Chem II

How much energy is required to vaporize 125 g of butane at its boiling point? The heat of vaporization for butane is 23.1 kJ/mol.

(125 g C4H10) / (58.1222 g C4H10/mol) x (23.1 kJ/mol) = 49.68 kJ

When a solid is placed in a container and heat is applied, a phase change occurs

-when a solid is converted to a liquid, heat is absorbed -temperature remains constant while all of a solid is converted to a liquid

-has a melting point of 1064 ∘C -is insoluble in water -conducts electricity as a solid, and is hard

Au

What volume will 125 g of butane occupy at 745 torr and 35 ∘C?

Convert g of butane to moles, then use the ideal gas law to find the volume. 125 g C4H10 x (1 mole C4H10 / 58.1 g C4H10) = 2.15 moles C4H10 PV = nRT V = nRT / P V = volume in L n = moles of C4H10 R = gas constant = 0.0821 L atm / K mole T = Kelvin temperature = 35 C + 273 = 308 K P = pressure in atm = 745 torr x (1 atm / 760 torr) = 0.980 atm V = (2.15)(0.0821)(308) / (0.980) = 55.5 L

-has a melting point of 1000 ∘C -is soluble in water -does not conduct heat as a solid, and is hard

NiCl2

A solution is prepared by dissolving 50.4 g sucrose (C12H22O11C12H22O11) in 0.332 kg of water. The final volume of the solution is 355 mL. For this solution, calculate the molarity

[Moles sucrose = 50.6 g / 342.3 g/mol = 0.1472 moles sucrose] -M = n/V = m×MW/V = 50.4 g/0.355 L × 1 mol/342.286 g = .415 M molarity: .415 M -molality = moles solute / kilogram solvent molality = 0.1472 mol / 0.332 kg = 0.4452 molal solution molality: 0.445 -percent by mass = msolute/msolution×100% = 50.4 g/332 g+50.4 g×100% = 13.2% percent by mass: 13.2 % -mole fraction = moles solute / (moles solute + moles solvent) moles solvent = 332 g H2O / 18.0 g/mol = 18.4 moles solvent mole fraction = 0.1472 / (18.4 + .15) = .00793531 mole fraction: 7.94x10^-3 -Mole % = mole fraction X 100 = 0.7935 mole %: 0.7935

What is the correct way to prepare a 1 molar solution of KNO3KNO3?

add 101.1 grams of KNO3KNO3 to a small volume of water in a 1-L volumetric flask, dissolve the KNO3, and then fill the flask up to the 1 L mark with water

Bragg's equation

d=nλ/2sin(θ)

If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporization of water at 100.0 ∘C is 40.7 kJ⋅mol−1. (Assume the density of water is 1.00 g/mL)

m = Q/delta H MM = 18 g/mol The enthalpy in mass: ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g Finally, solving for m: m = 970 / 2.261 = 429 g Converting this into volume: 429 g = 429 mL 429 / 1000 = 0.429 L of water

The heat energy associated with a change in phase at constant temperature is given by q=mΔH where q is heat in joules, m is mass in grams, and ΔH is the enthalpy in joules per gram.

q=m(delta)H

The heat energy associated with a change in temperature that does not involve a change in phase is given by q=msΔT where q is heat in joules, m is mass in grams, s is specific heat in joules per gram-degree Celsius, J/(g⋅∘C), and ΔT is the temperature change in degrees Celsius

q=ms(delta)T

An aqueous solution is saturated with both a solid and a gas at 5 ∘C. What is likely to happen if the solution is heated to 85 ∘C?

some gas will bubble out of solution and more solid will dissolve

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 112.0 ∘C? The heat capacities of ice, water, and steam are 2.09, 4.18, and 2.01 J⋅g−1⋅∘C−1, respectively. The enthalpy of fusion of ice is 6020 J⋅mol−1, and the enthalpy of vaporization of water is 40,700 J⋅mol−1.

specific heat of water is 4.186 kJ/kgC specific heat of ice is 2.06 kJ/kgC specific heat of steam is 2.1 kJ/kgK heat of fusion of ice is 334 kJ/kg heat of vaporization of water is 2256 kJ/kg Q1 = 13 * 2.09 * 14 = 271.7 J Q2 = 13 * 333 = 4329 J Q3 = 13 * 4.186 * 100 = 5441.8 J Q4 = 13 * 2260 = 29380 J Q5 = 13 * 2.01 * 14 = 337.68 J 271.7 + 4329 + 5441.8 + 29380 + 337.68 = 39760.18 J = 39.7 kJ

How much heat is evolved in converting 1.00 mol of steam at 160.0 ∘C to ice at -55.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C)

(1.00 mol H2O) x (18.01532 g H2O/mol) = 18.0153 g H2O (2.01 J/(g⋅∘C)) x (18.0153 g) x (160.0 - 100.0)∘C = 2172.645 J from cooling the steam to 100.0∘C (40660 J/mol) x (1.00 mol) = 40660 J from condensing the steam (75.37 J/(mol·°C) x (1.00 mol) x (100 - 0)∘C = 7537 J from cooling the condensed steam to its freezing point (6010. J/mol) x (1.00 mol) = 6010 J from freezing the water (2.09 J/(g⋅∘C)) x (18.0153 g) x (0 - (-55.0)∘C = 2070.8587 J from cooling the ice to -55.0 ∘C 2172.645 J + 40660 J + 7537 J + 6010 J + 2070.8587 J = 58450.5037 J = 58.5 kJ total

How much heat energy, in kilojoules, is required to convert 77.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ?

(2.09 J/(g⋅∘C)) x (77.0 g) x (0 - (-18)) ∘C = 2896.74 J to warm the ice to its melting point (334 J/g) x (77.0 g) = 25718 J to melt the ice (4.18 J/(g⋅∘C)) x (77.0 g) x (25.0 - 0) ∘C = 8046.5 J to warm the melted ice to 25.0∘C 2896.74 J + 25718 J + 8046.5 J = 36661.24 J = 36.7kJ total

Atomic solids Molecular solids Ionic solids

-Fe(s) He(s) -H2O(s) -K2O(s) He(s)He(s) and Fe(s)Fe(s) are atomic solids because helium (He)(He) and iron (Fe)(Fe) are atoms. H2O(s)H2O(s) is a molecular solid because water (H2O)(H2O) is a molecule that is composed of a nonmetal bonded to a nonmetal. K2O(s)K2O(s) is an ionic solid because potassium oxide (K2O)(K2O) is an ionic compound that is composed of a metal bonded to a nonmetal

Explain why the curve has two segments in which heat is added to the water but the temperature does not rise

-there are two horizontal lines in the heating curve because there are two endothermic phase changes -the heat that is added is used to change the phase from solid to liquid or from liquid to gas, and therefore there is no rise in temperature

X-rays with a wavelength of 1.58 Å scatter at an angle of 30.5∘ from a crystal. If n=1, what is the distance between planes of atoms in the crystal that give rise to this scattering?

1(5.8)/2sin(30.5)= 1.56Å

Which 1 MM solution would have the highest vapor pressure at a given temperature?

C6H12O6

Water and toluene are not miscible. Which of the following is most likely the formula of toluene?

C6H5CH3

How long would it take for 1.50 mol of water at 100.0 ∘C to be converted completely into steam if heat were added at a constant rate of 25.0 J/s ?

H(vap) = 40.67J/g*C 1.50mol = 27.0g amount of heat in joules is required to convert the water into steam at the boiling point 60750 J With each passing second, the water will gain 25.0 J of heat. How many seconds will it take to supply 60750 J of heat to the water sample? 2430 s = 0.68 hr = 40.5 min

The heat of vaporization of water at 100°C is 40.66 kJ/mol. Calculate the quantity of heat that is absorbed/released when 5.00 g of steam condenses to liquid water at 100°C

Q = moles steam x molar heat of vaporization Q = 5g / 18g/mole x 40.66kJ/mole = 11.3kJ released

What would be the boiling point of a 3.60 mole aqueous sucrose solution?

The boiling point elevation constant for water is 0.512 degrees Celsius per mole. The boiling point for a sucrose solution is 100.51 degrees. delta t = m x Kb x 1 = 3.60 x 0.512 = 1.84 C 1.84 + 100.51 = 102.35 solution boils at 102.35 C