Homework 24

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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:

Convert kJ/mol à J to keep units consistent.
Solve For energy of the photon
Subtract binding energy from the energy of the photon
Convert J --> eV

1.35*10^-19 kJ/mol = 1.35*10^-16 J/mol
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)
Energy of photon= binding energy + kinetic energy
1. 1.67*r10^-16 J=Kinetic energy
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 C₂H₆reacted?

C₂H₆ (g) ---> C₂H₂(g)+2H₂(g)

2.1 mol C₂H₆
1. mol C₂H_{6 -->} mol H₂
3.4 mol C₂H₆
1. mol C₂H_{6 -->} mol H₂
5.68 g C₂H₆
1. Convert g C₂H_6-->mol C₂H_{6 --->} mol H₂
2.35 g C₂H₆
1. Convert g C₂H_{6 -->} mol C₂H_{6 -->} 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 C₂H_{6 *}2 Mol H2/ 1 Mol C2H6= 4.2 mol H₂

2) 3.4 mol C₂H_{6 *}2 Mol H2/ 1 Mol C2H6= 6.8 mol H₂

3) 5.68 g C₂H_{6 *}1 Mol C2H6/ 30 Mol C2H6 = .1893 mol C₂H₆

.1893 mol C₂H_{6 *}2 Mol H2/ 1 Mol

4) 2.35 g C₂H_6*1 Mol C2H6/ 1 Mol C2H6₌ .0783 mol C₂H₆

.0783 mol C₂H_6*2 Mol H2/ 1 Mol C2H6 H₂