AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08 A. Properties of Solutions - homogeneous mixtures - may be a gas in a gas, solid in solid, gas in liquid, etc.
1. Solution composition a. Qualitative - dilute - little solute as compared to the amount of solvent - concentrated - a lot of solute compared to the amount of solvent b. Quantitative - molarity (M)- mol solute/Liter of solution Calculation hint : Density (g/mL) data can easily be converted into mass (g) by 骣mass solute - mass percent = 琪 x 100% multiplying by volume (mL) or converted 桫mass solution into volume (mL) by dividing by the mass (g) by the density given. nA n B - mole fraction XA = ; X B = m m n +n n +n D , m D x V, V A B A B V D - molality(m) - mol solute /Kg solvent - normality (N) equivalents/L
Advantage of molality over molarity?
Applications of molarity?
Applications of molality?
A solution is made by dissolving 2.50 g of lithium chloride in 50.0 g of water. Calculate the molality, mole fraction and mass percent (of LiCl).
What is the molarity of the above solution if it has a density of 1.02 g/mL?
- normality (N) - equivalents/Liter of solution - equivalent (eq) : - for an acid-base reaction an equivalent is the mass of an acid or a base that will provide one mole of H+ or OH- ions - for a redox reaction an equivalent is that mass of an oxidizing or reducing agent that can accept or donate one mole of electrons
1 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08 e.g Al(OH)3 = 78.01 g/mol = 3 eq/mol or one eq = 26.00 g/mol
0.10 M Al(OH)3 = ______N e.g. Mg(OH)2 = 58.33 g/mol = 2 eq/mol or one eq. = 29.17 g/mol
0.25 M Mg(OH)2 = ______N 2.2 M H2SO4 = _____N H2SO4
0.12 M HNO3 = _____ N HNO3
.0\25 M NaOH = _____ N NaOH
0.012 M Ca(OH)2 = _____ N Ca(OH)2
Redox example :
- 2+ MnO4 + 5e- + 8H+ ---> Mn + 4 H2O
Gram equivalent mass of KMnO4? (molar mass = 158.04 g/mol)
C2H5OH + 3 H2O ---> 2 CO2 + 12 H+ + 12 e-
C2H5OH (ethanol) is the reducing agent
Mass of an equivalent of ethanol?
Normality problem :
2.40 g of barium hydroxide are dissolved in enough water to form 50.0 mL of solution. What is the normality of this solution?
2. The Energies of Solution Formation a. Cardinal rule : "Like dissolves like."
Which of the following solutes will dissolve in hexane (C6H14).
____ CH4 ____ C2H5OH ____ NaCl
____ C6H12O6
Which of the following solutes will dissolve in water?
_____ CH3OH _____ NaI
_____ BaSO4 _____ CCl4 _____ C2H6
2 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
b. Entropy and enthalpy changes in the solution process : ΔG = ΔH – TΔS where : ΔG - Gibbs free energy ΔH - enthalpy (heat content) Diagram of hydration process :
ΔS - entropy A reaction is only spontaneous if ΔG is negative.
c. Three steps to solution formation : - Step 1 : break up of the solute into individual components (ions, atoms or molecules)
- endothermic - requires energy to overcome attractive forces of particles (ΔH1>0) - Step 2 : separation of the solvent N.B. : ΔH1 is opposite of - endothermic - intermolecular forces must be overcome(ΔH2>0) lattice energy - Step 3 : solute and solvent interact to form solution - often exothermic - bonding of solute and solvent results in a more stable, lower energy
arrangement(ΔH3<0)
d. enthalpy (or heat) of solution = ΔHsoln = ΔH1 + ΔH2 + ΔH3 - if ΔHsoln is negative (exothermic), energy will be released (magnitude of ΔH3 is greater than ΔH1 + ΔH2). - if ΔHsoln is positive (endothermic), energy is absorbed (ΔH3 is not exothermic or magnitude of ΔH3 is less than ΔH1 + ΔH2). e. Entropy - a measure of the disorder of a system - along with enthalpy changes determines if a solution will form or not - factors favoring solution formation :
- decrease in enthalpy (negative value, exothermic : ΔH3 > (ΔH1 + ΔH2)) - increase in entropy (positive value, increase in disorder) - factors inhibiting solution formation :
- increase in enthalpy (positive value, endothermic : ΔH3 < (ΔH1 + ΔH2)) - decrease in entropy (negative value, decrease in disorder)
3 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
Sample problem :
Solid cesium hydroxide has a lattice energy of -724 kJ/mol. If the heat of solution is -72 kJ/mol, what is the heat of hydration?
Summary of Energies and Solution Formation :
ΔH1 ΔH2 ΔH3 ΔHsoln Result Polar solvent, Large Large Large, Small Solution polar solute negative forms* Polar solvent, Small Large Small Large, No solution nonpolar solute positive forms Nonpolar solvent, Small Small Small Small Solution nonpolar solute forms* Nonpolar solvent, Large Small Small Large, No solution polar solute positive forms *Note "Like dissolves like," substances with similar intermolecular forces tend to be soluble in each other. 4 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
Rank the following alcohols in order of increasing solubility.
_____ CH3CH2CH2OH
_____ CH3CH2OH
_____ CH3OH
_____ CH3CH2)4OH
Substance Formula Solubility (mg/L) Melting point Boiling point Alkanes
Methane CH4 24 -183°C −164°C
Ethane C2H6 60 -183°C -88.5°C
Propane C3H8 62 -90°C -42.7°C
n-Butane C4H10 61 -138°C -0.5°C
n-Pentane C5H12 40 -130°C 36°C
n-Hexane C6H14 10 -95°C 68.95°C Alcohols
Methanol CH3OH Infinite -97°C 64.5 ºC
Ethanol C2H5OH Infinite -115°C 78.4°C
Propanol C3H7OH Infinite −126.5 °C°C 97°C
n-Butanol C4H9OH 90,000 -89.8°C 117°C
n-Penatanol C5H11OH 27,000 -78.9°C 137°C
n-Hexanol C6H13OH 6,000 -44.6°C 157°C For example, the melting point of n-butanol, iso-butanol, and tert-butanol (all of which have hydrogen bonding and -because they have the same molar mass, can be expected to have the same degree of London forces) are: -89C, -115C, and +25C respectively
Substance Electronegativity Melting Point Boiling point difference HF 1.9 -83.85 °C 20 °C HCl 0.9 -114.8 °C -85 °C HBr 0.7 -86.8 °C -66 °C HI 0.4 -50.76 °C -36°C
Substance Electronegativity Melting Point Boiling point difference
F2 0 -220 °C -188 °C
Cl2 0 -101 °C -34.6 °C
Br2 0 -7.2 °C 58.78 °C
I2 0 113.5°C 184.35 °C
Explain why the boiling point of Cl2 (nonpolar) is higher than the boiling point of HCl (polar).
Why is ethanol used (or methanol ) to remove water from gas tanks?
5 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
3. Factors Affecting Solubility a. Molecular structure - determines polarity e.g. - fat soluble vitamins(A,D,E and K) are nonpolar and so dissolve in nonpolar oils and fats - water soluble (B's and C) vitamins are polar and dissolve in water - An important implication of this is that we can store fat soluble vitamins in our fatty tissues. We ca also overdose on fat soluble vitamins since they can be stored. It would be very difficult to overdose on water soluble vitamins since they are easily excreted in urine, which also prevents their being stored.
Day 2 : b. Pressure effects - pressure effects the solubility of gases : - increased pressure increases the solubility of gases - really an equilibrium effect. An increase in pressure above a solvent will concentrates the gas particles above the solvent. To reach equal concentrations in and out of the solvent more gas particles will move into the solvent to form a more concentrated solution. - decreased pressure decreases the solubility of gases - opposite of above - Henry's Law : States that the solubility of a gas is directly proportional to the pressure of the gas above the solution. (Assuming no chemical interaction between the gas and the solvent). (P = kC, where : P = pressure, k = solubility constant, C = concentration of the dissolved gas). - pressure has little effect on the solubility of solids and liquids ** To remember the above think of a bottle of soda. If you open it (decreases pressure) the gases will leave (become less soluble), but the sugars (solids) will not leave(solubility unaffected).
_____ (T/F) An increase in pressure increases the solubility of a gas.
______law states that the amount of a gas dissolved in a solution is directly proportional to the partial pressure of the gas above the liquid.
Sample problem :
Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 5.0 atm over the liquid at 25 ºC. The Henry's law constant for CO2 in water at this temperature is 32 L·atm/mol.
6 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
c. Temperature Effects (for aqueous solutions) - the solubility of gases decreases with increases temperature and vice versa (thermal pollution of bodies of water decreases oxygen content and can lead to a die off of aquatic animals). - the solubility of solids generally increases with increase in temperature (sodium sulfate and cerium sulfate become less soluble as temperature rises). 4. Colligative properties - depend on the number of solute particles -not type of particles. - vapor pressure reduction - boiling point elevation - freezing point elevation - osmotic pressure
_____ (T/F) The solute must be nonvolatile for vapor pressure lowering to occur.
_____ (T/F) Dissolved oxygen will lower the vapor pressure of pure water.
_____ (T/F) One mole of glucose (C6H12O6) will lower the vapor pressure of a solution more than one mole of aluminum chloride(AlCl3).
Vapor pressure lowering: - solute particles block escape of solvent particles -solute particles bond to solvent particles weighing them down
a. The Vapor Pressure of Solutions : - The presence of a nonvolatile solute lowers the vapor pressure of a solvent. It does so by diluting the solvent, reducing the numbers of solvent particles at the surface and thereby reducing the numbers of solvent particles escaping and creating vapor pressure. The vapor pressure of a solution is then directly proportional to the mole fraction of the solvent present as stated in
Raoult's Law : Psoln = XsolventPºsolvent, and illustrated below.
7 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
Evidence for Raoult's law :
:-because aqueous solution has a lower vapor pressure,
equilibrium with pure solvent will never be reached
What will happen to the vapor pressure of a 5% solution of sodium chloride in an open beaker over time?
Raoult's law can be used to count molecules, or determine the number of moles present in a solution, and thereby determine the molar mass of a substance. A mass of the solute is first dissolved in a given amount of solvent(number of moles known). The vapor pressure can then be measured and the mole fractions determined. The moles of solute is then calculated and molar mass determined by dividing the mass of solute by moles of solute.
Sample Problem : Glycerin, C3H8O3 is a nonvolatile nonelectrolyte solute with a density of 1.26 g/mL at 25 ºC. Calculate the vapor pressure at 25 ºC of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25ºC is 23.8 torr.
Need mole fraction of solvent (water):
Moles glyc. = (50.0 mL)(1.26g/mL)(1 mol/92.1g)= .684 mol moles water=500.0 mL(1.00g/mL)(1 mol/18.02 g)=27.8 mol
XH2O = 7.8 mol/ (.684 mol + 27.8 mol) = 0.976
Raoult's law can also be used to determine the number of particles a solute dissociates into. If one mole of a substance is put into solution and the vapor pressure is lowered by an amount that
indicates three moles of solute then the solute dissociates into three particles.(e.g. MgCl2(s) + 2+ H2O(l) Mg (aq) + 2 Cl-(aq))
Sample 2- Ionic solute : Predict the vapor pressure of a solution made by mixing 35.0 grams of solid sodium sulfate (molar mass =142 g/mol) with 175 g of water at 25 ºC. The vapor pressure of pure water at this temperature is 23.8 torr.
Calculating mole fraction of solvent (water): mol water = 175 g/18.02 g/mol = 9.72 mol mol sodium sulfate = 35.0g/142g = .246 mol*
*- ionic compounds dissociate into ions in solution(if soluble) mole solute =.246 mol (3 mol ions/mol)= .738 mol ions
8 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
XH2O= 9.72 mol/(9.72 mol + .738 mol)= .929 No calculator problem :
What is the mole fraction of the solute of an aqueous solution that has a vapor pressure of 17.9 torr? (PºH2O = 23.8 at 25 ºC)
a. 0.25 b. 0.33 c. 0.50 d. 0.75 e. 0.90
What is the vapor pressure of an aqueous solution at 25 ºC formed by dissolving 90 grams of glucose (C6H12O6 ; fw = 180.18 g/mol) in 180 grams of water? (Pº H2O = 23.8 torr at 25 ºC) a. 15.5 torr b. 17.3 torr c. 19.8 torr d. 21.0 torr e. 22.7 torr
For solutions of two liquids : The solute contributes to the vapor pressure (i.e. solute is volatile) .
- modified Raoult's law : Psoln = PA + PB = XAPºA + XBPºB
Ideal solutions - solute-solvent, solute-solute, and solvent-solvent interactions are nearly identical.
Nonideal solutions - solute-solvent, solute-solute, and solvent-solvent interactions are not identical.
A solution is prepared by adding one mole of methanol (CH3OH) and 2 moles of propanol (C3H7OH). What is the composition of the vapor(in mole fractions) above the solution at 40 ºC? (vapor pressures of pure methanol and pure propanol at 40 ºC are 303 and 44.6 torr, respectively.
Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25°C the vapor pressures of pentane and hexane are 511 and 150. torr, respectively. The mole fraction of pentane is 0.40. a. What is the vapor pressure of the resulting solution?
b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?
9 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08 Benzene and toluene form an ideal solution. Consider a solution of benzene and toluene formed at 25 ºC. Asuming equal mole fractions of toluene and benzene in the vapor state, calculate the composition of the solution. At 25 ºC the vapor pressures of toluene and benzene are 95 and 28 torr respectively.
- a liquid-liquid solution that obeys Raoult's law is considered an ideal solution - in an ideal solution the solvent-solvent, solute-solute and solvent-solute interactions have to be very similar - in a solution where the solute-solvent bonds are greater than solute-solute or solvent- solvent bonds :
- energy will be released (ΔHsoln is negative, exothermic formation of solution) - there will be a negative deviation from Raoult's law (lower vapor pressure than expected) - in a solution where the solute-solvent bonds are weaker than the solute-solute or solvent- solvent bonds :
- energy will be absorbed (ΔHsoln is positive, endothermic formation of solution) - there will be a positive deviation from Raoult's law (higher vapor pressure than expected)
Which picture shows and ideal liquid-liquid solution? -a positive deviation from Raoult's law? - a negative deviation from Raoult's law? In which solution are the solute-solvent bonds greater than the solvent-solvent bonds?
Interactive forces between ΔHsoln ΔT for Solution Deviation from Example solute(A) and solvent (B) Formation Raoult's Law AA, BB =AB Zero Zero None (=ideal Benzene and toluene solution) (similar in size and polarity) AA, BB
b. Boiling-point elevation - a nonvolatile solute raises the boiling point of a solvent proportionally to the amount of solute present 10 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
- ΔTb = Kbmsolute where : Kb = molal boiling point constant - specific for each liquid msolute - molality of solute c. Freezing-point depression - a nonvolatile solute lowers the freezing point of a liquid - ΔTf = Kfmsolute where : Kf = molal boiling point constant - specific for each liquid msolute - molality of solute
Sample problem 1 : 17.84 g of glucose (molecular solid) are dissolved in 150.0 g of water. The boiling point is found to be 100.34 ºC. Calculate the molar mass of the glucose. (Kb= .512 ºC/m)
Sample 2 : 7.480 g of an organic compound is dissolved in 292 g of water. The compound has an empirical formula of C3H4O2N. If the solution freezes at -0.185 ºC, what is the molecular formula of the compound?
6. Osmotic Pressure a. Osmosis - the flow of a solvent through a semipermeable membrane (net movement of solvent is always toward the side with the greater concentration of solute) b. Osmotic pressure - the pressure created by the movement of the solvent through the membrane - equal to the minimum external force needed to stop the flow of the solvent - can be used to indicated the number of particles a solute dissociates into - can be used to determine the molar mass of a substance - very useful for this since a small amount of solute creates a large pressure - π = MRT where : π is osmotic pressure M is molarity R is the gas law constant(.08206 L · atm/K · mol) T is the Kelvin temperature
Why does osmosis occur?
11 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08 Sample problem : Calculate the molar mass of a protein in a solution in which 0.00100 g of the protein was dissolved in enough water to form 1.00 mL of solution. The osmotic pressure was found to be 1.12 torr at 25.0 ºC.
c. Dialysis - phenomenon in which small solute particles as well as solvent particles pass through a semipermeable membrane d. isotonic solutions - solutions with equal osmotic pressures e. hypertonic solution - a solution with a higher solute content and therefore a lower osmotic pressure than a reference solution (often cytoplasm) - hypertonic solutions can cause crenation (shriveling due to water loss) in red blood cells f. hypotonic solutions - a solution with a lower solute content and therefore a higher osmotic pressure than a reference solution (often cytoplasm) - hypotonic solutions can cause hemolysis (bursting due to too much water present) in red blood cells g. Reverse osmosis - the forcing of water through a semipermeable membrane against the flow of osmosis by the application of an external pressure greater than the osmotic pressure - used in the desalination of seawater to obtain freshwater 7. Colligative Properties of Electrolyte Solutions a. van't Hoff factor - factor equating moles of solute with moles of ions present
- e.g. The van't Hoff factor for magnesium nitrate ( Mg(NO3)2)would be three since three moles of ions would be formed in solution from one mole of magnesium nitrate. - the equation for changes in the boiling or freezing-point of a solution containing an electrolyte becomes :
ΔT = imK
Where : i is the van't Hoff factor m = molality of solute K = freezing-point or boiling-point constant
- the equation for osmotic pressure becomes :
π = iMRT
b. Ion Pairing - in any electrolyte solution there is some ion pairing - the joining of oppositely charged ions due to electrostatic attraction. This will reduce the expected number of effective solute particles causing colligative effects. The greater the charge on an ion the greater its tendency to pair in solution.
12 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08 Calculating deviations in the van't Hoff factor (i).
The observed osmotic pressure for a 0.10 M solution of Fe(NH4) 42(SO4)2 at 25 ºC is 10.8 atm. calculate the expected and experimental values of i.
8. Colloids a. characteristics of colloids - permanent mixtures (will not settle out over time) - have a particle size intermediate to solutions and suspensions - exhibit the Tyndall effect - permanence in mixture is due to the electrostatic repulsion of the colloidal particles (they all have the same charge) - structure of colloidal particles - surrounded by layers of like charged particles :
b. Coagulation - the destruction of a colloid by destroying electrostatic repulsion by : - heating - speeds up particles and collisions remove outer layers of ions - addition of an electrolyte- removes layers of ions c. Examples of colloids : Example Dispersing Medium Dispersed Substance Colloid Type Fog, aerosol sprays Gas Liquid Aerosol Smoke Gas Solid Aerosol Whipped cream, soap Liquid Gas Foam suds Milk, mayonnaise Liquid Liquid Emulsion Paint Liquid Solid Sol Butter, cheese Solid Liquid Solid emulsion
Dispersal systems solutions - permanent suspensions - sm. particles - temporary -cannot be filtered - lge particles -homogeneous - can be filtered colloids - heterogeneous - permanent - Tyndall effect - med particles - borderline filter - borderline homogeneous -Tyndall effect
Tyndall effect :
13 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08 Day
Review :
Steps to solution formation of NaCl in water?
1.
2.
3.
Enthalpy of solution formation : ΔHsoln = ΔH1 + ΔH2 + ΔH3
Remember : ΔH2 + ΔH3 = ΔHhyd and ΔHhyd is the opposite of lattice energy
In which of the following cases will a solution form?
Case 1 Case 2 Case 3
ΔH1 is large ΔH1 is small Lattice energy is large and negative ΔH2 is large ΔH2 is small ΔHhyd is larger ΔH3 is small ΔH3 is large
Compare the following in aqueous solution :
CH3OH C6H14
______ΔH1 ______
______ΔH2 ______
______ΔH3 ______
Raoult’s Law revisited :
14 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
Day :
Review :
Calculate the molarity and mole fraction of acetone in a 1.00 m solution of acetone (CH3COCH3 , gmm = 3 58.09 g) in ethanol (C2H5OH, gmm= 46.08 g) (density of acetone = 0.788 g/cm ; density of ethanol = 0.789 g/cm3). Assume volumes are additive.
Find the mass percent of CuSO4 in a solution whose density is 1.30 g/mL and whose molarity is 1.22 M.
What volume of oxygen gas is need to react with an excess of carbon disulfide to form 6 L of carbon dioxide gas? a. 6 L b. 8 L c. 12 L d. 18 L e. 24 L
Which of the following will increase in solubility in an acidic solution? a. NaOH b. Mg(OH)2 c. AlPO4 d. KNO3 e. two of the above
15 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08 Match the following : a. MgCr2O7 _____ purple aqueous solution b. CuCl2 _____ yellow aqueous solution c. FeCl3 _____ clear aqueous solution (basic) d. ZnCl2 _____ clear aqueous solution (neutral) e. NiCl2 _____ reacts with HCl to form a gas f. KMnO4 _____ blue aqueous solution g. Na2CO3 _____ green aqueous solution h. MgCrO4 _____ orange aqueous solution
Rank the following in order of increasing solubility.
______CH3CH2CH2CH3 ______CH3CH2-O-CH2CH3 ______CH3CH2OH ______CH3OH
Which of the following has the lowest conductivity?
a. 1.0 M MgSO4 b. 1.0 M CaCl2 c. 1.0 M HI
d. 1.0 M HNO2 e. 1.0 M NaOH
1.0 M HCl was reacted with a white solid. A gas was produced which was bubbled through a solution of calcium hydroxide (limewater), causing a precipitate to form. a. Which of the following may have been the white solid?
Mg(OH)2 MgSO3 MgSO4 Na2CO3 NaCl b. Write the net reaction for the gas and the lime water.
Sodium phosphate was produced during a neutralization reaction. Which of the following acid and base pair involved in the neutralization reaction?
a. NaCl and Na3PO4 b. H3PO4 and NaH c. H3PO4 and NaOH 3- d. Na+ and PO4 e. none of the above
What volume of water needs to be added to 50.0 mL of 1.0 M HCl to form a 0.50 M HCl solution? a. 100.0 mL b. 50.0 mL c. 25.0 mL d. 10.0 mL e. none of the above
16 AP Chemistry - Notes : Chapter 11 Mr. Ferwerda - Tecumseh High School 5/20/08
Which of the following favors the formation of a solution from water and an ionic solid?
a. large magnitude in the enthalpy of hydration b. small magnitude in lattice energy c. a large polarity of the solvent d. all of the above.
_____ (T/F) Solution formation is exothermic if the magnitude of the hydration energy is greater than the magnitude of the lattice energy.
The lattice energy of sodium iodide is - 686 kJ/mol and its heat of solution is -7.6 kJ/mol. Calculate the energy of hydration of NaI.
Solid KF has a lattice energy of -804 kJ/mol and a heat of solution (in water) of -15 kJ/mol. Solid RbF has a lattice energy of -768 kJ/mol and a heat of solution (in water) of -24 kJ/mol. Which salt forms stronger attractions for water?
The high melting points of ionic compounds indicates that a lot of energy is needed to separate the ions from each other (break lattice energy). How is it possible that the ions of an ionic solute separate when dissolved in water at room temperature?
Which of the following processes are endothermic? a. evaporation of water b. combustion of gasoline c. mixing of sulfuric acid in water d. freezing of water e. two of the above
Gas X has a solubility of 0.25 g/L under a pressure of 9.8 atm. What is its solubility under a pressure of 5.3 atm.?
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