Which finding would alert the nurse to suspect that a newborn is experiencing respiratory distress?
asymmetrical chest movement
respiratory rate of 50 breaths/minute
acrocyanosis
short periods of apnea (less than 15 seconds)
The Correct Answer is A
Asymmetrical chest movement is a sign of respiratory distress in the newborn, as it indicates unequal lung expansion or airway obstruction. A respiratory rate of 50 breaths/minute (choice B) is normal for a newborn, as is acrocyanosis (choice C), which is a bluish discoloration of the hands and feet due to immature peripheral circulation. Short periods of apnea (less than 15 seconds) (choice D) are also common and benign in newborns unless they are associated with bradycardia or cyanosis.
Choice B is not correct because a respiratory rate of 50 breaths/minute is within the normal range for a newborn.
Choice C is not correct because acrocyanosis is a normal finding in newborns and does not indicate respiratory distress.
Choice D is not correct because short periods of apnea (less than 15 seconds) are normal in newborns and do not indicate respiratory distress.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is C
Explanation
Place the infant on the mother's abdomen after birth. This will help the infant maintain an adequate body temperature by providing skin-to-skin contact with the mother, which reduces heat loss and promotes bonding. Skin-to-skin contact also stimulates the baby's natural feeding cues and helps initiate breastfeeding.
Choice A is not correct because turning up the temperature in the birth room may not be enough to prevent heat loss from the infant, especially if they are wet or exposed to cold surfaces. It may also make the mother uncomfortable or dehydrated.
Choice B is not correct because bathing the infant immediately after birth may increase heat loss from evaporation and conduction. It may also interfere with the baby's natural protective coating (vernix) and microbiome. Bathing should be delayed until at least 24 hours after birth.
Choice D is not correct because wrapping the infant in a warm, dry blanket may not provide the same benefits as skin-to-skin contact with the mother. It may also prevent the baby from smelling and seeing the mother's breast, which are important cues for breastfeeding initiation.
Correct Answer is B
Explanation
Jaundice in an infant who is 4-hr old. This is because jaundice is a yellow discoloration of the skin and eyes caused by high levels of bilirubin in the blood. Jaundice usually appears between the second and fourth day after birth and lasts for one to two weeks. Jaundice that appears within the first 24 hours of life is considered early-onset jaundice and may indicate a serious problem, such as an infection, a blood type mismatch, or a liver disorder. The nurse should notify the charge nurse of this finding and request a blood test to check the bilirubin level.
Choice A is wrong because a hematocrit of 60% in an infant who is 8-hr old is not abnormal. Hematocrit is the percentage of red blood cells in the blood. Newborns normally have higher hematocrit levels than older children and adults because they have more red blood cells at birth.
Choice C is wrong because a blood glucose fingerstick of 40 mg/dL for an infant who is 1-hr old is not abnormal.
Blood glucose is the amount of sugar in the blood. Newborns normally have lower blood glucose levels than older children and adults because they have less glycogen (stored sugar) at birth.
Choice D is wrong because acrocyanosis in an infant who is 2-hr old is not abnormal. Acrocyanosis is a bluish discoloration of the hands and feet caused by poor circulation. Newborns normally have acrocyanosis for the first few days of life because they are adjusting to the temperature outside the womb.
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