NCLEX-PN
Basic Care and Comfort NCLEX PN Questions Questions
Extract:
Question 1 of 5
The client is in skeletal traction with 20 lb of traction applied to a right lower leg fracture. Which intervention should the nurse perform at regular intervals?
Correct Answer: A
Rationale: A: Regular pin site care prevents infection. B: Weights are only removed in emergencies. C: Repositioning disrupts alignment. D: ROM is avoided to maintain traction.
Question 2 of 5
The client with diarrhea has had four bowel movements in the past eight hours, measuring 150 mL, 100 mL, 100 mL and 150 mL. The client is to receive one-to-one replacement with a bolus of IV 0.9% NaCl to be infused over the next two hours. How many mL of 0.9% NaCl will the nurse infuse each hour?
Correct Answer: 250
Rationale:
Total loss is 500 mL (150+100+100+150). One-to-one replacement means 500 mL over 2 hours, so 500/ 2 = 250 mL per hour.
Question 3 of 5
Which of the following foods present a problem for a client diagnosed with Celiac Disease?
Correct Answer: B
Rationale: Celiac disease, or celiac sprue, is a malabsorption disorder affecting the small intestine in which there is a problem with the ingestion of gluten, a protein normally found in grain products such as wheat, rye, oats, or barley. The other choices reflect substances that do not contain gluten and should not pose problems for a client with this disorder.
Question 4 of 5
The nurse provides a postoperative client with an analgesic medication and darkens the room before the client goes to sleep for the night. The nurse's actions:
Correct Answer: C
Rationale: Reduction of environmental stimuli (particularly light and noise) from the cerebral cortex (which can be an area of arousal) facilitates sleep. Sleep occurs when there is a decreased input into this area.
Question 5 of 5
The nurse is assessing the female client who is 65 inches tall and has a small body frame. Based on the information in the chart illustrated, what is the client's approximate ideal body weight?
Correct Answer: 117
Rationale: Height is 5'5†(65/ 12 = 5 remainder 5). Formula: 105 lb for 5 ft + 5 lb × 5 = 130 lb. Small frame subtracts 10% (130 × 0.1 = 13 lb). 130 − 13 = 117 lb.