Which assessment information obtained by the nurse when caring for a COPD patient receiving mechanical ventilation indicates the need for suctioning?
The pulse oximeter shows a Sp02 of 90%.
The patient has not been suctioned for the last 6 hours.
The respiratory rate is 32 breaths/min.
The lungs have occasional audible expiratory wheezes.
The Correct Answer is C
In a patient receiving mechanical ventilation, a high respiratory rate can indicate increased work of breathing and potential airway obstruction. COPD patients, in particular, may have excessive mucus production and airway inflammation, leading to mucus plugging and compromised airway clearance. Suctioning may be necessary to remove excessive secretions and maintain a patent airway.
A. The pulse oximeter shows a SpO2 of 90% in (option A) is incorrect because While a SpO2 of 90% is suboptimal and may require intervention, it does not specifically indicate the need for suctioning. Other interventions, such as adjusting oxygen delivery or ventilation settings, may be more appropriate.
B. The patient has not been suctioned for the last 6 hours in (option B) is incorrect because The duration since the last suctioning episode alone does not necessarily indicate the need for suctioning. The need for suctioning should be based on the patient's clinical presentation, such as signs of airway obstruction or excessive secretions.
D. The lungs have occasional audible expiratory wheezes in (option D) which is incorrect because Occasional audible expiratory wheezes may be common in patients with COPD and may not specifically indicate the need for suctioning. Wheezing is more commonly associated with narrowing of the airways, and suctioning is typically performed to clear secretions or maintain airway patency.
C. Therefore, in a COPD patient receiving mechanical ventilation, a high respiratory rate (C) is the assessment information that would indicate the need for suctioning to help remove excessive secretions and ensure a patent airway
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is A
Explanation
The characteristics described in the monitor strip analysis suggest ventricular tachycardia. The absence of a visible P wave and the wide and distorted QRS complex indicates that the electrical impulse is originating in the ventricles rather than the atria. The ventricular rate of 196 and regular R-R intervals further support the diagnosis of ventricular tachycardia.
B. Atrial fibrillation in (option B) is incorrect because it is characterized by irregularly irregular R-R intervals and the absence of discernible P waves. The QRS complex is typically narrow
C. Atrial tachycardia in (option C) is incorrect because it would have a rapid atrial rate with regular R-R intervals, and P waves may or may not be discernible. The QRS complex is typically narrow.
D. Ventricular fibrillation in (option D) is incorrect because it would present as a chaotic, rapid, and irregular electrical activity with no discernible P waves, QRS complexes, or regular R-R intervals. It is a life-threatening emergency that requires immediate defibrillation.
Therefore, based on the provided information, the nurse would interpret the patient's cardiac rhythm as ventricular tachycardia. However, it is important to note that an accurate interpretation should be made by a qualified healthcare professional, and the patient's clinical context should also be considered.
Correct Answer is C
Explanation
Heart rate: 72 beats per minute Stroke volume: 90 mL/contraction
Cardiac output = Heart rate × Stroke volume
Cardiac output = 72 beats/minute × 90 mL/contraction
To simplify the calculation, you can convert the units:
72 beats/minute × 90 mL/contraction = (72 × 90) beats/minute × mL/contraction
Now, perform the multiplication:
72 × 90 = 6,480
Therefore, the cardiac output is 6,480 mL per minute.
The correct answer is:
C. 6,480 mL
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