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Dominik [7]
3 years ago
8

1-butanol yields 1-bromobutane in the presence of concentrated sulfuric acid and an excess of sodium bromide. CH3CH2CH2CH2OH (l)

→ CH3 CH2CH2CH2Br (l) If 15.89 mL of 1-butanol produced 12.23 g of 1-bromobutane, the percentage yield of the product equals: (Assume the density of 1-butanol is 0.81 g/mL, the molar mass of 1-butanol is 74 g/mol, and the molar mass of 1-bromobutane is 137 g/mol.)
Chemistry
1 answer:
Goshia [24]3 years ago
4 0

Answer:

51.34 %

Explanation:

You first need to calculate the theoretical yield of Bromobutane taking into account the amount of initial Butanol doing some stoichiometric calculations:

15.89 mL Butanol*\frac{0.81 g Butanol}{1 mL}*\frac{1 mol Butanol}{74 g} *\frac{1 mol Bromobutane  Produced}{1 mol Butanol Consumed} *\frac{137g }{1 mol Bromobutane} =23.83 g Bromobutane Produced

Then you calculate the yield giving the real amount you produced:

Yield=\frac{12.23 g}{23.82 g} *100=51.34%

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<h3>Answer:</h3>

Al- [Ne] 3s²3p¹

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Explanation:

  • Electron configuration of an element shows the arrangement of electrons in the energy levels or orbitals in the atom.
  • Noble-gas configuration involves use of noble gases to write the configuration of other elements.
  • This is done by identifying the atomic number of the element and then identifying the noble gas that comes before that particular element on the periodic table.
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