A client is admitted with uncontrolled atrial fibrillation with a rate between 120's-130's. IV heparin therapy is prescribed. In addition, the nurse prepares the client for a transesophageal echocardiogram (TEE). What does the nurse understand about this diagnostic study?
The TEE evaluates if emboli are present if cardioversion is required
The study will use electric current to stop the abnormal conduction pathway
A TEE will help restore the client's underlying baseline normal rhythm,
A TEE is required prior to implantable defibrillator placement in clients with atrial fibrillation
The Correct Answer is A
A. The TEE evaluates if emboli are present if cardioversion is required: A transesophageal echocardiogram (TEE) provides detailed imaging of the heart’s chambers, particularly the left atrium and left atrial appendage, where clots often form in atrial fibrillation. It is used to rule out thrombi before cardioversion to reduce the risk of stroke during the procedure.
B. The study will use electric current to stop the abnormal conduction pathway: This describes electrical cardioversion, not TEE. TEE is a diagnostic imaging procedure and does not involve the delivery of electrical currents to modify the heart's rhythm or conduction.
C. A TEE will help restore the client's underlying baseline normal rhythm: TEE does not treat arrhythmias. Its role is diagnostic—to visualize cardiac structures, not to restore normal rhythm. Cardioversion or pharmacologic agents are required to correct atrial fibrillation.
D. A TEE is required prior to implantable defibrillator placement in clients with atrial fibrillation: TEE is not routinely required before implantable cardioverter-defibrillator (ICD) placement. ICD decisions are usually based on ejection fraction, rhythm stability, and risk of sudden cardiac arrest, not the presence or absence of atrial thrombi.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is ["A","B","D","E"]
Explanation
A. The P-P and R-R distances are equal and regular: Equal and regular spacing between P-P and R-R intervals indicates that both atrial and ventricular rhythms are regular. This is a fundamental aspect of rhythm interpretation, helping to distinguish between regular and irregular rhythms such as atrial fibrillation or sinus arrhythmia.
B. The rhythm rate using a 3-second strip: Assessing the heart rate using a 3-second or 6-second ECG strip helps determine whether the rhythm is bradycardic, tachycardic, or within normal limits, which is crucial for accurate rhythm classification.
C. The duration of the U waves: U waves are typically small and follow the T wave. Although their presence can suggest conditions like hypokalemia, they are not routinely assessed in basic rhythm identification. Evaluating U wave duration is more relevant in electrolyte imbalance analysis than in identifying rhythm type.
D. There is a QRS complex after each P wave: A consistent QRS following every P wave indicates effective conduction from the atria to the ventricles. Each atrial depolarization (P wave) should be followed by a ventricular depolarization (QRS complex) if the signal is being conducted properly through the AV node. This finding supports a diagnosis of sinus rhythm and helps rule out AV blocks, where conduction may be delayed or blocked entirely.
E. P waves are present, upright and rounded: P waves that are upright and rounded in lead II suggest the electrical impulse is originating from the SA node. Their presence and morphology are essential criteria for identifying sinus rhythm and differentiating it from atrial arrhythmias like flutter or fibrillation.
Correct Answer is ["125"]
Explanation
Calculate the total volume to be infused in drops.
Total drops = Total volume (mL) × Drop factor (gtts/mL)
= 1000 mL × 60 gtts/mL
= 60000 gtts
Calculate the total infusion time in minutes.
Total infusion time (minutes) = Total hours × 60 minutes/hour
= 8 hours × 60 minutes/hour
= 480 minutes
Calculate the IV flow rate in drops per minute (gtts/min).
IV rate (gtts/min) = Total drops / Total infusion time (minutes)
= 60000 gtts / 480 minutes
= 125
=125 gtts/min
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