CHEM 120 Unit 1 Quiz (GRADED A) Question and Answer Solutions | 100% VERIFIED.

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CHEM 120 Unit 1 Quiz 1. Question : (TCO 1) You have been growing the same variety of corn for over 5 years. In the past two years, you have noticed that the yield has decreased by 10% each... year. You grow corn on 10 acres of land and your yield should be 1,000 bushels per acre. Growing conditions have been consistent for the past 5 years. You want to identify another variety of corn that will produce 1,000 bushels per acre. You currently grow Variety ZQB. You want to test two new varieties, CAQ and TWX. You hypothesize that CAQ will produce 1,000 bushels per acre. Design an experiment to test this hypothesis. Include your control(s) (2 pts) and your experimental design (4 pts). Describe the experimental results that would support your hypothesis. (2 pts) Be specific. Student Answer: Fist I need to identify the question and decide which variety of corn that will produce 1,000 bushels per acre. My hypothesis would be that CAQ will produce 1,000 bushels per acre. My control group would be the variety that I am currently growing ZQB. ZQB which has produced 1,000 bushels per acre for a period of 5 years. To conduct the Experiment I will use 6 acres of land, plant CAQ variety on 3 and TWX variety on the other 3 acres on same day. Procedure: I will water all 10 acres of land at the same time every day. Each section of land will be fertilized land on same day. I will then conduct observations over the next 2 years and document accordingly. Post the second year I will take inventory of each variety of bushels and calculate an average for each acre. ZQB- produced 900 bushels per acre for year 1 ZQB- produced 850 bushels per acre for year 2. This averages to 875 bushels per acre per year. CAQ- produced 900 bushels per acre for year 1 CAQ- produced 1400 bushels per acre for year 2. This averages to 1150 bushels per acre per yea.r TWX- produced 650 bushels per acre for year 1 TWX- produced 750 bushels per acre for year 2. The average of both years is 700 bushels per acre per year. Just by looking at the numbers there is a substantial distance between the averages of bushels per acre per year. CAQ averaged 1150 bushels per each acre per year, which exceeded my hypothesis. In comparison to the average of TWX 700 and ZQB 875 bushels per acre per year. Instructor Explanation: Answers will vary. Chapter 1 and lecture. Points Received: 8 of 8 Comments: nice Question 2.Question : (TCO 1) How many meters are in 9 yards? Show your work. Student Answer: 1 yard=0.914 m 0.914m x 9y= 8.226 (8.23) meters Instructor Explanation: x meters = 9 yd x (0.914 m/1 yd) = 8.23 m. (Chapter 1; 1 pt for units, 1 point for conversion factor; 1 point for answer) Points Received: 3 of 3 Comments: Question 3.Question : (TCO 1) How many mm are in 8.7 inches? Show your work. Student Answer: 1 inch= 25.4mm 25.4mm x 8.7in= 220.98mm Instructor Explanation: x inches = 8.7 inch x (2.54 cm/1 in) x (10 mm/1 cm)= 220.98 mm (Chapter 1; 1 pt for units, 2 point for conversion; 1 point for answer) Points Received: 4 of 4 Comments: Question 4.Question : (TCO 1) What is the mass of 50.0 mL of an intravenous glucose solution with a density of 1.15 g/mL? Show your work. Student Answer: m=d x v m= 1.15 g/mL x50.0mL m= 57.5g Instructor Explanation: Density = mass in g/volume in mL; m = d * v; x g = (50.0 mL) x (1.15 g/mL) = 57.5 g (Chapter 1; 1 pt for units, 2 point for showing work; 1 point for answer) Points Received: 4 of 4 Comments: Question 5.Question : (TCO 2) Determine the number of protons, neutrons and electron for the element Sodium. The mass number for the isotope of this element is 25. You must show your work for each determination to receive credit. (2 pts per determination) Student Answer: The atomic # of Na is 11 from the periodic table. The number of protons and electrons is 11 each. The mass is 27.99. Number of neutrons= mass - protons neutrons= 25 - 11= 14 protons 11 Instructor Explanation: neutrons 14 electrons 11 Chapter 3; atomic number from Periodic Table of elements = 11. Therefore, # of protons = 11 (2 pts). # of protons = number of electron, thus = 11 electrons (2 pts); If mass number is 25, then there are 14 neutrons (# neutrons = mass number - # protons) = 25-11 = 14. Points Received: 6 of 6 Comments: Question 6.Question : (TCO 2) Consider a neutral atom with 30 protons and 34 neutrons. The atomic number of the element is Student Answer: 94. 30. 64. 32. 34. Instructor Explanation: Chapter 3; Atomic number = number of protons. Therefore, in this case the atomic number = 30. Points Received: 3 of 3 Comments: Question 7.Question : (TCO 2) No matter how much extra oxygen is available, 12 grams of carbon always combines with 32 grams of oxygen. This best illustrates the law of Student Answer: conservation of mass. multiple proportions. definite proportions. all of the above. none of the above. Instructor Explanation: Chapter 2 Points Received: 3 of 3 Comments: Question 8.Question : (TCO 2) Which type of radioactivity has essentially no mass? Student Answer: alpha beta gamma all of the above. none of the above. Instructor Explanation: Chapter 3 Points Received: 3 of 3 Comments: Question 9.Question : (TCO 2) The proton has Student Answer: the same mass and charge as the electron. the smaller mass and same charge as the electron. a smaller mass and opposite charge as the electron. a larger mass and opposite charge as the electron. none of the above. Instructor Explanation: Chapter 3 Points Received: 3 of 3 Comments: Question 10.Question : (TCO 2) Which subatomic particles have approximately the same mass? Student Answer: electrons and neutrons electrons and protons neutrons and protons electrons, neutrons and protons none of the above Instructor Explanation: Chapter 3 Points Received: 3 of 3 Comments: [Show More]

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