A student looks through a microscope and sees a cell that contains a cell membrane, cytoplasm, mitochondria, and lysosomes. Whic
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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.
Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
