Table of contents

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

9. Solutions

Osmolarity

GOB Chemistry

9. Solutions

Osmolarity

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Multiple Choice

How many milligrams of nitride ions are required to prepare 820 mL of 0.330 M Ba₃N₂ solution?

120 mg

320 mg

560 mg

760 mg

7600 mg

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Verified step by step guidance

First, understand that the molarity (M) of a solution is defined as the number of moles of solute per liter of solution. Here, you are given a 0.330 M solution of Ba3N2, which means there are 0.330 moles of Ba3N2 per liter of solution.

Convert the volume of the solution from milliliters to liters. Since 1 L = 1000 mL, 820 mL is equivalent to 0.820 L.

Calculate the number of moles of Ba3N2 in 0.820 L of a 0.330 M solution using the formula: \( \text{moles of Ba3N2} = \text{Molarity} \times \text{Volume in liters} \).

Determine the number of moles of nitride ions (N^{3-}) in the solution. Since each formula unit of Ba3N2 contains 2 nitride ions, multiply the moles of Ba3N2 by 2 to find the moles of nitride ions.

Convert the moles of nitride ions to milligrams. Use the molar mass of the nitride ion (N^{3-}), which is approximately 14.01 g/mol, to convert moles to grams, and then multiply by 1000 to convert grams to milligrams.