The nurse is caring for an infant with Hirschsprung’s disease who just returned from surgery for a laparoscopically assisted endorectal pull-through procedure.
The nurse must monitor which of the following orders?
NPO until bowel sounds return.
Apply barrier cream to diaper area.
Assess surgical site every 2 hours.
Monitor rectal temperature every 8 hours.
The Correct Answer is A
Choice A rationale
After a laparoscopically assisted endorectal pull-through procedure for Hirschsprung’s disease, it is important to monitor for the return of bowel sounds. This indicates that the gastrointestinal tract is starting to recover and function again after surgery. Until bowel
sounds return, indicating that peristalsis has resumed, the child should remain NPO (nothing by mouth) to prevent vomiting and aspiration.
Choice B rationale
Applying barrier cream to the diaper area is generally good practice for infants to prevent diaper rash, but it is not specifically related to the care of a child who has just undergone surgery for Hirschsprung’s disease.
Choice C rationale
While it is important to assess the surgical site regularly to check for signs of infection or complications, this does not need to be done every 2 hours. A more typical frequency might be every 4 to 8 hours, depending on the child’s condition and hospital policy.
Choice D rationale
Monitoring rectal temperature every 8 hours is not specifically related to the care of a child who has just undergone surgery for Hirschsprung’s disease. While it is important to monitor for fever as a sign of infection, temperature can usually be measured less invasively.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is D
Explanation
Answer: D. Pulse of 150, poor skin turgor, oliguria, and an admission weight of 29.5 lbs. Rationale:
Choice A: A pulse of 160 is slightly elevated for a 3-year-old (normal range is 90-140 bpm), but not significantly so. Oliguria (decreased urine output) is a common sign of dehydration. However, sunken fontanels are typically seen in infants with severe
dehydration and not necessarily present in moderate cases. An admission weight of 30 lbs is irrelevant for assessing dehydration.
Choice B: A pulse of 180 is significantly elevated and suggests potential tachycardia, which can be a sign of severe dehydration or other underlying conditions. While a furrowed tongue can occur with dehydration, it's not specific enough. Mottled skin can be caused by various factors and isn't a definitive sign of moderate dehydration.
Admission weight alone isn't indicative of dehydration.
Choice C: A pulse of 120 is within the normal range for a 3-year-old. Dry mucous membranes are a common sign of dehydration. However, pale skin can be caused by various factors and isn't specific to dehydration. Admission weight alone isn't indicative of dehydration.
Choice D: A pulse of 150 is slightly elevated, potentially indicating dehydration or other factors. Poor skin turgor is a key assessment finding in moderate dehydration. It refers to the skin's slow return to its original shape after being pinched, indicating a lack of fluid. Oliguria is a common sign of dehydration, suggesting decreased fluid intake or output. An admission weight of 29.5 lbs might be slightly lower than the child's baseline weight due to dehydration, but weight alone shouldn't be used to diagnose dehydration.
Therefore, considering the combination of a slightly elevated pulse, poor skin turgor, oliguria, and a potential decrease in weight, choice D presents the most likely assessment findings in a child with moderate dehydration.
It's important to remember that a comprehensive assessment, including history, physical examination, and potentially laboratory tests, is crucial for diagnosing and classifying the severity of dehydration in children.
Correct Answer is ["69"]
Explanation
Step 1 is: To calculate the total body surface area (TBSA) involved in burns, we use the rule of nines or a modified version for children. For a 7-year-old child, the head accounts for 8.5% (front) + 8.5% (back), and each leg accounts for 6.5% (front) + 6.5% (back)3.
Step 2 is: Therefore, the total percentage of TBSA involved is (8.5% + 8.5%) + 2 * (6.5% + 6.5%) = 17% (head) + 26% (each leg) * 2 = 17% + 52% = 69%. So, the calculated percentage of total body surface area involved is 69%.
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