HESI RN
HESI 799 RN Exit Exam Quizlet
1. After checking the fingerstick glucose at 1630, what action should be implemented?
- A. Notify the healthcare provider.
- B. Administer 8 units of insulin aspart SubQ.
- C. Give an IV bolus of Dextrose 50% 50 ml.
- D. Perform quality control on the glucometer.
Correct answer: B
Rationale: Administering insulin aspart (rapid-acting insulin) is the appropriate action to manage the elevated glucose level of 1630. Choice A, notifying the healthcare provider, is not the immediate action needed for this glucose level. Choice C, giving an IV bolus of Dextrose 50%, would exacerbate hyperglycemia instead of treating it. Choice D, performing quality control on the glucometer, is not relevant to the management of the patient's glucose level at this time.
2. During a clinic visit, a client with a kidney transplant asks, 'What will happen if chronic rejection develops?' Which response is best for the nurse to provide?
- A. Dialysis would need to be resumed if chronic rejection becomes a reality.
- B. Immunosuppressive therapy would be intensified.
- C. A second transplant would be scheduled immediately.
- D. We would monitor your kidney function closely.
Correct answer: A
Rationale: The best response for the nurse to provide is that dialysis would need to be resumed if chronic rejection becomes a reality. Chronic rejection of a transplanted kidney can lead to kidney failure, necessitating the need for dialysis until another transplant is possible. Choice B is incorrect because although immunosuppressive therapy may be adjusted, the immediate concern is the potential need for dialysis. Choice C is incorrect because scheduling a second transplant immediately is not typically the first step following chronic rejection. Choice D is also incorrect as close monitoring of kidney function is essential but does not address the immediate need for dialysis if chronic rejection occurs.
3. A nurse is caring for four clients. The nurse should observe which of the following clients for a risk of vitamin B6 deficiency?
- A. A client who has cystic fibrosis
- B. A client who has chronic alcohol use disorder
- C. A client who takes phenytoin for a seizure disorder
- D. A client who is prescribed rifampin for tuberculosis
Correct answer: B
Rationale: Chronic alcohol use disorder can lead to vitamin B6 deficiency due to impaired absorption and increased excretion of the vitamin. While clients with cystic fibrosis may be at risk for fat-soluble vitamin deficiencies, such as vitamins A, D, E, and K, they are not specifically at high risk for vitamin B6 deficiency. Clients taking phenytoin are at risk for folate deficiency, not vitamin B6. Clients prescribed rifampin for tuberculosis are at risk for vitamin D deficiency, not vitamin B6.
4. A client with chronic obstructive pulmonary disease (COPD) is receiving oxygen therapy. Which finding indicates that the therapy is effective?
- A. The client is able to expectorate secretions easily.
- B. The client's oxygen saturation is 92%.
- C. The client's respiratory rate is 20 breaths per minute.
- D. The client's arterial blood gases show a pH of 7.35.
Correct answer: C
Rationale: In a client with COPD receiving oxygen therapy, an effective response is indicated by a respiratory rate of 20 breaths per minute. This suggests that the client is effectively oxygenating while maintaining an appropriate respiratory rate. Choices A, B, and D are incorrect because expectorating secretions easily, having an oxygen saturation of 92%, and arterial blood gases showing a pH of 7.35 are not specific indicators of the effectiveness of oxygen therapy in COPD. Oxygen saturation of 92% may still be suboptimal in COPD, and arterial blood gases showing a pH of 7.35 may not necessarily reflect the overall effectiveness of oxygen therapy.
5. When preparing a class on newborn care for expectant parents, what content should be taught concerning the newborn infant born at term gestation?
- A. Milia are white marks caused by sebaceous glands and typically resolve within 2 to 4 weeks.
- B. Meconium is the first stool and is typically dark green or black in color.
- C. Vernix caseosa is a white, cheesy substance mainly found in skin folds, providing a protective layer.
- D. Pseudostrabismus in newborns usually self-resolves without the need for intervention.
Correct answer: C
Rationale: Vernix caseosa is a white, cheesy substance that acts as a protective barrier on the skin of newborns, particularly present in skin folds. It helps to prevent dehydration and protect the delicate skin of the newborn from the amniotic fluid in utero. Educating expectant parents about the presence and function of vernix caseosa can help them understand the importance of its preservation during the immediate postnatal period. Choices A, B, and D are incorrect as they do not directly relate to the protective function of vernix caseosa in newborns. Milia are small, white bumps on the skin due to blocked oil glands, meconium is the first stool of a newborn and is typically dark green or black in color, and pseudostrabismus refers to false appearance of misalignment of the eyes, which usually resolves on its own without intervention.