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 A
Explanation
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.
Correct Answer is D
Explanation
This is because lochia rubra is the first stage of lochia, the vaginal discharge after giving birth. It comprises blood, shreds of fetal membranes, decidua, vernix caseosa, lanugo, and membranes. It is red in color because of the large amount of blood it contains. It lasts 1 to 4 days after birth.
Choice A is not correct because lochia alba is the last stage of lochia. It is whitish or yellowish-white in color and contains fewer red blood cells and more leukocytes, epithelial cells, cholesterol, fat, mucus, and microorganisms. It lasts from the second through the third to sixth weeks after delivery.
Choice B is not correct because there is no such thing as lochia normal. Lochia has three stages: lochia rubra, lochia serosa and lochia alba.
Choice C is not correct because lochia serosa is the second stage of lochia. It is brownish or pink in color and contains serous exudate, erythrocytes, leukocytes, cervical mucus, and microorganisms. It lasts for 4 to 12 days after delivery.
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