CHEM 120 W8 Exam3 and W8 Lab Practicum at Chamberlain University
Access The Exact Questions for CHEM 120 W8 Exam3 and W8 Lab Practicum at Chamberlain University
💯 100% Pass Rate guaranteed
🗓️ Unlock for 1 Month
Rated 4.8/5 from over 1000+ reviews
- Unlimited Exact Practice Test Questions
- Trusted By 200 Million Students and Professors
What’s Included:
- Unlock Actual Exam Questions and Answers for CHEM 120 W8 Exam3 and W8 Lab Practicum at Chamberlain University on monthly basis
- Well-structured questions covering all topics, accompanied by organized images.
- Learn from mistakes with detailed answer explanations.
- Easy To understand explanations for all students.
Free CHEM 120 W8 Exam3 and W8 Lab Practicum at Chamberlain University Questions
Which of the following would be considered an application of radioactive materials? Select all that apply.
-
CT scan
-
Tracking of leaks in pipes
-
Fuel additive for cars
-
Smoke detectors
-
Hand sanitation
Explanation
A. CT scan
Certain CT scans and other imaging techniques, such as PET scans, use radioactive tracers (radioisotopes) to visualize organ function, detect cancer, and study metabolic activity within the body. This is a key medical application of radioactive materials.
B. Tracking of leaks in pipes
Radioactive isotopes are used in industrial tracing to locate leaks in pipelines or measure the flow rate of liquids. The radiation emitted helps identify where leaks occur without the need for excavation or disruption.
D. Smoke detectors
Most smoke detectors contain a small amount of americium-241, a radioactive isotope that ionizes air particles, allowing the device to detect smoke particles and trigger an alarm — a common household use of radioactive materials.
A long chain of glucose molecules forms. Which of the following terms would best describe the resulting biomolecule?
-
Monosaccharide
-
Polysaccharide
-
Sugar
-
Disaccharide
Explanation
A polysaccharide is a complex carbohydrate composed of many glucose (monosaccharide) units bonded together through glycosidic linkages. Examples include starch, glycogen, and cellulose. These large molecules serve important biological functions such as energy storage (starch in plants, glycogen in animals) and structural support (cellulose in plant cell walls).
During transcription, a DNA molecule has the partial sequence: GGT-AAA-AAC. What mRNA sequence is made?
-
GGT-AAA-AAC
-
TTT-TTG-CCA
-
GGU-AAA-AAC
-
CCA-UUU-UUG
Explanation
During transcription, the mRNA strand is formed complementary to the DNA template strand. Base-pairing rules are as follows:
DNA G pairs with RNA C
DNA C pairs with RNA G
DNA A pairs with RNA U
DNA T pairs with RNA A
So, if the DNA template sequence is GGT-AAA-AAC, the complementary mRNA sequence will be:
CCA-UUU-UUG
You have a chunk of radioactive iron. This isotope has a half-life of about 2.5 years. How many grams of an 80-gram sample of this radioactive iron will remain intact after 5 years?
-
80 g
-
40 g
-
20 g
-
35 g
Explanation
A half-life is the time it takes for half of a radioactive substance to decay. After one half-life (2.5 years), 80 g becomes 40 g. After another half-life (5 years total), half of the remaining 40 g decays, leaving 20 g. Therefore, after two half-lives, only one-quarter of the original sample (20 g) remains intact.
When glucose and galactose react, a glycosidic linkage is formed. What are the products of this reaction?
-
Water and a disaccharide
-
Disaccharide
-
Water and a polysaccharide
-
Polysaccharide
Explanation
When glucose and galactose undergo a dehydration synthesis (condensation) reaction, a glycosidic bond forms between them, releasing one molecule of water as a byproduct. The resulting disaccharide is lactose, commonly known as milk sugar. This process links two monosaccharides to form larger carbohydrate molecules while producing water due to the removal of hydrogen (H) and hydroxyl (OH) groups from the reacting sugars.
Why does the correct Lewis structure of CO2 involve a double bond between each of the oxygen atoms and the carbon atom?
-
To complete the octet for carbon
-
Because the oxygens would otherwise have more than eight electrons
-
Because carbon is more electronegative than oxygen
-
To emphasize that it's a covalent bond
Explanation
In the Lewis structure of CO₂, carbon needs four additional electrons to complete its octet, while each oxygen needs two. By forming two double bonds—one between carbon and each oxygen—both the carbon and oxygen atoms achieve stable octets. This arrangement minimizes formal charges and accurately represents the molecule’s bonding, showing that CO₂ is a linear molecule with strong covalent bonds between atoms.
In the redox reaction below, determine the reducing agent:
CaO + H₂ → Ca + H₂O
-
H₂ is the reducing agent
-
H₂O is the reducing agent
-
CaO is the reducing agent
-
Ca is the reducing agent
Explanation
In the reaction:
CaO + H₂ → Ca + H₂O
- Oxidation states:
- In CaO: Ca is +2, O is –2.
- In H₂: H is 0.
- In Ca: Ca is 0.
- In H₂O: H is +1, O is –2.
- H₂ changes from 0 to +1 (oxidation), so H₂ is oxidized and is the reducing agent.
- Ca in CaO changes from +2 to 0 (reduction), so CaO is reduced, and CaO (or more precisely the Ca²⁺ in it) is the oxidizing agent.
Thus, H₂ is the reducing agent.
Which of the following best describes the terms "independent variable" and "dependent variable"?
-
The independent variable is the cause, and the dependent variable is the effect
-
The dependent variable is the cause, and the independent variable is the effect.
-
The independent variable is held constant as a positive control, and the dependent variable is the negative control.
-
There is no relationship between the independent variable and the dependent variable.
Explanation
The independent variable is the one that the researcher manipulates or changes to observe its effect on another variable. The dependent variable is the measurable outcome that responds to these changes. In other words, the independent variable causes the dependent variable to vary. For example, in testing plant growth, the amount of sunlight (independent) affects the height of the plant (dependent).
Which form of radiation is able to penetrate most deeply into the human body?
-
Alpha decay
-
Gamma ray emission
-
Skin would block all of these types of radiation
-
Beta decay
Explanation
Gamma rays are high-energy electromagnetic waves with no mass and no charge, allowing them to penetrate deeply through human tissue and even several centimeters of lead or concrete. Because of their penetrating power, they pose a significant internal and external radiation hazard.
The enzyme OREOase acts on chocolate sandwich cookies to break the cookies apart to get to the creamy center. Occasionally, a vanilla cookie also fits into the OREOase active site, but the enzyme cannot act on the vanilla cookie so the reaction halts. Eventually, the vanilla cookie leaves the active site. What is the best description of the vanilla cookie in this scenario?
-
Competitive inhibitor
-
Denaturing of the protein
-
Cofactor
-
Coenzyme
-
Non-competitive inhibitor
Explanation
A competitive inhibitor is a substance that resembles the enzyme’s normal substrate and competes for binding at the active site. In this case, the vanilla cookie fits into OREOase’s active site just like the chocolate cookie (the true substrate) but does not undergo a reaction. This temporarily blocks the enzyme’s activity until the inhibitor leaves, allowing the correct substrate to bind again.
How to Order
Select Your Exam
Click on your desired exam to open its dedicated page with resources like practice questions, flashcards, and study guides.Choose what to focus on, Your selected exam is saved for quick access Once you log in.
Subscribe
Hit the Subscribe button on the platform. With your subscription, you will enjoy unlimited access to all practice questions and resources for a full 1-month period. After the month has elapsed, you can choose to resubscribe to continue benefiting from our comprehensive exam preparation tools and resources.
Pay and unlock the practice Questions
Once your payment is processed, you’ll immediately unlock access to all practice questions tailored to your selected exam for 1 month .