NS 101 HW 9 PDF
Document Details
2024
Marc Ruiz
Tags
Related
- (all answers correct)Atoms, ions and molecular bonds - Homework Quiz_ BMS 1004 - Intro Biomd Sci_Princ Drug Ac1 (PAE) (DE).pdf
- (all answers correct)Carbohydrate Structures Homework Quiz_ BMS 1004 - Intro Biomd Sci_Princ Drug Ac1 (PAE) (DE).pdf
- UNIT 1 HOMEWORK (1).pdf
- Unit 1 Introduction To Analytical Chemistry PDF
- Chemistry Homework Chapters 7 & 8 PDF
- Homework Chapter 5 & 6 PDF
Summary
This assignment is a chemistry homework for a course called NS 101, due October 7, 2024. The assignment requires students to calculate the molar mass and percentage composition of various chemical compounds and elements.
Full Transcript
Marc Ruiz 14/14 October 07,2024 NS 101 HW 9 Mrs. Liban Answer the following questions: (14 pts.) Solve the following Problems: 2 points each A. Calculate the molar mass of the following: 1. N2O - 2(14)...
Marc Ruiz 14/14 October 07,2024 NS 101 HW 9 Mrs. Liban Answer the following questions: (14 pts.) Solve the following Problems: 2 points each A. Calculate the molar mass of the following: 1. N2O - 2(14) + 1(16) > 28 + 16 = 44 amu 2. PbS - 1(207) + 1(32) > 207 + 32 = 239 amu 3. MgSO4 - 1(24) + 1(32) + 4(16) > 24 + 32 + 64 = 120 amu 4. HgCl2 - 1(201) + 2(35) > 200 + 72 = 271 amu B. Calculate the percentage composition of the following: 5. K2CO3 Mass of K = 2(39.09) = 78.18 g/mol Mass of C = 12.01 g/mol Mass of O = 3(16.00) = 48.00 g/mol Percentage of K = (78.18 / 138.21) × 100% = 56.57% Percentage of C = (12.01 / 138.21) × 100% = 8.69% Percentage of O = (48.00 / 138.21) × 100% = 34.74% 6. CaSO4 Mass of Ca = 40.08 g/mol Mass of S = 32.06 g/mol Mass of O = 4(16.00) = 64.00 g/mol Percentage of Ca = (40.08 / 136.14) × 100% = 29.44% Percentage of S = (32.06 / 136.14) × 100% = 23.55% Percentage of O = (64.00 / 136.14) × 100% = 47.01% 7. KNO3 Mass of K = 39.09 g/mol Mass of N = 14.01 g/mol Mass of O = 3(16.00) = 48.00 g/mol Percentage of K = (39.09 / 101.11) × 100% = 38.65% Percentage of N = (14.01 / 101.11) × 100% = 13.86% Percentage of O = (48.00 / 101.11) × 100% = 47.49%
