ATI Pharmacology (ASSURED A) ATI Pharmacology IMPORTANT LAB VALUES • Sodium (Na): 136-145 mEq/L • Calcium: 9.0-10 mg/dL • Chloride: 98-106 mEq/L • Bicarb HCO3: 21-28 mEq/L • Potassium: 3.5-5.0 mg/L • Phosphorus PO4 : 3.0-4.5 mg/dL • Magnesium: 1.3- 2.1 mEq/L • Cholesterol: o Total <200mg/dL o LDL ("bad") <100 o HDL ("good) >40 o Triglycerides <150mg/dL • Liver enzymes o ALT/SGPT 8-20 units/L o AST/SGOT 5-40 units/L o ALP 42-128 units/L o ...

CHEM 120 Final Exam 1. Question: (TCO 8) 35.0 mL of 0.25 M NaOH is neutralized by 23.6 mL of an HCl solution. The molarity of the HCl solution is (show your work): (Pts. : 5) 2. Question: (TCO 1) How many mL are in 3.5 pints? Show your work. (Pts. : 5) 3. Question: (TCO 3) What is the name of the following compound: Zn3P2? (Pts. : 5) 4. Question: (TCO 3) What is the name of the following compound: AgNO3? (Pts. : 5) ...

CHEM 120 Week 8 Final Exam A+ GRADED 6. (TCO 6) A gas at a temperature of 95 degrees C occupies a volume of 165 mL. Assuming constant pressure, determine the volume at 25 degrees C. Show your work. (Points : 5) Using Charles’ Law, (V1/T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t1 +273) Thus, T1 = 95 + 273 = 368. We have V1 (165 mL) & T2 = (25 + 273) = 298. V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL. 7. (TCO 6) A sample of helium gas occupies ...

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