# Homework 24

- Page ID
- 28939

## 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
- h= 6.626* 10^-34 (Planck’s Constant)
- c=2.99*1010 10^-16 m/s (speed of light)
- E=Energy
- wavelength (convert to meters)

- Energy of photon= binding energy + kinetic energy
- 1.67*r10^-16 J=Kinetic energy

- 1.167*10^-16 J* * 1 eV/1.603*10^-19 J
- =728.5 eV

## 8.37

How many moles of H_{2} form for each amount of C_{2}H_{6 }reacted?

C_{2}H_{6} (g) ---> C_{2}H_{2}(g)+2H_{2}(g)

- 2.1 mol C
_{2}H_{6}- mol C
_{2}H_{6 -->}mol H_{2}

- mol C
- 3.4 mol C
_{2}H_{6}- mol C
_{2}H_{6 -->}mol H_{2}

- mol C
- 5.68 g C
_{2}H_{6}- Convert g C
_{2}H_{6--> }mol C_{2}H_{6 --->}mol H_{2}

- Convert g C
- 2.35 g C
_{2}H_{6}- Convert g C
_{2}H_{6 -->}mol C_{2}H_{6 -->}mol H_{2}

- Convert g C

(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 C_{2}H_{6 * }2 Mol H2/ 1 Mol C2H6= 4.2 mol H_{2}

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

3) 5.68 g C_{2}H_{6 *}1 Mol C2H6/ 30 Mol C2H6 = .1893 mol C_{2}H_{6 }

.1893 mol C_{2}H_{6 *}2 Mol H2/ 1 Mol

4) 2.35 g C_{2}H_{6*}1 Mol C2H6/ 1 Mol C2H6_{=} .0783 mol C_{2}H_{6}

.0783 mol C_{2}H_{6*}2 Mol H2/ 1 Mol C2H6 H_{2}