A nurse is assessing a newborn following a vaginal delivery. Which of the following findings should the nurse report to the provider?
Heart rate 136/min
Nasal flaring
Transient strabismus
Overlapping of sutures
The Correct Answer is B
A. Heart rate 136/min is a normal finding for a newborn. The normal range of heart rate for a newborn is 100 to 160/min.
B. Nasal flaring is an abnormal finding for a newborn. Nasal flaring indicates respiratory distress and may be caused by conditions such as pneumonia, meconium aspiration, or congenital heart defects.
C. Transient strabismus is a normal finding for a newborn. Transient strabismus is a temporary misalignment of the eyes that occurs due to weak eye muscles and poor coordination. It usually resolves by 3 to 6 months of age.
D. Overlapping of sutures is a normal finding for a newborn. Overlapping of sutures is caused by molding of the skull during delivery and allows the head to fit through the birth canal. It usually resolves within a few days after birth.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is A
Explanation
A. Correct. The nurse should avoid including raw fruits in the client's diet because they can harbor bacteria and fungi that can cause infection in a client who has neutropenia, which is a low white blood cell count.
B. Incorrect. The nurse should limit visits from anyone who is sick or has been exposed to an infection, but there is no need to restrict visits from young children specifically, as long as they are healthy and follow proper hand hygiene.
C. Incorrect. The nurse should measure the client's temperature at least every 4 hr, or more frequently if indicated, because fever is a sign of infection in a client who has neutropenia and requires prompt intervention.
D. Incorrect. The nurse should use disposable gloves from a box inside the client's room, not outside, to prevent cross-contamination and protect the client from exposure to pathogens.
Correct Answer is ["18"]
Explanation
The client weighs 198 lb, which is equivalent to (198 ÷ 2.2 = 90kg.
Therefore, the amount of mannitol for the test dose is 0.2 g/kg x 90 kg = 18 g.
The nurse should administer 18 g of mannitol IV bolus over 5 min as a test dose to the client who has severe oliguria.
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