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NBRC Mock TMC Exam Review 2024 Questions and Answers 2024 / 2025 (Verified Answers by Expert) £10.36   Add to cart

Exam (elaborations)

NBRC Mock TMC Exam Review 2024 Questions and Answers 2024 / 2025 (Verified Answers by Expert)

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NBRC Mock TMC Exam Review 2024 Questions and Answers 2024 / 2025 (Verified Answers by Expert)

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  • August 9, 2024
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NBRC Mock TMC Exam Review.pdf file:///C:/Users/HP/Desktop/TYPA%20NEW/NBRC%20Mock%20




NBRC Mock TMC Exam Review



1. A patient is receiving O2 from an E cylinder at 4 L/min through a nasal cannula.

The cylinder pressure is 1900 psig. How long will the cylinder run until it is empty?


A. 47 min

B. 1.7 h

C. 2.2 h

D. 3.6 h E cylinder = 0.28

1900x0.28/4 = 133/60 = 2.21


Answer C. 2 hours, 21 minutes




2. After the Respiratory Therapist sets up a nonrebreathing mask on a patient at a flow

rate of 10 L/min, the reservoir bag collapses before the patient finishes inspiring. The RT

should do which of the following?


A. Change to a simple mask at a flow rate of 10 L/min





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B. Remove the one-way valve from the exhalation port.

C. Place the patient on continuous positive airway pressure (CPAP)

D. Increase the flow rate to 15 L/min

ANS D. Increase the flow rate to 15 L/min



3. A patient with carbon monoxide (CO) poisoning can best be treated with which of

the following therapies?


A. Nasal Cannula at 6 L/min

B. Simple O2 mask at 10 L/min

C. CPAP and 60% O2

D. Nonrebreathing mask

ANS D. Nonrebreathing mask



4. The following blood gas levels have been obtained from a patient using a 60% aerosol

mask.


pH 7.47

PaCO2 31 mmHg





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PaO2 58 mmHg


What should the RT recommend at this time?


A. Place the patient on CPAP

B. Increase the O2 to 70%

C. Intubate and place the patient on mechanical ventilation

D. Change to a nonrebreathing mask

ANS A. Place the patient on CPAP









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5. Given the following data, what is the patient's total arterial O2 content?


Ph 7.41

PaCO2 37 mmHg

PaO2 88 mmHg

HCO3 26 mEq/L

SaO2 95%

Hb 14 g/dL


A. 12 mL/dL

B. 14 mL/dL

C. 16 mL/dL

D. 18 mL/dL

(Hb x 1.36 x SaO2) + (0.003 x PaO2) (14 x 1.36 x 0.95) +

(0.003 x 88)

18.088 + 0.264

= 18.352


Answer






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