The total magnification achieved using a 10× objective lens with a 10× eyepiece lens is 20×.
True or False?
True
False
The Correct Answer is B
The total magnification of a microscope is calculated by multiplying the magnification of the objective lens and the magnification of the eyepiece lens. Therefore, using a 10× objective lens with a 10× eyepiece lens would result in a total magnification of 10×10 = 100×, not 20×.
To achieve a total magnification of 20×, one would need to use a 2× objective lens with a 10× eyepiece lens, or a 10× objective lens with a 2× eyepiece lens. Therefore, the statement that the total magnification achieved using a 10× objective lens with a 10× eyepiece lens is 20× is false.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is D
Explanation
Choice A rationale: An isotonic solution has the same solute concentration as the plant cells, which means it has the same water potential. Water will move in and out of the plant cells at the same rate, causing them to remain unchanged in size and shape. This will not help the lettuce to become less wilted.
Choice B rationale: A hypertonic solution has a higher solute concentration than the plant cells, which means it has a lower water potential. Water will move out of the plant cells by osmosis, causing them to shrink and become plasmolysed. This will make the lettuce more wilted and not crisper.
Choice C rationale: Soaking the lettuce in water would have an effect, depending on the relative solute concentration of the water and the plant cells. Water will move across the cell membrane by osmosis, either into or out of the plant cells, causing them to change in size and shape.
Choice D rationale: A hypotonic solution has a lower solute concentration than the plant cells, which means it has a higher water potential. Water will move into the plant cells by osmosis, causing them to swell and become turgid. This will make the lettuce crisper and more appealing.
Choice E rationale: A hypotonic solution will cause the plant cells to gain water and become turgid, not more wilted. Wilted lettuce is caused by the loss of water from the plant cells, which makes them flaccid and soft.
Correct Answer is D
Explanation
Choice A reason: Biuret test is a test for the presence of proteins or peptides in a solution. It involves adding Biuret reagent, which is a blue solution of copper (II) sulfate and sodium hydroxide, to the solution. If proteins or peptides are present, they form a complex with the copper (II) ions, which changes the color of the solution to violet or pink. ³
Choice B reason: Gram's iodine test is a test for the presence of starch in a solution. It involves adding iodine solution, which is a brown solution of iodine and potassium iodide, to the solution. If starch is present, it forms a complex with the iodine molecules, which changes the color of the solution to blue-black. ⁴
Choice C reason: Ninhydrin test is a test for the presence of amines or amino acids in a solution. It involves adding ninhydrin reagent, which is a purple solution of ninhydrin, to the solution. If amines or amino acids are present, they react with ninhydrin to produce a purple color, often called Ruhemann's purple. ¹
Choice D reason: Brown paper test is a test for the presence of lipids, such as fats or oils, in a solution. The test involves placing a drop of the solution on a piece of brown paper and letting it dry. If lipids are present, they leave a translucent spot on the paper, which can be seen by holding the paper against a light source. The test is based on the fact that lipids are nonpolar and do not dissolve in water, but can dissolve in organic solvents and stain the paper. ²
Choice E reason: Benedict's test is a test for the presence of reducing sugars, such as glucose or maltose, in a solution. The test involves adding Benedict's reagent, which is a blue solution of copper (II) sulfate, sodium carbonate, and sodium citrate, to the solution and heating it in a water bath. If reducing sugars are present, they reduce the copper (II) ions to copper (I) ions, which form a red, orange, or green precipitate of copper (I) oxide. ⁵
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