Nursing Elites

1. When preparing a site for the insertion of an IV catheter, how should excess hair at the site be treated?

• A. Leaving the hair intact
• B. Shaving the area
• C. Clipping the hair in the area
• D. Removing the hair with a depilatory

Correct answer: C

Rationale: The correct answer is to clip the hair in the area. Excess hair at the site of IV catheter insertion should be removed because it can be a potential source of infection. Clipping the hair is preferred over shaving because shaving can cause skin abrasions, increasing the risk of infection. Using depilatories is not recommended as they can irritate the skin, which is undesirable when preparing a clean site for an invasive procedure. Therefore, clipping the hair in the area is the most appropriate and safe method to prepare the site for IV catheter insertion.

2. A patient with diabetes insipidus is admitted to the intensive care unit after a motor vehicle accident that resulted in head trauma and damage to the pituitary gland. Diabetes insipidus can occur when there is a decreased production of which of the following?

• A. ADH
• B. Estrogen
• C. Aldosterone
• D. Renin

Correct answer: A

Rationale: The correct answer is A: ADH. Diabetes insipidus is characterized by a deficiency in antidiuretic hormone (ADH), leading to excessive urine output and thirst. In this scenario, the head trauma and damage to the pituitary gland can result in decreased production or release of ADH. Estrogen (Choice B) is not directly related to diabetes insipidus. Aldosterone (Choice C) is a hormone that regulates sodium and potassium levels, not water balance like ADH. Renin (Choice D) is an enzyme involved in the regulation of blood pressure and fluid balance but not directly related to diabetes insipidus.

3. Third spacing occurs when fluid moves out of the intravascular space but not into the intracellular space. Based on this fluid shift, the nurse will expect the patient to demonstrate:

• A. Hypertension
• B. Bradycardia
• C. Hypervolemia
• D. Hypovolemia

Correct answer: D

Rationale: In the scenario of third-spacing fluid shift, where fluid moves out of the intravascular space but not into the intracellular space, the patient is expected to demonstrate hypovolemia. Hypertension (Choice A) is unlikely as hypovolemia typically leads to decreased blood pressure. Bradycardia (Choice B) is not a common manifestation of hypovolemia, as the body often tries to compensate by increasing heart rate. Hypervolemia (Choice C) indicates an excess of fluid, which is the opposite of what occurs in third spacing.

4. A client with a serum potassium of 7.5 mEq/L and cardiovascular changes needs immediate intervention. Which prescription should the nurse implement first?

• A. Prepare to administer sodium polystyrene sulfate (Kayexalate) 15 g by mouth.
• B. Provide a heart-healthy, low-potassium diet.
• C. Prepare to administer dextrose 20% and 10 units of regular insulin IV push.
• D. Prepare the client for hemodialysis treatment.

Correct answer: C

Rationale: In a client with a serum potassium level of 7.5 mEq/L and cardiovascular changes, the priority intervention is to lower the potassium level quickly to prevent life-threatening complications like arrhythmias. The correct answer is to prepare to administer dextrose 20% and 10 units of regular insulin IV push. This combination helps shift potassium from the extracellular to the intracellular space, reducing serum potassium levels rapidly. Administering sodium polystyrene sulfate (Kayexalate) by mouth may take several hours to work, making it a less effective immediate intervention. Providing a heart-healthy, low-potassium diet is important for long-term management but is not the most urgent action in this situation. While hemodialysis is a definitive treatment for hyperkalemia, it is not the first-line intervention for acute management of high potassium levels with cardiovascular manifestations.

5. You are an emergency-room nurse caring for a trauma patient. Your patient has the following arterial blood gas results: pH 7.26, PaCO2 28, HCO3 11 mEq/L. How would you interpret these results?

• A. Respiratory acidosis with no compensation
• B. Metabolic alkalosis with a compensatory alkalosis
• C. Metabolic acidosis with no compensation
• D. Metabolic acidosis with a compensatory respiratory alkalosis

Correct answer: D

Rationale:

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