The nurse is caring for a client diagnosed with acute respiratory failure as a result of right middle and lower lobe pneumonia. To optimize ventilation and secretion removal the nurse should position the client:

in the prone position.

in high-Fowler's position.

on the left side.

on the right side.

Answer and Explanation

The Correct Answer is D

A. In the prone position:
The prone position has been shown to be beneficial in certain respiratory conditions, particularly in acute respiratory distress syndrome (ARDS), where it can help improve oxygenation by redistributing blood flow in the lungs. However, prone positioning is typically not the first choice for pneumonia, especially when it is localized to specific lobes of the lung. It is more commonly used in cases of diffuse bilateral lung injury or severe hypoxemia. Therefore, while prone positioning can improve oxygenation in ARDS, it is not specifically targeted for secretion removal in localized pneumonia.

B. In high-Fowler's position:
The high-Fowler's position (sitting up at a 60-90 degree angle) can help with dyspnea and promote lung expansion in conditions like heart failure or dyspneic states. However, for pneumonia, it is not as effective as lateral positioning for facilitating secretion drainage from specific lung lobes. The high-Fowler's position may be useful for promoting overall comfort and reducing dyspnea, but it is not the best position for improving secretion removal from the right middle and lower lobes.

C. On the left side:
Positioning the patient on the left side is not ideal for right middle and lower lobe pneumonia, as it would not optimize drainage from the affected lobes. The right middle and lower lobes are better drained when the patient is positioned on the right side, as gravity can help move the secretions from the affected lobes toward the larger airways for easier clearance.

D. On the right side: In the case of right middle and lower lobe pneumonia, positioning the client on the right side can help optimize ventilation and promote better secretion removal from the affected areas of the lung. This position allows gravity to assist in draining secretions from the right middle and lower lobes toward the larger airways, where they can be more easily cleared by coughing or suctioning. This positioning can improve oxygenation and facilitate secretion management, which is crucial for improving respiratory function in pneumonia.

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Correct Answer is A

Explanation

A. pH 7.36, PaO2 98 mmHg, PaCO2 27 mmHg, HCO3 16 mEq/L, O2 sat 99%: This set of ABG results is consistent with fully compensated metabolic acidosis. pH 7.36: This is within the normal range (7.35-7.45), indicating that compensation has occurred, as the pH has returned to normal levels. PaCO2 27 mmHg: The PaCO2 is low, suggesting that the respiratory system has compensated for the metabolic acidosis by increasing ventilation to excrete CO2, thus reducing the acid load. HCO3 16 mEq/L: The bicarbonate level is low, which is consistent with metabolic acidosis as the primary disturbance. The PaO2 and O2 saturation are normal, indicating adequate oxygenation. Since the pH is within the normal range and the PaCO2 and HCO3 levels reflect the compensatory changes needed to correct the metabolic acidosis, this is a case of fully compensated metabolic acidosis.

B. pH 7.47, PaO2 91 mmHg, PaCO2 52 mmHg, HCO3 30 mEq/L, O2 sat 96%:

This result indicates alkalosis rather than acidosis. The pH is alkalotic (7.47), and PaCO2 is elevated (52 mmHg), which suggests respiratory acidosis as the primary disturbance. The HCO3 is also high (30 mEq/L), which is consistent with metabolic compensation for respiratory acidosis, not for metabolic acidosis. Therefore, this is not consistent with fully compensated metabolic acidosis.

C. pH 7.45, PaO2 86 mmHg, PaCO2 56 mmHg, HCO3 28 mEq/L, O2 sat 94%:

The pH is normal, but PaCO2 is elevated (56 mmHg), indicating respiratory acidosis rather than metabolic acidosis. The HCO3 is also elevated (28 mEq/L), which is consistent with compensation for respiratory acidosis, not metabolic acidosis. This result suggests respiratory acidosis with compensated metabolic alkalosis rather than metabolic acidosis.

D. pH 7.32, PaO2 88 mmHg, PaCO2 54 mmHg, HCO3 29 mEq/L, O2 sat 94%:

The pH of 7.32 indicates acidosis, but it is not within the normal range, so this is not fully compensated. The PaCO2 is elevated (54 mmHg), indicating respiratory acidosis, and the HCO3 is elevated (29 mEq/L), showing metabolic compensation. However, since the pH has not yet returned to normal (it remains acidotic), this is an example of partially compensated respiratory acidosis, not fully compensated metabolic acidosis.

respiratory acidosis, not fully compensated metabolic acidosis.

Correct Answer is ["3.9"]

Explanation

Step-by-Step Solution:

Convert the toddler's weight from pounds to kilograms.

1 pound is =0.453592 kilograms.

So, 26 pounds = 26 x 0.453592 = 11.793592 kilograms.

Calculate the total daily dose of prednisolone.

The prescribed dose is 2 mg/kg/day.

The toddler weighs 11.793592 kg.

The total daily dose = 2 mg/kg/day x 11.793592 kg = 23.587184 mg/day.

Determine the dose per administration.

The total daily dose is divided into two equal doses (every 12 hours).

The dose per administration = 23.587184 mg/day / 2 = 11.793592 mg per dose.

Calculate the volume of prednisolone syrup to administer.

The available prednisolone syrup is 15 mg/5 ml.

We need to administer 11.793592 mg per dose.

To find the volume, we can set up a proportion:

15 mg / 5 ml = 11.793592 mg / x ml

Cross-multiplying:

15x = 5 x 11.793592

Solving for x:

x = (5 x 11.793592) / 15 = 3.931197 ml

Round the answer to the nearest tenth.

3.9 ml.

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The nurse is caring for a client diagnosed with acute respiratory failure as a result of right middle and lower lobe pneumonia. To optimize ventilation and secretion removal the nurse should position the client:

Nursing Test Bank

Naxlex Comprehensive Predictor Exams

Correct Answer is A

Explanation

A client is receiving a continuous infusion of IV insulin at 8 mL/hour. Pharmacy makes an infusion of 75 units of regular insulin in 150 mL of Normal Saline for this institution. How many units per hour is the client receiving? (Round to the nearest whole number)

Correct Answer is ["3.9"]

Explanation

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