-CH2-O-CH2-CH3). Ethers are common in organic chemistry and pervasive in biochemistry, as they are common linkages in carbohydrates and lignin. Question 13 of 16 5.0/ 5.0 Points Describe how DNA replication occurs. DNA replication ( occurs from 5'----------------------3')  -occurs in the S phase and in the nucleus which is where DNA is  3 enzymes are important  helicase  primase  ligase  Polymerase  *helicase unwinds the strands of DNA to produce two separate strand.  *Primase acts as an enzyme that produces RNA primer which serves as a template in which new DNA is synthesized *Polymerase helps for DNA synthesis  *LIgase joins the two strands together after synthesis  Other enzymes are topoisomerase e.t.c  Question 14 of 16 5.0/ 5.0 Points What kinds of molecules are involved in smell? Organic Molecules are involved in the sense of smell; specifically esters, amines, ketones, and even aldehydes.

Part 5 of 5 - 30.0/ 30.0 Points Question 15 of 16 15.0/ 15.0 Points A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine:  C6H8O6(aq) + I2(aq) →    C6H6O6(aq) + HI(aq) (a) A 25.0 mL sample of juice requires 11.8 mL of a 0.0164 M  I2 solution for reaction. how many moles of ascorbic acid are in the sample? (b) What is the molarity of the acid? (c) If a person wanted to consume the FDA recommended 60 mg of Vitamin C, how many ounces of juice would that person need to consume? (4 qts = 128 fluid oounces; 1L = 1.057 qt) The balanced equation shows that the molar ratio between ascorbic acid and I2 is one to one.  Moles I2 = Molarity times volume in Liters  Moles I2 0.0164 molar times 0.0118 liters = l.93 x l0-4 moles I2 and also moles ascorbic acid in the sample.  Molarity of acid = moles acid over volume in liters = l.93 x l04 moles over .025 liters =7.7 x l0^-3rd molar acid  c 60mg = .060 grams. Moles vitamin C = .060 grams over 176 grams/mole = 3.4 x l0^-4 moles.  From previous work we know that 25 ml of juice contains l.93 x l0^-4 moles vitamin C  so we can scale up or down.  25 ml juice over l.93 x l0^-4 moles = 3.4 x l0^-4 moles vitamin C needed over X ml juice  Cross multiplying and solving for X ml juice.  We get 44.2 ml. Then divide by 29.6 ml/fluid ounce = 1.49 fluid ounces if juice Question 16 of 16 15.0/ 15.0 Points  CO(g) + NO(g) → CO2(g) + N2(g) (a) How many grams of N2(g) will be formed from 5.19 x 1022 molecules of CO(g) at STP? (b) Name each chemical. When you balance the reaction, it becomes: 2 CO + 2 NO  2 CO2 + N2 Using Avagadro's Number, convert molecules of CO to mol of CO: 5.19 x 10^22 molecules divided by Avogadro's number (6.022 x 10^23 molecules per mol) yields 0.0862 mol CO According to the balanced reaction, for every 2 mol of CO consumed, 1 mol of N2 is produced.  Assuming NO(g) is in excess and all CO is consumed to produce the maximum theoretical yield, there should be 0.0431 mol of N2 producd. Converting mol of N2 to grams of N2 by multiplying 0.0431 mol of N2 by the molecular weight of N2 (which is 28.02 g/mol) to convert units into grams, yielding the answer to be 1.21 grams of N2 produced. b)   CO is Carbon Monoxide         NO is Nitrogen Monoxide    CO2 is Carbon Dioxide    N2 is Elemental/Natural Nitrogen

Part 1 of 5 - 10.0/ 10.0 Points Please show as much work as possible for full credit. Question 1 of 16 5.0/ 5.0 Points The octet rule is extremely important in science. Answer the following regarding the octet rule. What is it? Why is it important relative to what we learned in this lesson? Are there any exceptions to this rule? If so, explain in detail. Be sure to base your answer on what we discussed in THIS class and use the terminology from the course materials. The Octet rule states that atoms will react with each other in a way that will give them eight electrons in their valence shell (Outer most electron shell). It makes them stable. Usually octet rule refers to electrons in s and p orbitals. For example take a look at Carbon. It has four valence electrons and if we were to draw its Lewis dot structure, we would see one electron on each side of the Carbon. For that reason, Carbon is highly reactive where it can bond with almost anything! And this rooted the separate branch of chemistry namely "Organic Chemistry". So as it determines the type of bonding, its really important in Science and we have learned about different types of bonds in this lesson and thats the relevant part here about Octet rule.  There are three general exceptions to the octet rule:  (1) Molecules, with an odd number of electrons : Because most molecules or ions that consist of s- and p-block elements contain even numbers of electrons, their bonding can be described using a model that assigns every electron to either a bonding pair or a lone pair. There are, however, a few molecules containing only p-block elements that have an odd number of electrons. (2) Molecules in which one or more atoms possess more than eight electrons, such as SF6 : The most common exception to the octet rule is a molecule or an ion with at least one atom that possesses more than an octet of electrons.  (3) Molecules, in which one or more atoms possess less than eight electrons, such as BCl3 : Molecules with atoms that possess less than an octet of electrons generally contain the lighter s and p block elements, especially beryllium, typically with just four electrons around the central atom, and boron, typically with six.  Feedback: This is discussed in chapters 9 and 10 Comment: good! Question 2 of 16 5.0/ 5.0 Points What are the full and abbreviated electron configurations for the following elements. Be sure to explain how you arrived at this configuration.  Explain it to me--don't just give me the configuration. a) Arsenic b) Tungsten a) Arsenic:  Atomic number of Arsenic is 33, Full Electronic Configuration : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 So total electrons in the first, second , third and fourth subshells are 2,,8,18,5, which means it has 5 valence shells.  Abbreviated Electronic Configuration : As the nearest noble gas is Argon (Stable), the abbreviated electronic configuration can be given as :   [Ar] 3d10 4s2 4p3   b) Tungsten:  Atomic number of Tungsten is 74, Full Electronic Configuration : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d4 6s2 So total electrons in the first, second , third,  fourth, fifth and sixth subshells are 2,,8,18,32,12 and 2 which means it has 2 valence shells. Abbreviated Electronic Configuration : As the nearest noble gas is Xenon (Stable), the abbreviated electronic configuration can be given as :  [Xe] 4f14 5d4 6s2 Model Short Answer: chapter 9 Feedback: chapter 9

Part 2 of 5 - 15.0/ 20.0 Points Question 3 of 16 0.0/ 5.0 Points Which compound will have a higher boiling point, HF or HBr? Explain how you know and why, in detail, using the information from the course materials. What is the most prominent intermolecular force present? HF will have the higher boiling point because of its capability to hydrogen bond with itself. HBr and HF both have dipole-dipole attractions, but only HF can hydrogen bond. They both also have dispersion forces, but the hydrogen bonding will predominate over the increased dispersion forces in HBr due to the increase in molecular weight. Feedback: Intermolecular forces can be found in Chapter 12. Comment: This was copied from my answers on a previous quiz, which is the same as copying from your book. You need to put answers in your own words. Question 4 of 16 5.0/ 5.0 Points If I have a tank that is holding 7.50 L of elemental nitrogen gas with a mass of 250.0 g at a temperature of 58.0 degrees C, what is the pressure? You must show all work to receive credit.   From Ideal Gas Equation, PV = nRT  P = nRT/V  V = 7.50  L  T = 58 C =  331 K  mass = 250 g Molar mass = Mass/n, so n = 250/28 = 8.929 moles P = 8.929*(0.0821)*331/7.50 P = 32.4 atm.  Feedback: Gas laws are in Ch 11 Comment: good Question 5 of 16 5.0/ 5.0 Points You have a gas that occupies 9.50L at 55.0*C and 12.5 atm. Calculate the final volume after the temperature is decreased to 30*C and the pressure is decreased to 2.50 atm. You must show all work to receive credit. As the relation between, the pressure, velocity and temperature is given by,   Here, from the question, we can see that P1= 12.5 atm V1= 9.50 L T1= 55 C = 328.15 K  P2 = 2.5 atm T2 = 30 C = 303.15 K By plugging in these values into the equation and solving for V2  so,   so, V2 = 12.5*9.5*303.15 divided by (328.15*2.5) = 43.881228 L so 43.88 L rounded to two decimals.    Feedback: Gas laws are in Ch 11 Comment: Good! Question 6 of 16 5.0/ 5.0 Points A gas initially occupies 38.0 L at 25*C and 3.15 atm. Calculate the volume after the temperature is increased to 68*C and the pressures is decreased to 1 atm. You must show all work to receive credit and don't forget units in your answer. As the relation between, the pressure, velocity and temperature is given by,   Here, from the question, we can see that P1= 3.15 atm V1= 38 L T1= 25 C = 298.15 K  P2 = 1 atm T2 = 68 C = 341.15 K By plugging in these values into the equation and solving for V2  so,   so, V2 = 3.15*38*341.15 divided by (298.15*1) = 136.9634 L so 136.96 L rounded to two decimals.    Feedback: Gas laws are in Ch 11 Comment: good job! You would round to the closest significant figures, which is 3 and the answer would be 137 L.

Part 3 of 5 - 10.0/ 20.0 Points Question 7 of 16 5.0/ 5.0 Points Calculate the number of grams of sodium bicarbonate, NaHCO3 in each of the following solutions. You must show your work to receive credit. a) 25.0 mL of 0.975M  b) 1.30 L of 2.50 M c) 625 mL of 6.80 M d) 3.00 L of 0.025 M Number of moles = Molarity * Volume in litres a) 25.0 mL of 0.975M  n = 0.975 * 0.025 = 0.024375 moles  1 mole of NaHCO3 weighs 84 grams. As molar mass is 84 grams 0.024375 moles weighs  0.024375*84 = 2.0475 grams   b) 1.30 L of 2.50 M n = 2.5 * 1.30 = 3.25 moles  1 mole of NaHCO3 weighs 84 grams. As molar mass is 84 grams 3.25 moles weighs  3.25*84 = 273 grams   c) 625 mL of 6.80 M n = 6.80 * 0.625 = 4.25 moles  1 mole of NaHCO3 weighs 84 grams. As molar mass is 84 grams 4.25 moles weighs  4.25*84 = 357 grams   d) 3.00 L of 0.025 M n = 0.025 * 3 = 0.075 moles  1 mole of NaHCO3 weighs 84 grams. As molar mass is 84 grams 0.075 moles weighs  0.075*84 = 6.3 grams Comment: good Question 8 of 16 0.0/ 5.0 Points Explain the difference between a strong and weak acid AND between a strong and weak base. Use terminology and concepts from this lesson and provide sufficient detail. Acids are molecular substances. Strong acids completely ionize in water, whereas weak acids do not. Generally, weak acids do not ionize well in water and do not produce H+ nearly as easily or frequently as strong acids.  Example Acetic acid is week because only 1 in 10,000 acid molecules break up to give the H+ ion in solution. HCl is a strong acid because it completely dissociates into H+ and Cl-   Strong bases are bases that are fully ionic compounds, although some may not readily dissolve or ionize in water. Strong bases produce hydroxide ions in water, whereas weak bases do not. For example, ammonia does not produce hydroxide ions in water, but can act as a hydrogen acceptor and form ammonium, which in turn will form some hydroxide (not much). A weak base does not completely convert into hydroxide ions. Example NaOH is a stronger base than diethylamine.  Comment: You examples are good, however the text of your answer matches my response to this question previously. If you can provide a reason, I will award points. Question 9 of 16 5.0/ 5.0 Points Part 1: Determine the milliliters of concentrated sulfuric acid (12.0M) you would need to make each of the following solutions. You must show your work to receive credit. a) 5.00 L of 0.25M b) 25.0 mL of 2.50 M c) 775 ml of 8.50 M d) 7.50 L of 5.00 M Part II:  How would you actually make the solutions above?  Describe the steps you would take in the lab, in detail using the terminology discussed in this lesson.   Using the dilution formula,  M1V1=M2V2 a) 5.00 L of 0.25M M1 =12.0 M V2 = 5.00 L = 5000 mL M2 = 0.25 M so 12.0V1=5000*0.25 V1 = 104.166 = 104.17 mL Part II : This means you want to add 104.17 mL of the 12.0 M H2SO4 into a 200-mL volumetric flask and then fill the rest with distilled water up to the etch line in the flask and then will mix 4 L of distilled water into this solution using the same volumetric flask.    b) 25.0 mL of 2.50 M M1 =12.0 M V2 = 25.0 mL  M2 = 2.50 M so 12.0V1=25*2.5 V1 = 5.208 = 5.21 mL Part II : This means you want to add 5.21 mL of the 12.0 M H2SO4 into a 25-mL volumetric flask and then fill the rest with distilled water up to the etch line in the flask.  c) 775 ml of 8.50 M M1 =12.0 M V2 = 775 mL M2 = 8.50 M so 12.0V1=775*8.5 V1 = 548.958 = 548.96 mL Part II : This means you want to add 548.96 mL of the 12.0 M H2SO4 into a 1000-mL volumetric flask and then fill the rest with distilled water up to  775mL in the flask.    d) 7.50 L of 5.00 M M1 =12.0 M V2 = 7.50 L = 7500 mL M2 =5.00 M so 12.0V1=7500*5.00 V1 = 3125 = 3125 mL Part II : This means you want to add 1000 mL of the 12.0 M H2SO4 into a 1-L volumetric flask thrice and then add 125mL of it into a 1-L Volumetric flask,  rest with distilled water up to the etch line in the flask and then will mix 7.5 L of distilled water into this solution using the same volumetric flask.    Comment: good! Question 10 of 16 0.0/ 5.0 Points You titrate 25.0 mL of vinegar (acetic acid) with 8.50 mL of 1.01 M NaOH. Determine the concentration (M) of the acetic acid.   M = mol/L   HC2H3O2 + NaOH --> NaC2H3O2 + H2O You must show your work to receive credit. Using the dilution formula,  M1V1=M2V2 M1 = ? V1= 25.0 mL V2 = 8.50 L = 8500 mL M2 = 1.01 M so M1*25.0=8500*1.01 M1 = 343.4 M   Comment: You cannot use M1V1=M2V2 for this calculation. This is a titration and not a dilution.

Part 4 of 5 - 13.75/ 20.0 Points Question 11 of 16 5.0/ 5.0 Points For each compound below, write the chemical formula and the structural formula. You must show your work and discuss/explain how you reached your answer to receive credit.  (a) 2-heptene , (b) 1-pentene, (c) 3-octyne Hept represents 7 carbons, the -ene means there's at least one double bond, and the 2 tells you the double bond occurs from carbon 2 to 3. (a) C7H14       -     H3C-CH=CH-CH2-CH2-CH2-CH3  Pent represents 5 carbons, the -ene means there's at least one double bond, and the 1 tells you the double bond occurs from carbon 1 to 2.  (b) C5H10      -      H2C=CH-CH2-CH2-CH3    Oct represents 8 carbons, the -yne means there's at least one triple bond, and the 3 tells you the double bond occurs from carbon 3 to 4.  (c) C8H14       -      H3C-CH2-C≡C-CH2-CH2-CH2-CH3    Feedback: Organic chemistry is discussed in Ch. 18 Comment: Good! Question 12 of 16 2.5/ 5.0 Points Explain the terms "organic" and "inorganic" in respect to their usage in classifying compounds.   Historically, what was the difference between organic and inorganic?  How has that changed today in terms of compounds?  Include the terminology and content from this lesson and give examples as necessary. By 18th century, Chemists divided the compounds into two broad categories as organic (Obtained from living beings) and inorganic(Obtained from Earth). They observed that organic compounds are easily decomposable where as the inorganic are not. They also observed that Organic cannot be synthesized in the laboratories where as many inorganic compounds were easily synthesized. These properties of the organic compounds made them to postulate that they are unique to living organisms. They assumed that the living organisms employed a vital force and this belief rooted to Vitalism. This has been completely changed today.    Feedback: Organic chemistry is discussed in Ch. 18 Comment: Good! How has the definition been changed today and how do we define inorganic compounds today? Question 13 of 16 2.5/ 5.0 Points Name the following compounds according to what you learned in this Lesson. You must spell all names correctly and explain how you arrived at your answers in order to receive credit:   a)   b)   c)    a) From the given  image we can see that the Chemical formula C6H14O which means we have hexa--6 carbons, and as per lowest sum rule of IUPAC, it would be 3-Hexanol. b) Ethoxy Ethane and its common name is Diethyl Ether. chemical formula is C4H10O c) Pentanoic acid, because of the presence of 5 carbons and due to presence of carboxylic acid COOH. Feedback: Organic chemistry is discussed in Ch. 18 Comment: You're on the right track. For (a) your explanation doesn't explain how you got the -ol portion of the name. For (b), I am surprised that you came up with ethoxy ethane. Your explanation doesn't provide reasoning for either that or diethyl ether. For (c) yes, the COOH is a carboxylic acid group which means the name ends in -oic acid. Question 14 of 16 3.75/ 5.0 Points (1) What is meant by the primary structure of a protein molecule? How does it differ from the secondary structure of a protein? (2) How are secondary protein structures held together? (3) Discuss the importance of the tertiary and quaternary structures of proteins. Do all proteins have primary, secondary, tertiary, and quaternary structures? Explain. You must provide sufficient detail in your answer and incorporate information from the course materials to receive credit. Primary structure is simply the sequence of residues making up the protein. Thus primary structure involves only the covalent bonds linking residues together. Secondary structure describes the local folding pattern of the polypeptide backbone and is stabilized by hydrogen bonds between N-H and C=O groups. Tertiary structure describes how regions of secondary structure fold together, that is, the 3D arrangement of a polypeptide chain, including a helices, b sheets, and any other loops and folds. In quaternary, Some proteins are composed of more than one polypeptide chain. Comment: Good information, however you did not answer the last question about what types of structure do all proteins have.

Part 5 of 5 - 18.0/ 30.0 Points Question 15 of 16 3.0/ 15.0 Points A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine:  C6H8O6(aq) + I2(aq) →    C6H6O6(aq) + HI(aq) (a) A 25.0 mL sample of juice requires 11.8 mL of a 0.0164 M  I2 solution for reaction. how many moles of ascorbic acid are in the sample? (b) What is the molarity of the acid? (c) If a person wanted to consume the FDA recommended 60 mg of Vitamin C, how many ounces of juice would that person need to consume? (4 qts = 128 fluid oounces; 1L = 1.057 qt) (a)   (11.8mL)*(0.0164 M) = 1.9352x10^-4 moles of I2 for reaction to occur.  reaction is balanced (sort of should be 2HI, but regardless)  the I2 reacts with ascorbic acid at a 1:1 molar ratio.  Therefore, if there is 1.9352x10^-4 moles of I2 there are the same number of moles of ascorbic acid. <round to 3 > (b)  This I'm not 100% on but it seems to make sense to me.  if we have 1.9352x10^-4 moles of the acid, divide that by total volume of solution.  25.0mL of juice + 11.8mL of Iodine = 36.8 mL total volume.    (1.9352x10^-4 moles)/36.8mL = 5.26x10^-3 M (c)  (' x' L of juice)*(5.26x10^-3 mole/liter)*(176.12 g/mole) = 60 mg of Vit C Algebra is your friend and you should get something like, 0.064767 L of juice Convert using giving ratios for L to Qts., Qts. to Oz. 2.19 oz  Comment: You have the right idea.  In (a), you forgot to convert mL to L since Molarity is mol/L. Yes, you do have to balance the reaction. Molarity is mol/L. For (b) take what you calculated in a for moles and divide by the volume given for ascorbic acid in Liters.  For C, you need to convert from the mg needed to moles using mg-->g and molar mass for moles. Once you have moles, you can use Molarity to get liters and then calculate fluid ounces. Question 16 of 16 15.0/ 15.0 Points  CO(g) + NO(g) → CO2(g) + N2(g) (a) How many grams of N2(g) molecules will be formed from 5.19 x 1022 molecules of CO(g) at STP? (b) Name each chemical. a) When you balance the reaction, it becomes: 2 CO + 2 NO  2 CO2 + N2 Using Avagadro's Number, convert molecules of CO to mol of CO: 5.19 x 10^22 molecules divided by Avogadro's number (6.022 x 10^23 molecules per mol) yields 0.0862 mol CO According to the balanced reaction, for every 2 mol of CO consumed, 1 mol of N2 is produced.  Assuming NO(g) is in excess and all CO is consumed to produce the maximum theoretical yield, there should be 0.0431 mol of N2 producd. Convert mol of N2 to grams of N2 by multiplying 0.0431 mol of N2 by the molecular weight of N2 (which is 28.02 g/mol) to convert units into grams, yielding the answer to be 1.21 grams of N2 produced. b)   CO is Carbon Monoxide         NO is Nitrogen Monoxide    CO2 is Carbon Dioxide    N2 is Elemental/Natural Nitrogen Comment: excellent