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

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