HESI RN
HESI Medical Surgical Specialty Exam
1. A client with acute glomerulonephritis (GN) is being evaluated by a nurse. Which manifestation should the nurse recognize as a positive response to the prescribed treatment?
- A. The client has lost 11 pounds in the past 10 days.
- B. The client’s urine specific gravity is 1.048.
- C. No blood is observed in the client’s urine.
- D. The client’s blood pressure is 152/88 mm Hg.
Correct answer: A
Rationale: A weight loss of 11 pounds in the past 10 days indicates fluid loss, a positive response to treatment for acute glomerulonephritis. It signifies that the glomeruli are functioning adequately to filter excess fluid. A urine specific gravity of 1.048 is high, indicating concentrated urine, which is not a positive response in this context. Blood in the urine is not a typical finding in glomerulonephritis, so its absence is expected and does not indicate a positive response to treatment. A blood pressure of 152/88 mm Hg is elevated and may suggest kidney damage or fluid overload, which are not positive responses to treatment.
2. A client in the emergency department is severely dehydrated and is prescribed 3 L of intravenous fluid over 6 hours. At what rate (mL/hr) should the nurse set the intravenous pump to infuse the fluids? (Record your answer using a whole number.)
- A. 500 mL/hr
- B. 400 mL/hr
- C. 550 mL/hr
- D. 600 mL/hr
Correct answer: A
Rationale: To calculate the rate of the intravenous pump, divide the total volume of fluid (3 L = 3000 mL) by the total time in hours (6 hours), which equals 500 mL/hr. The correct answer is A. Choice B (400 mL/hr) is incorrect as it would result in a slower infusion rate. Choice C (550 mL/hr) and Choice D (600 mL/hr) are incorrect as they would result in a faster infusion rate, exceeding the prescribed amount of fluid to be infused over 6 hours.
3. A client’s baseline vital signs are temperature 98°F oral, pulse 74 beats/min, respiratory rate 18 breaths/min, and blood pressure 124/76 mm Hg. The client suddenly spikes a fever to 103°F. Which of the following respiratory rates would the nurse anticipate as part of the body’s response to the change in client status?
- A. 12 breaths/min
- B. 16 breaths/min
- C. 18 breaths/min
- D. 22 breaths/min
Correct answer: D
Rationale: When a client experiences a fever, there is an increase in body temperature, leading to a higher metabolic rate and oxygen demand. As a result, the respiratory rate typically increases to meet the body's increased oxygen needs. Therefore, in response to the fever spike from 98°F to 103°F, the nurse would anticipate a higher respiratory rate. Choices A, B, and C are incorrect because a decrease in body temperature, not an increase as seen in fever, would lead to a decrease in respiratory rate to conserve energy and oxygen consumption.
4. A client is scheduled for a barium swallow (esophagography) in 2 days. The nurse, providing preprocedure instructions, should tell the client to:
- A. Eat a regular supper and breakfast
- B. Remove all metal and jewelry before the test
- C. Expect diarrhea for a few days after the procedure
- D. Take all oral medications as scheduled with milk on the day of the test
Correct answer: B
Rationale: The correct answer is B: 'Remove all metal and jewelry before the test.' Before a barium swallow procedure, the client should fast for 8 to 12 hours to ensure the stomach and intestines are empty for optimal visualization. Removing all metal and jewelry is essential to prevent any interference with x-ray imaging. Choice A is incorrect because the client should fast, not eat supper and breakfast, before the test. Choice C is incorrect as diarrhea is not an expected outcome of a barium swallow. Choice D is incorrect as the client should not take any oral medications with milk on the day of the test to ensure accurate test results.
5. The client is being educated by the nurse on home blood glucose monitoring. Which of the following blood glucose measurements indicates hypoglycemia?
- A. 59 mg/dL (3.3 mmol/L)
- B. 75 mg/dL (4.2 mmol/L)
- C. 108 mg/dL (6 mmol/L)
- D. 119 mg/dL (6.6 mmol/L)
Correct answer: A
Rationale: A blood glucose level of 59 mg/dL (3.3 mmol/L) is considered hypoglycemia, which is an abnormally low blood sugar level. This level requires immediate attention as it can lead to symptoms such as confusion, shakiness, and even loss of consciousness if left untreated. Choices B, C, and D have blood glucose levels within the normal range or slightly higher, indicating euglycemia or normal blood sugar levels, and not hypoglycemia.
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