Chem 120 week 8 final exam Study guides, Class notes & Summaries

CHEM120 WEEK 8 FINAL EXAM/ CHEM 120 FINAL EXAM LATEST ACTUAL EXAM 2024/ QUESTIONS AND CORRECT DETAILED ANSWERS|A+GRADE 1. The name of the compound FES04 is iron sulfate 2. Peptide bonds link the amino acids together to make a primary polypeptide sequence. The bonds are AMIDE

CHEM120 Final Exam (Version-2, Updated-2023-2024)/ CHEM 120 Final Exam / CHEM120 Week 8 Final Exam / CHEM 120 Week 8 Final Exam: Chamberlain College of Nursing CHEM120 Week 8 Final Exam (Version-2, Updated-2023-2024)/ CHEM 120 Week 8 Final Exam / CHEM120 Final Exam / CHEM 120 Final Exam: Chamberlain College of Nursing

CHEM 120 WEEK 8 FINAL EXAM 2 LATEST VERSIONS ACTUAL EXAM 250 QUESTIONS AND CORRECT DETAILED ANSWERS WITH RATIONALES (VERIFIED ANSWERS) |ALREADY GRADED A+

CHEM120 WEEK 8 FINAL EXAM/ CHEM 120 FINAL EXAM LATEST ACTUAL EXAM 2024/ QUESTIONS AND CORRECT DETAILED ANSWERS|A+GRADE

CHEM 120 WEEK 8 FINAL EXAM, STUDY GUIDE AND REVIEW COMPILATION/ACTUAL EXAMS WITH CORRECT DETAILED ANSWERS AND WORKINGS/A+ GRADE/LATEST UPDATE 2024

CHEM 120 WEEK 8 FINAL EXAM, STUDY GUIDE AND REVIEW COMPILATION/ACTUAL EXAMS WITH CORRECT DETAILED ANSWERS AND WORKINGS/A+ GRADE/LATEST UPDATE 2024

CHEM 120 WEEK 8 Final Exam 2023 with correct answers Already graded A+ molarity = moles solute / liters solution 0.25 M = moles NaOH / 0.035 L moles NaOH = 0.00875 moles NaOH 1. (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): (Points : 5) 2. (TCO 1) How many mL are in 3.5 pints? Show your work. (Points : 5) 3.5 pints is equivalent to 1656.116 1 pint = 473.176 ml 3.5 pints* 473.176mL = 1656.116mL 3. (TC...

CHEM120 Final Exam (Version-1, Latest-2022)/ CHEM 120 Final Exam / CHEM120 Week 8 Final Exam / CHEM120 Unit 8 Final Exam: Chamberlain College of Nursing |Verified and 100% Correct Q & A|

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. 0 7 Short 16 7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant tempera...

CHEM 120 WEEK 8 FINAL EXAM (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): (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine the number of grams of H3PO4 needed to prepare 0.75L of a 0.25M H3PO4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.25 M H3PO4 solution can be prepared by diluting 50 mL of a 2.5M H3PO4 solution?