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Chemistry LibreTexts

Homework 24

  • Page ID
    28939
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    3.78

    A photon of wavelength 0.787 nm strikes a surface. The surface has a binding energy of 1.35 *1010 10^-19 kJ/mol. What is the kinetic energy of the emitted electron in eV?

    Solution

    Strategy:

    1. Convert kJ/mol à J to keep units consistent.
    2. Solve For energy of the photon
    3. Subtract binding energy from the energy of the photon
    4. Convert J --> eV
    1. 1.35*10^-19 kJ/mol = 1.35*10-16 J/mol
    2. Energy=(Planck's Constant)(Speed of light)/ wavelength
      1. h= 6.626* 10^-34 (Planck’s Constant)
      2. c=2.99*1010 10^-16 m/s (speed of light)
      3. E=Energy
      4. wavelength (convert to meters)
    3. Energy of photon= binding energy + kinetic energy
      1. 1.67*r10^-16 J=Kinetic energy
    4. 1.167*10^-16 J* * 1 eV/1.603*10^-19 J
      1. =728.5 eV

    8.37

    How many moles of H2 form for each amount of C2H6 reacted?

    C2H6 (g) ---> C2H2(g)+2H2(g)

    1. 2.1 mol C2H6
      1. mol C2H6 --> mol H2
    2. 3.4 mol C2H6
      1. mol C2H6 --> mol H2
    3. 5.68 g C2H6
      1. Convert g C2H6--> mol C2H6 ---> mol H2
    4. 2.35 g C2H6
      1. Convert g C2H6 --> mol C2H6 --> mol H2

    (Hint: In the equation, the number before the molecule is the amount of moles that specific molecule expresses. If it doesn’t have a number, that molecule has 1 mole)

    Solution

    (Hint: When converting from grams (g) to moles, divide by the number in the denominator (molar mass) to cancel out grams. Use the periodic table.)

    1) 2.1 mol C2H6 * 2 Mol H2/ 1 Mol C2H6= 4.2 mol H2

    2) 3.4 mol C2H6 * 2 Mol H2/ 1 Mol C2H6= 6.8 mol H2

    3) 5.68 g C2H6 *1 Mol C2H6/ 30 Mol C2H6 = .1893 mol C2H6

    .1893 mol C2H6 *2 Mol H2/ 1 Mol

    4) 2.35 g C2H6*1 Mol C2H6/ 1 Mol C2H6= .0783 mol C2H6

    .0783 mol C2H6*2 Mol H2/ 1 Mol C2H6 H2


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