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The volume shown in the graduated cylinder according to the image would 28 mL. Thus, the correct option would be C.
In the cylinder, the top of the liquid formed a concave meniscus with the walls of the cylinder. Thus, the bottom of the meniscus would be read as this is the normal rule.
The rule is such that for liquids that form concave meniscus with the walls of cylinders, the bottom of the meniscus is read while it is the top of the meniscus for liquids that form convex meniscus with the walls of cylinders.
Hence, looking at the image, the bottom of the concave is on the 28 mL mark on the cylinder.
More on concave and convex meniscus can be found here: brainly.com/question/17022999
Answer:
<h2> 162g/mol</h2>Explanation:
The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:
<em>Nicotine has the formula </em><em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>
- <em>1.0 mol of CO₂</em>
- <em>0.70 mol of H₂O</em>
- <em>0.20 mol of NO₂</em>
<em>Assume that all the atoms in nicotine are present as products </em>
<h2>Solution</h2>To find the empirical formula you need to find the moles of C, H, and N in each of the compound.
- 1.0 mol of CO₂ has 1.0 mol of C
- 0.70 mol of H₂O has 1.4 mol of H
- 0.20 mol of NO₂ has 0.20 mol of N
Thus, the ratio of moles is:
- C: 1.0
- H: 1.4
- N: 0.20
Divide all by the smallest number: 0.20
- C: 1.0 / 0.20 = 5
- H: 1.4 / 0.20 = 7
- N: 0.20 / 0.20 = 1
Hence, the empirical formula is C₅H₇N
Find the mass of 1 mole of units of the empirical formula:
- C: 5mol × 12g/mol = 60g
- H: 7mol × 1g/mol = 7 g
- N: 1 mol × 14g/mol = 14g
Total mass = 60g + 7g + 14g = 81g
Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.
Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.