A compound microscope has an ocular micrometer calibrated to 12μm for each observable division in its scale.The widest part of an ellipsoidal microorganism observed under 4x magnification spans to the 18th division on the ocular microscope. What is the actual size of themicroorganism?12 μm18 um72 μm180 µmO216 μmQUESTION 4Q When viewed using the scanning objective lens (4x magnification), the field of view diameter measures 5.25mm across. What is the field of view diameter (in μm) when viewedunder the highest power objective lens (100x)?210 μmO131,250 μm7.62x10-6 μm4.76x10-3 μm4.76x10-3 μm Will the entire organism described in question 3 be observable under the highest power magnification (100x)? Provide a justification for your answer.OYes, because the field of view diameter at 100x magnification is less than the diameter of the organismO No, because the field of view diameter at 100x magnification is less than the diameter of the organismYes, because the field of view diameter at 100x magnification is greater than the diameter of the organismO No, because the field of view diameter at 100x magnification is greater than the diameter of the organism

A high-resolution microscope that employs two sets of lenses to provide a two-dimensional image of…

A compound microscope has an ocular micrometer calibrated to 12μm for each observable division in its scale. The widest part of an ellipsoidal microorganism observed under 4x magnification spans to the 18th division on the ocular microscope. What is the actual size of the microorganism? 12 μm 18 um 72 μm 180 μm 216 μm

A compound microscope has an ocular micrometer calibrated to 12μm for each observable division in its scale. The widest part of an ellipsoidal microorganism observed under 4x magnification spans to the 18th division on the ocular microscope. What is the actual size of the microorganism? 12 μm 18 um 72 μm 180 μm 216 μm

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter12: Gravimetric Methods Of Analysis

Section: Chapter Questions

Problem 12.2QAP

See similar textbooks

Similar questions

0.4ml to minims
Please create a caption for this table. Solution NaCl Conc. (%) Osmolality (mOsm) % transmittance Absorbance % hemolysis % crenation C distilled 0 0 0.001029 4.987584625 100 0.03354 1 0.177179111 54.61 0.001551 4.809388202 96.42720001 0.05837 2 0.297126222 91.58 0.01012 3.994819487 80.09527231 0.08444 3 0.442542222 136.4 3.849 1.414652089 28.3634704 0.134 4 0.590164444 181.9 64.8 0.188424994 3.777880643 0.2125 5 0.74752 230.4 95.64 0.019360433 0.388172513 0.3368 6 0.89644 276.3 99.56 0.001915112 0.038397585 0.5336 7 1.095648889 337.7 99.98 8.68676E-05 0.001741676 0.9834 8 1.336711111 412 100 0 0 2.1 9 1.755568889 541.1 100 0 0 7.9 10 2.674395556 824.3 100 0 0 57.83 11 4.490211111 1384 100 0 0 99.72
Define chemiluminescence.
A cellulose sponge, 7.0cm x 12cm x 2.5cm, weights 12.0 grams. Calculate the specific gravity of the sponge.
Is 10 ppm 400% less than 50 ppm?
When something cannot be directly obeserved, it can be represented by a Group of answer choices Model Graph Hypothesis Theory
Consider the following data that was collected for a protein that contain 0.76 % tryptophan amino acids by weight (Molecular weight of tryptophan amino acids is 204 g/mol). Absorbance Concentration (mg/L) 0.0250 0.100 0.150 0.226 0.259 0.400 0.600 0.964 1.14 1.78 1.45 2.35 1.78 2.87 How many mg of the protein present in 1.00 L solution that shows an absorbance of 1.20?
Define the following terms :(i) Kraft temperature(ii) Peptization(iii) Electrokinetic potential
Why HPLC is an important technique in the fields of chemistry . explain with different reasons?
Let me tell you about their task: The students were provided with 5 drinks (sports/soft drinks) and they were asked to determine the glucose concentration in these drinks in the units of g/100mL. (Why these units? Well, once the students have the concentrations in g/100mL they will be able to compare their values with the nutritional values given on the drink bottles’ labels). The samples of the five drinks were all diluted 1/100 (i.e. by a factor of 100). This was an essential step in the method because, without it, the machine used to analyse the glucose concentration (spectrophotometer) would have given an error as the concentration would have been too high for accurate detection. What this means for the students is that the dilution factor will have to be taken into consideration in their calculations (remember the aim is to calculate the concentration in the original drink and not in the diluted drink). The students decided to save time and form five groups where each group was…
2.Dscribe The History of Forensic Chemistry? 3.Enumerate and discuss the functions of forensic Chemist. 4.Discussed the Purposed of the Golden rules in the practice of Forensic Chrmistry. 5.What is Toxicology. 6.Discussed the History of Toxicology. 7.Discussed Tbr collection of Sample for toxicology. 8.Discussed The importance of toxicology. 9.Discussed the Types of poison. 10. Discussed the entrance and elimination of poisons in Human body....
Differentiate between the dispersed phase and the continuous phase of (a) soap suds; (b) milk; (c) airborne pollen grains; (d) margarine.