How does the emergency department nurse caring for a critically ill with septic shock recognize that severe tissue hypoxia is present?
Partial thromboplastin time 64 seconds
Lactate 9.0 mmol/L
Potassium 2.8 mEq/L (2.8 mmol/L)
PaCO2 58 mm Hg
The Correct Answer is B
Lactate is a by-product of anaerobic metabolism that accumulates when there is insufficient oxygen supply to meet cellular metabolic demands. In the context of severe tissue hypoxia, such as in septic shock, the body may resort to anaerobic metabolism, leading to increased lactate production and elevated lactate levels in the blood.
Elevated lactate levels, typically above 4.0 mmol/L, are indicative of tissue hypoxia and inadequate oxygenation at the cellular level. Higher lactate levels, such as 9.0 mmol/L, suggest more severe tissue hypoxia and increased anaerobic metabolism.
A. Partial thromboplastin time (PTT) 64 seconds in (option A) is incorrect because: PTT is a laboratory test that evaluates the intrinsic pathway of the coagulation cascade. While coagulation abnormalities may occur in septic shock, PTT alone does not specifically indicate severe tissue hypoxia.
C. Potassium 2.8 mEq/L (2.8 mmol/L) (option C) is incorrect because Low potassium levels (hypokalemia) can be a concern in septic shock, but it does not directly indicate severe tissue hypoxia.
D. PaCO2 58 mm Hg in (option D) is incorrect because: PaCO2 refers to the partial pressure of carbon dioxide in arterial blood and is a measure of the respiratory status. While an elevated PaCO2 can be a sign of respiratory acidosis, it is not specific to severe tissue hypoxia.
Therefore, in a critically ill patient with septic shock, an elevated lactate level, such as 9.0 mmol/L, indicates severe tissue hypoxia and inadequate oxygenation at the cellular level
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is ["A","B","C","D"]
Explanation
These conditions can lead to fluid loss, either through increased gastrointestinal output (diarrhea, vomiting, lower GI bleeding) or accumulation of air in the pleural space (tension pneumothorax), resulting in a decrease in blood volume and subsequent hypovolemic shock.
E. Diabetes insipidus in (option E) is incorrect because it is not directly associated with hypovolemic shock. Diabetes insipidus is a condition characterized by excessive thirst and the production of large volumes of dilute urine due to insufficient production or response to antidiuretic hormone (ADH). While diabetes insipidus can lead to dehydration and potential hypovolemia, it is not a direct cause of hypovolemic shock.
F. Valvular stenosis in (option F) is incorrect because it is a condition characterized by the narrowing or obstruction of one or more heart valves. While it can cause problems with cardiac output and circulation, it is not specifically related to hypovolemic shock, which is caused by a decrease in blood volume.
Therefore, the conditions that can cause hypovolemic shock include diarrhea, vomiting, lower GI bleeding, and tension pneumothorax.
Correct Answer is ["18mL\/hr"]
Explanation
Step 1: Convert the patient's weight from pounds to kilograms. 130 pounds ÷ 2.205 (1 pound = 0.453592 kilograms) ≈ 58.97 kilograms
Step 2: Calculate the total dosage of Dobutamine required per hour based on the weight-specific dose. 2.5 mcg/kg/min × 58.97 kg = 147.425 mcg/min
Step 3: Calculate the infusion rate (mL/hr) using the concentration of Dobutamine in the prepared solution. The solution contains 250 mg of Dobutamine in 500 mL, which means there are 250,000 mcg of Dobutamine in 500 mL. To determine the mL/hr, divide the required dosage (147.425 mcg/min) by the amount of Dobutamine in 500 mL (250,000 mcg) and multiply by 500 mL (volume of the solution).
(147.425 mcg/min ÷ 250,000 mcg) × 500 mL ≈ 0.295 mL/min
To get the mL/hr, we convert the rate from minutes to hours (60 minutes = 1 hour):
0.295 mL/min × 60 min/hr ≈ 17.7 mL/hr
Round the answer to the nearest whole number:
Approximately 18 mL/hr of Dobutamine should be administered to the patient.
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