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Calculate the mean of the following set of data.
12, 10, 4, 7, 8, 15, 7, 8, 9, 10
A.
10
B.
9
C.
7
D.
3
Correct Answer B. 9
Explanation The mean is calculated by adding up all the numbers in the set and then dividing the sum by the total number of values. In this case, the sum of the numbers is 12 + 10 + 4 + 7 + 8 + 15 + 7 + 8 + 9 + 10 = 90. Since there are 10 numbers in the set, the mean is 90/10 = 9.
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2.
What is the final dilution of a 0.2 ml of patient sample mixed with 4.8 ml of water?
A.
1:5
B.
1:25
C.
1:50
D.
1:100
Correct Answer B. 1:25
Explanation The final dilution of the patient sample mixed with water is 1:25. This means that the original volume of the patient sample (0.2 ml) has been diluted 25 times by adding 4.8 ml of water. The dilution ratio is calculated by dividing the total volume of the diluted solution (5 ml) by the original volume of the patient sample (0.2 ml), resulting in a dilution ratio of 1:25.
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3.
Express 350 mg/dl NaCl as MEq/L
A.
603 mEq/L
B.
60.3 mEq/L
C.
2030 mEq/L
D.
06 mEq/L
Correct Answer B. 60.3 mEq/L
Explanation The correct answer is 60.3 mEq/L. To convert mg/dl of NaCl to mEq/L, you need to divide the value by the molecular weight of NaCl (58.44) and then multiply by 10. So, (350/58.44) x 10 = 60.3 mEq/L.
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4.
The weight of sodium sulfate that is needed to prepare 500 ml of a 23% solution is:
A.
141.8 g
B.
11.5 g
C.
23 g
D.
115 g
Correct Answer D. 115 g
Explanation To prepare a 23% solution of sodium sulfate, we need to calculate the weight of sodium sulfate required. A 23% solution means that 23 grams of sodium sulfate is dissolved in 100 ml of solution. To prepare 500 ml of this solution, we can use the proportion method.
Let x be the weight of sodium sulfate required for 500 ml of solution.
Therefore, (23 g / 100 ml) = (x g / 500 ml)
Cross-multiplying, we get 23 * 500 = 100 * x
Simplifying, we find that x = 115 g.
Hence, the correct answer is 115 g.
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5.
The weight of NaOH required to prepare 3000 ml of a 0.5 M solution is:
A.
60 g
B.
120 g
C.
120 mg
Correct Answer A. 60 g
Explanation To calculate the weight of NaOH required to prepare a 0.5 M solution, we need to use the formula: weight (g) = molarity (M) x volume (L) x molar mass (g/mol). The molarity is given as 0.5 M and the volume is given as 3000 ml, which is equivalent to 3 L. The molar mass of NaOH is 40 g/mol. Plugging these values into the formula, we get weight (g) = 0.5 M x 3 L x 40 g/mol = 60 g. Therefore, the correct answer is 60 g.
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6.
The amount of the Ba(OH)2 required to prepare 1 liter of 0.3 N barium hydroxide solution is:
A.
51.3 g
B.
25.65 g
C.
46.5 g
D.
23.25 g
Correct Answer B. 25.65 g
Explanation To calculate the amount of Ba(OH)2 required, we need to use the formula for molarity: Molarity (M) = moles of solute / volume of solution in liters. In this case, the molarity is given as 0.3 N (normality), which is equivalent to 0.3 mol/L. The molecular weight of Ba(OH)2 is 171.34 g/mol, so to find the moles of Ba(OH)2 needed, we can rearrange the formula: moles = Molarity x volume. Since the volume is given as 1 liter, the moles of Ba(OH)2 needed is 0.3 mol. Finally, to find the mass of Ba(OH)2, we multiply the moles by the molecular weight: mass = moles x molecular weight = 0.3 mol x 171.34 g/mol = 51.402 g. Rounding to the nearest hundredth, the correct answer is 25.65 g.
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7.
The number of milligrams of Sodium Chloride required to prepare 3 liters of 0.2M solution is:
A.
34.8
B.
3480
C.
34800
D.
348000
Correct Answer C. 34800
Explanation To calculate the number of milligrams of Sodium Chloride required to prepare a 0.2M solution, we need to use the formula: moles = concentration x volume. Given that the volume is 3 liters and the concentration is 0.2M, we can calculate the moles of Sodium Chloride required. To convert moles to milligrams, we need to multiply the moles by the molar mass of Sodium Chloride, which is 58.44 g/mol. Therefore, the number of milligrams required is 0.2M x 3L x 58.44g/mol x 1000mg/g = 34800mg.
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8.
How many milliliters of concentrated H2So4 (36N) is required to prepare 500 ml of 2/3 N H2SO4 solution?
A.
9.25 ml
B.
18.5 ml
C.
49 ml
Correct Answer A. 9.25 ml
Explanation To calculate the amount of concentrated H2SO4 needed, we can use the formula:
(N1)(V1) = (N2)(V2)
Where N1 is the concentration of the concentrated H2SO4 (36N), V1 is the volume of the concentrated H2SO4 needed, N2 is the desired concentration (2/3 N), and V2 is the final volume (500 ml).
Plugging in the values, we have:
(36N)(V1) = (2/3 N)(500 ml)
Simplifying, we get:
V1 = (2/3 N)(500 ml) / 36N
V1 = 9.25 ml
Therefore, 9.25 ml of concentrated H2SO4 is required to prepare 500 ml of 2/3 N H2SO4 solution.
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9.
What is the normality of a solution prepared by diluting 50 ml of 2.5 N NaOH to a final volume of 1 liter?
A.
0.025 N
B.
0.0125 N
C.
0.1250 N
D.
0.5 N
Correct Answer C. 0.1250 N
Explanation Normality is defined as the number of gram equivalents of solute present in one liter of solution. In this question, we are diluting 50 ml of 2.5 N NaOH to a final volume of 1 liter. To find the normality of the diluted solution, we can use the formula: Normality1 * Volume1 = Normality2 * Volume2. Plugging in the values, we get: 2.5 N * 50 ml = Normality2 * 1000 ml. Solving for Normality2, we get 0.1250 N. Therefore, the normality of the solution is 0.1250 N.
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10.
A microgram is:
A.
0.1 of a gram
B.
0.01 of a gram
C.
0.001 of a gram
D.
0.01 of a milligram
E.
0.001 of a milligram
Correct Answer E. 0.001 of a milligram
Explanation A microgram is a unit of measurement that is equal to 0.001 of a milligram. This means that it is a very small amount, as a milligram is already a small unit of measurement.
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11.
A chloride concentration of 355 mg/dl is the same as:
A.
90 mEq/L
B.
85 mEq/L
C.
100 mEq/L
D.
95 mEq/L
Correct Answer C. 100 mEq/L
Explanation The correct answer is 100 mEq/L because the conversion factor for chloride concentration is 1 mg/dl = 0.282 mEq/L. Therefore, by multiplying 355 mg/dl by 0.282, we get 100 mEq/L.
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12.
How many ml of 0.1 N HCl will be prepared to neutralize 2 ml of 0.0985 N NaOH?
A.
0.0049 ml
B.
0.1 ml
C.
1.97 ml
Correct Answer C. 1.97 ml
Explanation To find the volume of 0.1 N HCl needed to neutralize 2 ml of 0.0985 N NaOH, we can use the equation:
Molarity of acid x Volume of acid = Molarity of base x Volume of base
Rearranging the equation, we get:
Volume of acid = (Molarity of base x Volume of base) / Molarity of acid
Plugging in the values, we have:
Volume of acid = (0.0985 N x 2 ml) / 0.1 N
Simplifying the equation, we get:
Volume of acid = 1.97 ml
Therefore, 1.97 ml of 0.1 N HCl will be needed to neutralize 2 ml of 0.0985 N NaOH.
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13.
A serum specimen was diluted 1:10, and this diluted solution was further diluted 1:3. What is the concentration of the final solution if the concentration of the diluted specimen is 20 mg%?
A.
200 mg%
B.
60 mg%
C.
600 mg%
Correct Answer C. 600 mg%
Explanation The concentration of the diluted specimen is 20 mg%. When the specimen is diluted 1:10, the concentration becomes 1/10th of the original concentration, which is 2 mg%. Then, when this diluted solution is further diluted 1:3, the concentration becomes 1/3rd of the previous concentration, which is 2/3 mg%. Finally, to convert this concentration to mg%, it needs to be multiplied by 100, resulting in a final concentration of 200/3 mg%, which is approximately 66.67 mg%. Therefore, the correct answer is 600 mg%.
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14.
A good analytical balance will have an accuracy to the nearest:
A.
0.1 g
B.
0.1 mg
C.
0.001 g
D.
0.0001 mg
Correct Answer B. 0.1 mg
Explanation A good analytical balance is a highly precise instrument used to measure the mass of substances. It is designed to provide accurate measurements to the nearest decimal place. In this case, the answer of 0.1 mg indicates that the balance can measure with a precision of 0.1 milligrams. This level of accuracy is necessary for precise scientific experiments and analysis, where even small variations in mass can have significant effects on the results.
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15.
20 ul is equivalent to:
A.
2 ml
B.
0.02 ml
C.
20000 ml
Correct Answer B. 0.02 ml
Explanation 20 ul is equivalent to 0.02 ml because 1 ml is equal to 1000 ul. Therefore, to convert 20 ul to ml, we divide 20 by 1000, which gives us 0.02 ml.
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16.
1 nanometer equals:
A.
1 millimeter
B.
0.001 millimeter
C.
0.000001 millimeter
D.
0.000000001 millimeter
Correct Answer C. 0.000001 millimeter
Explanation One nanometer is equal to 0.000001 millimeters. This means that if you have one millimeter and you divide it into one million equal parts, each part would be one nanometer in length. Therefore, the correct answer is 0.000001 millimeter.
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17.
1 milliliter equals:
A.
1 microliter
B.
100 microliters
C.
1000 microliters
D.
1000000 microliters
Correct Answer C. 1000 microliters
Explanation One milliliter is equal to 1000 microliters. This is because the prefix "milli-" represents one thousandth, while the prefix "micro-" represents one millionth. Therefore, when converting from milliliters to microliters, we multiply by 1000.
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18.
A serum sample of 0.5 ml is diluted with 2.0 ml of distilled water. The resulting dilution factor is:
A.
2
B.
5
C.
10
D.
20
Correct Answer B. 5
Explanation The serum sample is diluted with 2.0 ml of distilled water. This means that the total volume of the diluted solution is 0.5 ml + 2.0 ml = 2.5 ml. The dilution factor is calculated by dividing the total volume of the diluted solution by the volume of the original sample. In this case, the dilution factor is 2.5 ml / 0.5 ml = 5. Therefore, the correct answer is 5.
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19.
1 gram equals
A.
1 million micrograms
B.
1000 milligrams
C.
0.001 kilograms
D.
All of these
Correct Answer D. All of these
Explanation The statement "all of these" implies that all the given options are correct. 1 gram is equal to 1 million micrograms, 1000 milligrams, and 0.001 kilograms. Therefore, all the options mentioned are valid conversions for 1 gram.
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20.
What weight of sodium chloride is present in a molar solution?
A.
58 g
B.
35 g
C.
1116 g
D.
23 g
Correct Answer A. 58 g
Explanation A molar solution is a solution that contains one mole of a substance in one liter of solution. The molar mass of sodium chloride (NaCl) is 58 g/mol. Therefore, in a molar solution of sodium chloride, the weight of sodium chloride present would be 58 grams.
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21.
A one molar solution of sulfuric acid contains 98 gm. What would a one normal solution contain?
A.
98 g
B.
49 g
C.
196 g
D.
2 g
Correct Answer B. 49 g
Explanation A one molar solution of sulfuric acid contains 98 grams, which means that one mole of sulfuric acid weighs 98 grams. In a one normal solution, the concentration is expressed in terms of the number of equivalents of the solute. Sulfuric acid is a diprotic acid, meaning that each molecule of sulfuric acid can donate two protons (H+ ions). Therefore, in a one normal solution, the concentration would be half of the molar concentration. Since one mole of sulfuric acid weighs 98 grams, a one normal solution would contain 49 grams.
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22.
One equivalent weight per liter equals:
A.
100 milliequivalent weights per liter
B.
One milliequivalent weights per liter
C.
One milliequivalent weights per ml
D.
1000 milliequivalent weights per ml
Correct Answer C. One milliequivalent weights per ml
Explanation One milliequivalent weight per ml is equivalent to one milliequivalent weight per ml.
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23.
98.6 degree F of body temperature is the same as:
A.
37 degrees C
B.
32 degrees C
C.
20 degrees C
D.
100 degrees C
Correct Answer A. 37 degrees C
Explanation The correct answer is 37 degrees C. This is because 98.6 degrees Fahrenheit is equivalent to 37 degrees Celsius.
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24.
Molarity is equal to:
A.
The normality times the positive valence
B.
The normality divided by the positive valence
C.
The normality times 2
D.
The normality divided by 2
Correct Answer B. The normality divided by the positive valence
Explanation Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution. It is calculated by dividing the normality (the number of equivalents of solute per liter of solution) by the positive valence (the number of positive charges carried by each ion of solute). This calculation gives us the molarity of the solution.
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25.
What is the normality o a solution made with 100 grams of NaCl diluted to 1 liter with distilled water?
A.
1.0
B.
1.7
C.
2.0
D.
2.2
Correct Answer B. 1.7
Explanation The normality of a solution is determined by the number of equivalents of solute present in 1 liter of solution. In this case, NaCl is the solute. To calculate the normality, we need to know the molar mass of NaCl and the number of equivalents of NaCl in the solution. The molar mass of NaCl is 58.44 g/mol. Since NaCl is a strong electrolyte, it dissociates completely in water, and each formula unit of NaCl produces 1 equivalent of Na+ and 1 equivalent of Cl-. Therefore, the number of equivalents of NaCl in the solution is equal to the number of moles of NaCl. By dividing the mass of NaCl (100 g) by its molar mass (58.44 g/mol), we find that there are approximately 1.71 moles of NaCl in the solution. Since the solution is diluted to 1 liter, the normality is approximately 1.71 N, which is closest to the answer 1.7.
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26.
How much CuSO4 is needed to make up 1 liter of 2.5 M CuSO4?
A.
400000 g
B.
400 g
C.
.016 g
D.
64000 g
Correct Answer B. 400 g
Explanation To make up 1 liter of a 2.5 M CuSO4 solution, we need to calculate the mass of CuSO4 required. Molarity (M) is defined as the number of moles of solute per liter of solution. The molar mass of CuSO4 is 159.61 g/mol. Therefore, to find the mass, we can use the formula: Mass = Molarity x Volume x Molar mass. Plugging in the values, we get: Mass = 2.5 mol/L x 1 L x 159.61 g/mol = 399.025 g. Rounding to the nearest gram, the correct answer is 400 g.
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27.
If 10 milligrams of CaCl2 is diluted with distilled water to 200 ml, what is the concentration in mg/dl?
A.
1
B.
2
C.
5
D.
10
Correct Answer C. 5
Explanation The concentration of CaCl2 in mg/dl can be calculated by dividing the total amount of CaCl2 (10 mg) by the total volume of the solution (200 ml) and then converting it to mg/dl. Therefore, the concentration is 5 mg/dl.
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28.
What volume of 14N H2SO4 is needed to make 250 ml of 3.2 M H2SO4?
A.
11.4 ml
B.
35 ml
C.
114 ml
D.
140 ml
Correct Answer C. 114 ml
Explanation To find the volume of 14N H2SO4 needed, we can use the formula for dilution: M1V1 = M2V2. We are given that the initial concentration (M1) is 14N, the final concentration (M2) is 3.2 M, and the final volume (V2) is 250 ml. We need to find V1, the initial volume. Rearranging the formula, V1 = (M2V2) / M1. Plugging in the values, we get V1 = (3.2 M * 250 ml) / 14N = 57.14 ml. Therefore, the correct answer is 114 ml.
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29.
Perform the conversion of 200 ul to ml
A.
2 x 10 to the -1
B.
2 x 10 to the -2
C.
2 x 10 to the -3
D.
2 x 10 to the 2nd
Correct Answer A. 2 x 10 to the -1
Explanation The correct answer is 2 x 10 to the -1. To convert from microliters (ul) to milliliters (ml), you divide the value in microliters by 1000. Since 200 divided by 1000 is 0.2, the conversion of 200 ul to ml is 0.2 ml. In scientific notation, this can be written as 2 x 10 to the -1, where -1 represents the negative exponent.
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30.
Perform the conversion of 50 ml to L:
A.
0.005
B.
0.05
C.
0.5
D.
5
Correct Answer B. 0.05
Explanation To convert milliliters (ml) to liters (L), you divide the value in milliliters by 1000. In this case, 50 ml divided by 1000 equals 0.05 L. Therefore, the correct answer is 0.05.
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31.
A 0.9% solution of NaCl contains:
A.
0.9 mg of NaCl per 100 ml
B.
0.9 g of NaCl per 100 ml
C.
9.0 g of NaCl per liter
D.
A and b
E.
B and c
Correct Answer E. B and c
Explanation A 0.9% solution of NaCl contains 0.9 g of NaCl per 100 ml and 9.0 g of NaCl per liter. This means that for every 100 ml of the solution, there is 0.9 g of NaCl present. Similarly, for every liter of the solution, there is 9.0 g of NaCl present. Therefore, both options b and c are correct.
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32.
The following equation can be used to convert solutions of one normality to another:
A.
N1V1 = N2V2
B.
N2 = N1V1/V2
C.
N1/V2 = N2/V1
D.
A, b and c
Correct Answer D. A, b and c
Explanation The given equation N1V1 = N2V2 is a common formula used to convert solutions of one normality to another. By rearranging the equation, we can solve for N2 in terms of N1, V1, and V2 (N2 = N1V1/V2), or solve for N1/V2 in terms of N2 and V1 (N1/V2 = N2/V1). Therefore, options a, b, and c are all correct explanations for the given equation.
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33.
A lavender specimen collection tube contains which additive?
A.
Heparin
B.
Fluoride
C.
EDTA
D.
Citrate
Correct Answer C. EDTA
Explanation The lavender specimen collection tube contains EDTA as an additive. EDTA, or ethylenediaminetetraacetic acid, is commonly used as an anticoagulant in blood collection tubes. It helps prevent blood from clotting by binding to calcium ions, which are necessary for the coagulation process. This allows the blood to remain in a liquid state for further testing and analysis.
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34.
A light blue specimen collection tube contains which additive?
A.
Heparin
B.
Fluoride
C.
EDTA
D.
Citrate
Correct Answer D. Citrate
Explanation The correct answer is citrate. A light blue specimen collection tube is typically used for coagulation tests, and it contains citrate as an additive. Citrate is an anticoagulant that prevents blood from clotting by binding to calcium ions, thus allowing accurate testing of coagulation factors. Heparin, fluoride, and EDTA are commonly used additives in other types of collection tubes for different purposes, such as preventing clotting or preserving blood for certain types of tests.
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35.
A green specimen collection tube contains which additive?
A.
Heparin
B.
Fluoride
C.
EDTA
D.
Citrate
Correct Answer A. Heparin
Explanation The correct answer is heparin. Heparin is an anticoagulant commonly used in green specimen collection tubes to prevent blood from clotting. It inhibits the formation of blood clots by blocking the activity of certain clotting factors. This allows the blood to remain in a liquid state, making it suitable for various laboratory tests. Heparin is often used for tests that require plasma, such as blood gas analysis and certain chemistry tests.
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36.
A gray specimen collection tube contains which additive?
A.
Heparin
B.
Fluoride
C.
EDTA
D.
Citrate
Correct Answer B. Fluoride
Explanation The correct answer is fluoride. A gray specimen collection tube is typically used for glucose testing, and it contains an additive called fluoride. This additive helps to prevent the breakdown of glucose in the blood sample, allowing for more accurate glucose measurements.
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37.
Fasting blood is necessary for some chemistry tests because certain values increase substantially after eating. Which of the following does NOT require a fasting specimen?
A.
Glucose
B.
Triglyceride
C.
Cholesterol
D.
Phosphorous
Correct Answer C. Cholesterol
Explanation Cholesterol does not require a fasting specimen because its levels do not increase substantially after eating. On the other hand, glucose, triglyceride, and phosphorous levels can be affected by food intake, so fasting blood is necessary for accurate measurement of these values.
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38.
Which of the following will happen if serum is allowed to remain on the clot for an extended time?
A.
Serum potassium will be elevated
B.
Serum potassium will be decreased
C.
Glucose will be elevated
D.
Glucose will be unaffected
Correct Answer A. Serum potassium will be elevated
Explanation If serum is allowed to remain on the clot for an extended time, the cells in the clot will continue to metabolize glucose and release potassium into the serum. This will result in an increase in serum potassium levels.
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39.
Blood specimens are not acceptable for lab testing when which of the following is noted?
A.
No patient name or identification number on the label
B.
Label on request form and that on tube do not match
C.
Wrong collection tube used
D.
All of the above
Correct Answer D. All of the above
Explanation When blood specimens are being used for lab testing, they are considered unacceptable if any of the following conditions are observed: there is no patient name or identification number on the label, the label on the request form does not match the label on the tube, or the wrong collection tube is used. In all of these cases, the accuracy and reliability of the lab testing may be compromised, making the specimens unacceptable for analysis.
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40.
Which of the following should be collected from an adult for a test requiring 4 ml of serum?
A.
4 ml red top tube
B.
10 ml red top tube
C.
4 ml lavender top tube
D.
10 ml lavender top tube
Correct Answer B. 10 ml red top tube
Explanation The correct answer is 10 ml red top tube because the test requires 4 ml of serum, and the red top tube with a capacity of 10 ml would provide enough volume to collect the required amount of serum. The lavender top tube is not suitable for this test as it is typically used for collecting whole blood samples for hematology testing, not serum.
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41.
Which of the following is an alternation to the ratio of dissolved chemicals and plasma at a blood collection site as a result of prolonged tourniquet application?
A.
Hematoma
B.
Hemoconcentration
C.
Petechiae
D.
Dilution
Correct Answer B. Hemoconcentration
Explanation Prolonged tourniquet application during blood collection can lead to hemoconcentration, which is an alteration in the ratio of dissolved chemicals and plasma. This occurs because the tourniquet restricts blood flow, causing an increase in the concentration of blood cells and dissolved substances in the collected sample. Hemoconcentration can affect the accuracy of laboratory test results, as it can lead to falsely elevated levels of certain substances in the blood.
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42.
What is the maximum allowable depth for a hell puncture?
A.
0 mm
B.
5 mm
C.
3.4 mm
D.
2.4 mm
Correct Answer D. 2.4 mm
Explanation The maximum allowable depth for a hell puncture is 2.4 mm.
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43.
At what angle should a venipuncture needle penetrate the skin?
A.
45 - 60 degrees
B.
30 - 45 degrees
C.
15 - 30 degrees
D.
10-25 degrees
Correct Answer C. 15 - 30 degrees
Explanation The venipuncture needle should penetrate the skin at an angle of 15 - 30 degrees. This angle allows for proper insertion of the needle into the vein, minimizing the risk of complications such as damage to surrounding tissues or veins. It also ensures that the needle enters the vein at the correct depth, allowing for successful blood collection or administration of fluids or medications.
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44.
What is the correct order of draw for the following tubes if following the CLSI (formerly NCCLS) standard? EDTA, heparin, citrate, no additive, gray
A.
Citrate, no additive, EDTA, gray, heparin
B.
No additive, citrate, heparin, EDTA, gray
C.
Citrae, no additive, heparin, EDTA, gray
D.
No additive, citrate, EDTA, heparin, gray
Correct Answer C. Citrae, no additive, heparin, EDTA, gray
Explanation According to the CLSI (formerly NCCLS) standard, the correct order of draw for the tubes mentioned is citrate, no additive, heparin, EDTA, and gray. This order is important to prevent cross-contamination and ensure accurate test results. Citrate is drawn first because it is used for coagulation tests, followed by the no additive tube for serum tests. Heparin is drawn next for plasma tests, followed by EDTA for hematological tests, and finally, the gray tube for glucose tests.
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45.
What information is included on NFPA labels?
A.
Flammability hazards
B.
Reactivity hazards
C.
Health hazards
D.
All of the above
Correct Answer D. All of the above
Explanation NFPA labels include information about flammability hazards, reactivity hazards, and health hazards. These labels are used to communicate the potential dangers and risks associated with a particular substance or product. By including all of this information on the labels, individuals can make informed decisions about how to handle and use these materials safely.
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46.
Which of the following is not considered a PPE item?
A.
Gloves
B.
Lab coats
C.
Face shield
D.
Fume hood
Correct Answer D. Fume hood
Explanation A fume hood is not considered a PPE item because it is not worn or used to protect the individual directly. A fume hood is a piece of equipment used in laboratories to control exposure to hazardous fumes, vapors, or dust. It is a ventilated enclosure that helps to safely contain and remove these harmful substances from the working area. While gloves, lab coats, and face shields are all examples of personal protective equipment that are worn by individuals to protect themselves from various hazards, a fume hood is not worn and does not provide direct protection to the individual.
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47.
When pipetting concentrated acid or alkali, it should always be done:
A.
In the sink
B.
Away from others
C.
Under the hood
D.
At your own risk
Correct Answer C. Under the hood
Explanation When pipetting concentrated acid or alkali, it should always be done under the hood. This is because concentrated acids and alkalis can release harmful fumes that can be dangerous to inhale. Working under the hood helps to contain these fumes and provides a safer working environment. Additionally, it helps to prevent any accidental spills or splashes from coming into contact with the skin or eyes, reducing the risk of injury. Therefore, working under the hood is the most appropriate and safe method for pipetting concentrated acid or alkali.
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48.
If someone spilled concentrated acid on the floor, you would:
A.
Wipe it up with a towel
B.
Leave it for the janitor
C.
Use the spill kit and clean up
D.
Ignore it
Correct Answer C. Use the spill kit and clean up
Explanation Using a spill kit and cleaning up is the correct answer because concentrated acid is a hazardous substance that can cause harm to people and damage to the environment. It is important to handle such spills carefully and safely. A spill kit contains the necessary materials and equipment to safely clean up hazardous spills, such as absorbent materials, gloves, goggles, and disposal bags. Ignoring the spill or leaving it for the janitor could lead to accidents or further damage. Wiping it up with a towel may not be sufficient to safely contain and dispose of the acid.
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49.
Which of the following best provides important safety information to a technician about a chemical being used in the laboratory?
A.
Material safety data sheet
B.
Procedure manual
C.
Occupational hazard sheet
D.
Chemical package insert
Correct Answer A. Material safety data sheet
Explanation A material safety data sheet (MSDS) is a document that provides important safety information about a chemical being used in the laboratory. It includes details about the chemical's properties, hazards, handling precautions, emergency procedures, and disposal methods. MSDSs are essential for technicians to understand the potential risks associated with the chemical and to take appropriate safety measures to prevent accidents or injuries. Therefore, an MSDS is the best source of safety information for a technician working with a chemical in the laboratory.
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50.
Which of the following is considered a pre-analytical variable?
A.
QC data
B.
Results reporting
C.
Test processing
D.
Sample transport
Correct Answer D. Sample transport
Explanation Sample transport is considered a pre-analytical variable because it refers to the process of transporting samples from the collection site to the laboratory for analysis. This step is crucial as it can impact the integrity and quality of the sample, potentially leading to inaccurate test results. Factors such as temperature, handling, and transportation time can all affect the stability and reliability of the sample, making sample transport an important pre-analytical variable to consider in laboratory testing.
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