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CHEM Quiz1 to Quiz 5- Multiple Versions, CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.€24,87
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CHEM Quiz1 to Quiz 5- Multiple Versions, CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.
CHEM Quiz1 to Quiz 5- Multiple Versions, CHEM 120: Introduction to General, Organic & Biological Chemistry with Lab, Verified and Correct Answers, Chamberlain College of Nursing.
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 Answers will vary. Chapter 1 and lecture.
Explanation:
, 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 x meters = 9 yd x (0.914 m/1 yd) = 8.23 m. (Chapter 1; 1 pt for units, 1
Explanation: 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 x inches = 8.7 inch x (2.54 cm/1 in) x (10 mm/1 cm)= 220.98 mm
Explanation: (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 Density = mass in g/volume in mL; m = d * v; x g = (50.0 mL) x (1.15
Explanation: 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
neutrons 14 electrons 11
Instructor Chapter 3; atomic number from Periodic Table of elements =
Explanation: 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 Chapter 3; Atomic number = number of protons. Therefore, in this case the
Explanation: 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 Chapter 2
Explanation:
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 Chapter 3
, Explanation:
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 Chapter 3
Explanation:
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 Chapter 3
Explanation:
Points Received: 3 of 3
Comments:
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