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enyata [817]
3 years ago
11

Determine the mass of each of t he following:

Chemistry
1 answer:
EastWind [94]3 years ago
3 0

Explanation:

Mass of compounds = Moles of compound × Molecular mass of compound

a) Moles of LiCl = 2.345 mol

Molecular mass of LiCl = 42.5 g/mol

Mass of 2.345 moles of LiCl = 2.345 mol × 42.5 g/mol = 99.6625 g

b) Moles of acetylene = 0.0872 mol

Molecular mass of acetylene= 26 g/mol

Mass of 0.0872 moles acetylene= 0.0872 mol × 26 g/mol = 2.2672 g

c) Moles of sodium carbonate= 3.3\times 10^{-2}mol

Molecular mass of sodium carbonate= 106 g/mol

Mass of  3.3\times 10^{-2}mol sodium carbonate

=  3.3\times 10^{-2}mol\times 106 g/mol= 3.498 g

d) Moles of fructose = 1.23\times 10^3 mol

Molecular mass fructose= 180 g/mol

Mass of  1.23\times 10^3 mol fructose

=  1.23\times 10^3mol\times 180 g/mol = 221,400 g

e) Moles of  FeSO_4(H_2O)_7=0.5758 mol

Molecular mass of FeSO_4(H_2O)_7 =278 g/mol

Mass of  1.23\times 10^3 mol fructose

=  0.5758 mol\times 278 g/mol= 160.0724 g

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Answer:

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Explanation:

<u>1. Number of moles of gasoline</u>

a)  Convert 60.0 liters to grams

  • density = 0.77kg/liter
  • density = mass / volume
  • mass = density × volume
  • mass = 0.77kg/liter × 60.0 liter = 46.2 kg

  • 46.2kg × 1,000g/kg = 46,200g

b) Convert 46,200 grams to moles

  • molar mass of C₈H₁₈ = 114.2 g/mol
  • number of moles = mass in grams / molar mass
  • number of moles = 46,200g / (114.2 gmol) = 404.55 mol

<u>2. Number of moles of carbon dioxide, CO₂ produced</u>

a) Balanced chemical equation (given):

  • C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)

b) mole ratio:

  • 1 mol C₈H₁₈ / 8 mol CO₂ = 404.55 mol C₈H₁₈ / x

Solve for x:

  • x = 404.55mol C₈H₁₈ × 8 mol CO₂ / 1mol C₈H₁₈ = 3,236.4 mol CO₂

<u> 3. Convert the number of moles of carbon dioxide to volume</u>

Use the ideal gas equation:

  • pV = nRT
  • V = nRT/p
  • p = 1 atm
  • T = 298.15K
  • n = 3,236.4 mol
  • R = 0.08206 (mol . liter)/ (K . mol)

Substitute and compute:

  • V =3,236.4 mol × 0.08206 (mol . liter) / (K . mol) 298.15K / 1 atm
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If this same quantity of energy were transferred to 2.5 kg of water at it's boiling point what fraction of the water would be va
garik1379 [7]

Answer:

Fraction of water that can be vaporized = 0.0961 or 9.61%

<em>Note : The question is incomplete. The complete question is given below:</em>

<em>A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body.</em>

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Explanation:

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Since 1 kcal = 4.18 kJ, 130 kcal = 130 * 4.18 kJ = 543.4 kJ

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Fraction of water that can be vaporized by 543.4 kJ energy = 543.4/5650

Fraction of water that can be vaporized = 0.0961 or 9.61%

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