Chem 1314 Exam 3 2015

Ideal Gas

100

A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass in the sample. (1 atm, 0C)

n = PV / RT n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ] ======n = 2.50866 mol g = 100 g

200

At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?

T = PV / nR T = (1.95 atm) (12.30 L) / (0.654 mol) (0.0821) =====T = 447 K

300

96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight?

n = PV / RT n = [ (.92atm) (48.0 L) / (0.0821) (293.0 K) =====n = 1.8388 mol ---> 96.0 g / 1.8388 mol =52.2 g/mol

400

A 30.6 g sample of gas occupies 22.414 L at STP. What is the molecular weight of this gas?

n = PV/RT --> (1Atm)(22.414L)/(.0821)(273) = 1 mm = g/mol --> 30.6g/1mol --> =====30.6 g/mol

500

5.600 g of solid CO2 is put in an empty sealed 4.00 L container at a temperature of 300 K. When all the solid CO2 becomes gas, what will be the pressure in the container?

P = nRT/V 5.600 g / 44.009 g/mol = 0.1272467 mol p = (0.1272467 mol) (0.0821) (300 K) / (.4L) --> P = 0.7831 atm

Thermodynamics

100

What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in Joules? Cs =4.18 J/g C

q = mcΔT q = (25 g)x(4.18 J/g·°C)x(100 °C) ----> q = 10450 J

200

How much energy would be needed to heat 450 grams of copper metal from a temperature of 25.0ºC to a temperature of 75.0ºC? The specific heat of copper at 25.0ºC is 0.385 J/g ºC.

q = mcΔT q = (450 g) (0.385 J/g ºC) (50.0ºC) ----> 8700 J

300

A 1.000 g sample of octane (C8H18) is burned in a bomb calorimeter containing 1200 grams of water at an initial temperature of 25.00ºC. After the reaction, the final temperature of the water is 33.20ºC. The heat capacity of the calorimeter (also known as the “calorimeter constant”) is 837 J/ºC. The specific heat of water is 4.184 J/g ºC. Calculate the heat of combustion of octane in kJ/mol.

q = (Cw x mw x ΔT) + (Ccal x ΔT) ---> q = (4.184) x (1200g) x (8.2 Co) + (837 x 8.8 Co) = 47994.6 J ------> 47.99 KJ / .00877 mols → -5486 Kj/m

400

A 245.7g sample of metal at 75.2 degrees Celsius was placed in 115.43g water at 22.6 degrees Celsius. The final temperature of the water and metal was 34.6 Celsius. If no heat was lost to the surroundings what is the specific heat of the metal? (Cs of water is 4.184 J/(g)(C))

-q = q -> mcΔT = - (mcΔT) -> Cmetal = - (Cw m ΔT) / (m ΔT) --> (4.184) x (115.43g) x (12 Co) / (245.7 g) (-40.6) ---> .5826 J/gC

500

A 32.5g cube of aluminum initially at 45.8ºC is submerged into 105.3g of water at 15.4ºC. What is the final temperature of both substances at thermal equilibrium? Cs = .902

-q = q --> mcΔT = - (mcΔT) --> -m C (Tf-Ti)= m C (Tf-Ti) --> -(32.5g) x (0.902J/goC) x (Tf-45.8 Co) = (105.3g) x (4.18J/goC) x (Tf-15.4 Co) --> Distribute terms to get -(29.315 Tf - 1342.67) = 440.64 Tf - 6785.96 --> 8128.63 = 469.96 Tf --> Tf = 17.296 Co

Light

100

Light with a frequency of 7.26 x 10^14 Hz lies in the violet region of the visible spectrum. What is the wavelength of this frequency of light? Answer in units of nm.

C/v = λ → (3 x 10^8) / (7.26 x 10^14) → 4.13 x 10^-7 m → 413 nm

200

What is the frequency of light with a wavelength of 900nm?

C/ λ = v → (3 x 10^8) / (900 x 10^-9 m) → 3.333 x 10^14 hz

300

When an electron beam strikes a block of copper, x-rays of frequency 1.07 x 10^19Hz are emitted. What is the wavelength of these x-rays? Answer in units of nm.

C/v = λ → (3 x 10^8) / ( 1.07 x 10^19Hz) → 2.80 x 10^-11 m → .028 nm

400

Light with a wavelength of 525 nm is green. Calculate the energy in joules for a green light photon?

∆E= hc/λ → ( 6.63 x 10^-34 J•s) x (3 x 10^8) / (525 x 10^-9 m) → 3.788 x 10^-19 J

500

4. Calculate the energy of the red light emitted by a neon atom with a wavelength of 703.2 nm

ΔE = hc/λ → (6.63 x 10^-34 J•s) x (3 x 10^8 m/s) / (703.2 x 10^-9) → 2.83 × 10^-19 J

Photo Electric

100

We need to determine the maximum kinetic energy of an electron ejected from a silver foil (Threshold =6,9×10^−19 )by ultraviolet radiation (250 nm)

KE = (hc/ λ) – Φ (6.63×10^−34)(3×10^8)(250×10^−9] / (6.9×10^−19)

200

If we were to shine the same ultraviolet radiation (f=1,2×10^15 Hz) on a gold foil (work function = 8,2 × 10^−19 J ) would any electrons be emitted from the surface of the gold foil?

hf=(6.63×10^−34) (1.2×10^15)= 7.96×10^−19 J The energy is lower then the required energy/threshold energy there would be no particles ejected.

300

If we were to shine the same ultraviolet radiation (f =1.2×10^15 Hz) on a gold foil (work function/Threshold = 8.2 × 10^−19J ) would any electrons be emitted from the surface of the gold foil?

KE = hv - Φ → (6.626 x 10^-34 J/s) (1.2×10^15 Hz) – ( 8.2 × 10^−19J ) → -9.6 x 10^-20 J --> no electrons ejected since there is not enough energy to accede the threshold

400

If I shine ultraviolet light with a wavelength of 288 nm onto some aluminium foil( 6.9×10^−19 J), what would the kinetic energy of the emitted electrons be?

KE = (hc/ λ) – Φ → (6,63×10^−34 m2kg·s−1) (3×10^8 m·s) / (288×10^−9 m) - (6.9×10^−19 J) =6.25×10^−22 J