×
Log in to StudySoup
Get Full Access to Chemistry - 12 Edition - Chapter 2 - Problem 2.5
Join StudySoup for FREE
Get Full Access to Chemistry - 12 Edition - Chapter 2 - Problem 2.5

Already have an account? Login here
×
Reset your password

Answer: Describe the contributions of the following

Chemistry | 12th Edition | ISBN: 9780078021510 | Authors: Raymond Chang; Kenneth Goldsby ISBN: 9780078021510 98

Solution for problem 2.5 Chapter 2

Chemistry | 12th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Chemistry | 12th Edition | ISBN: 9780078021510 | Authors: Raymond Chang; Kenneth Goldsby

Chemistry | 12th Edition

4 5 1 357 Reviews
25
1
Problem 2.5

Describe the contributions of the following scientists to our knowledge of atomic structure: J. J. Thomson, R. A. Millikan, Ernest Rutherford, James Chadwick.

Step-by-Step Solution:
Step 1 of 3

Week 8­10; Day 22­30 Solution Terms​: ● Solvent: the liquid in which another substance is dissolved (greatest amount) (water). ● Solute: a substance dissolved in a liquid (smaller amount) (salt). ● Heterogeneous​ : a non uniform mixture, regions of different compositions (ex: oil + water, soil sample, sand) (can see). Different particles within it. ● Homogeneous​ : a uniform mixture, same composition throughout (ex: blood, alcohol, + water, brewed coffee) (can’t see). ● Solubilit: the maximum amount of substance (solute) that will dissolve in a given solvent. ● Insoluble not much dissolves, forms a precipitate. Solubilit​Rules: ● Ionic compound with group 1 (cations and NH3) are always soluble. ● All nitrates are soluble. ● All Cl solutions are soluble except Pb, Mg, and Ag. ● Water moves to salts and sugars. Avogadro's​ Number​ = 6.022x10^23 ● % W/V​ = g solute/g solution x 100OR g solute/100g solution. ● % W/V​ = g solute/mL solution x 100 OR g solute/100mL solution. ● Molarity (M) = moles/liters. Example​: What is the molarity (M) of 250.0 mL of solution containing 60.0 g of NaOH M = 60.0 g/250.0 mL x 1 mole/39.998 g x 1000 mL/1 L ​ M​ Na = 22.99 O = 16 H = 1.008 = 39.998 * Get grams to moles and mL to liters * Salts: ● Usually is an element from column 1 combined with an element from column 7. ● Salt of a weak acid = removes the H+, forming a base. ● Salt of a weak base = accepts the H+, forming an acid. ● Normality (N) = number of equiv solute/liter. ● Equivalence (Eq)​= total charge. Example:​ How many equivalents are in each of the following H2PO4 = 1 x 2 = 2. H2 = 1+ charge. PO4 = 2­ charge. The negative does not matter so eliminate it. H3PO4 = 1 x 3 = 3. H3 = 1+ charge. PO4 = 3­ charge. The negative does not matter so eliminate it. Example:​ What is the normality of a 3.5% solution of AlCl3 3.5 g/1000 mL → eq/L 3.5 g/1000 mL x 1000 mL/1 L x 1 mole AlCl3/133.3 g x 3 equiv/1 mole Al​0.1 N. Al = 26.98 g Cl = 35.45(3) g = 106.35 g = 133.3 g. Equivalence = 3 → Al has a 3+ charge. Cl has a 1­ charge. 3 x 1 = 3. ● Osmoles​ (OSM) = number of particles. Remember: OSM​ = add charges. N = multiply charges. Example​: In whiskey, 87.5% v/v ethanol water: 87.5% = ethanol. The solvent is ethanol and the solute is water. Example​: In beer, 3.5% v/v ethanol water: 3.5% = ethanol. The solvent is water and the solute is ethanol. ● Osmotic Pressure:​ water moves from low concentration to high concentration. ● Hypertonic Solution​ having osmolarity greater than the surrounding blood, plasma, or cells → water to go out → cell shrivels/shrinks (crenation) → crenation (greater salt concentration) (lose wateMore solute, less water. ● Hypotonic Solution: having osmolarity less than the surrounding blood, plasma, or cells → water to go in → cell bursts → (lower salt concentration than blood) → hemolysis; red blood cells swelLess solute, more water. ● Isotonic Solutio: organism remains the same if concentration of solvent is the same as solute → water goes in and out (no change). ● Freezing Point: more particles are more energy, the larger it is, the lower the freezing point needs to be. ● Boiling Poin: has to knock the particles out of the way to evaporate, the larger it is, the higher the boiling point needs to be. ● Osmolarity​ ○ Freezing And Boiling Points. ● Osmolarity = osmoles (number of particles)/Liter. ● Osmoles ​(OSM) = total number of particles. Example​: How many osmoles are in the following (H3) (PO4) = H3 = 3 PO4 = 1 3 + 1 =4 osmoles​. (H2) (SO4) = H2 = 2 SO4 = 1 2 + 1 =3 osmoles​. Example​: What is the osmolarity of 12.0 M H2SO4 osm/L = 12 mole/L H2SO4 = H2 = 2 + SO4 = 1 2 + 1 = 3 osmoles. 12.0 mole/1 liter x 3 osmoles/1 mole H2SO4​6 osmoles/L. Equations: ● pH = ­log[H+]. ● pOH = ­log[OH­]. ● [H+] = antilog(­pH). ● [OH­] = antilog(­pOH). ● pH + pOH = 14. ● [H+] = radical (Ka) x [acid]. ● [OH­] = radical (Kb) x [base]. ● [ ] = concentration (of molarity) of species. ● Kw = 1 x 10^­14 = [H=][OH­] at 20 degrees celsius. ● Equilibrium Constant = Keg​→ their concentrations don’t change. ○ Pure solids and liquids are not used. ○ Keg = [products]/[reactants​ ○ If there is a coefficient in the equation, place it as a power in the equation above (outside of brackets). Example​ : H (aq) + CaCO3 (s) → CaO (aq) + CO2 (g). Reactants Products Keg = [CaO] [CO2] / [H+]​ * Can’t use CaCO3 because it is a solid. You cannot use solids/liquids * Aq​ = aqueous: of or containing water; typically as a solvent or medium. Example​ : CO (g) + 2H2 (g) → CH3OH (g). Reactants Products Keg = [CH3OH] / [CO] [H2]^2.​ * H2 has a coefficient of 2, so place that two outside of the brackets for H2 * Acids/Bases:​ ● Acids: ○ A proton donor; more H+. ○ High H+ = low pH value (acidic). ○ Gives away H+. ● Bases​: ○ A proton acceptor; more OH­. ○ Low H+ = high pH value (basic). ○ Accepts H+. pH Scale: * Weak acids and weak bases have Ka or Kb value * Acid (0­6) Base (8­14) ● Strong Acid​: 100% dissociated and does not have a Ka value. Begins with an H. ● Strong Base​: 100% dissociated and does not have a Kb value. Has an OH. ● Weak Acid​: Has a Ka value. Begins with an H. ● Weak Base: Has a Kb value. Has an OH. ● Conjugate​ Base​: forms when an acid loses a hydrogen ion. Proton acceptor. ● Conjugate​ Acid: forms when a base gains a hydrogen ion. Proton donor. Base = high pH. Acid = low pH. Example​: Base​ (when base adds an H+) Conjugate Acid NH3 ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­> NH4+. OH­ ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­> HOH. H2O ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­> H3O+. CO3 2­ ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­> HCO3­ Example​: H2CO3 + H2O H3O + HCO3­ Acid Base Base Acid The two bases gained → conjugate acid. The two acids lost → conjugate base. List Of Strong Acid (memorize for exam): ● Sulfuric Acid (H2SO4). ● Hydriodic Acid (HI). ● Hydrobromic Acid (HBr). ● Hydrochloric Acid (HCl). ● Nitric Acid (HNO3). ● Perchloric Acid (HClO4). Weak Acid Example​ : What is the [H+] of a 0.0150% m/v solution of phosphoric acid H3PO4 Ka = 7.5 x 10^­3. 0.0150 g/100 mL x 1000 mL/1 L x 1 mole/97.994 g H3PO4 ​.00153 mole/L​(molarity). [H+] = radical (Ka) x [acid molarity]. [H+] = radical (0.00153)(7.5x10^­3) [H+] 3.39 x 10^­3 H+ . Weak Base Example​ : What is the pH, pOH, [H+], and [OH­] of a 0.095% m/v solutions of NH3 Kb = 1.8 x 10^­5. 0.095 g/100 mL x 1000 mL/1 L x 1 mol/17 g NH3​.05577 mole/L (molarity). [OH­] = radical (Kb) x [base molarity]. [OH­] = radical (0.05577)(1.8 x 10^­5) [OH­] =​.001 OH­ M​ pOH = ­log[OH­]. pOH = ­log(0.001)​3 pOH.​ pH + pOH = 14. 14 ­ 3 ​1 pH​. [H+] = antilog(­pH). [H+] = antilog(­11) [H+] = 10^(­11) =​1 x 10^­11 H+ M. Strong Acid Example​ : What is the [H+], pH, pOH, and [OH­] of 3.34x10^­5 M HCl solution [H+] = 3.34x10^­5 H+ M. pH = ­log[H+]. pH = ­log(3.34 x 10^­5) =4.48 pH.​ pH + pOH = 14. 14 ­ 4.48 =​9.52 pOH​. [OH­] = antilog(­pOH). [OH­] = antilog(­9.52) [OH­] = 10^(­9.52) =3.02 x 10^­10 OH­ M. ​ Strong Base is the same as the strong acid example! ● KSP​ : the solubility of the products concentration. ○ Concentration should always be in molarity. ○ Solids and liquids are not included in KSP. ○ “KSP expression”. Example​ : PbF2 (s) + H2O (l) → Pb (aq) + 2F (aq) Reactants Products KSP of PbF2 = 7.12 x 10^­7. What is the KSP equilibrium concentration KSP = [Pb] [2F]^2 In the equation it is 2F, but when you write out the KSP equation, you have to put the coefficient in the brackets before the letter of the symbol AND outside of the brackets as a power. KSP = [x] [2x]^2 7.12 x 10^­7 = [x] [2x]^2 7.12 x 10^­7 = x x 4x^2 7.12 x 10^­7 = 4x^3 X = 0.0056. Example​ : Given the KSP = 1.46 x 10^­10 for KF, what would be the amount in grams that could be dissolved in 1.00 L of water KF → K + F KSP = [K] [F] 1.46 x 10^­10 = [x] [x] Radical 1.46 x 10^­10 = Radical x^2 X = 1.208 x 10^­5. 1.208 x 10^­5 mole/L x 58.1 g/1 mole =7.018 x 10^­4 g. Example​ : What is the KSP of the CuCl if a saturated solution containing [Cl^­1] = 2.07 x 10^­4 M and [Cu^+1] = 8.30 x 10^­4 M CuCl → Cu + Cl KSP = [Cu][Cl] KSP = 8.30 x 10^­4 x 2.07 x 10^­4. KSP = ​1.718 x 10^­7​ LeChatelier’s Principle​ ● Increase Pressure​ : shift to the area with least gas. ● Decrease Pressure: shift to the area with most gas. ● Add Solution​ : shift away (left) from side you added to. ● Take Away​ : shift towards (right) the side you took away from. * Pure solids and liquids have no effect * Example​ : CaCO3 (s) → CO2 (g) + CaO (s) By adding CO2 to the equation, it wi​lhift le t. ● Buffers​: solutions that resist changes in pH upon addition of acid or base. ○ Formed with a work acid/base and its salt. ○ pH = pKa + log [salt]/[acid]. ○ pKa = ­log(Ka). ○ Will be able to withstand more change if more of its opposite. Example​ : What is the ratio of KC2H3O2 and HC2H3O2 in a buffer with a pH of 5 Ka = 1.8 x 10^­5. Given: 5 pH. pKa = ­log(Ka). pKa = ­log(1.8 x 10^­5) = 4.74. 5 = 4.74 + log [KC2H3O2] / [HC2H3O2] 0.26 = log [KC2H3O2] / [HC2H3)2] 10^(0.26) = 1.8197 2(1.8197)/1 = [KC2H3O2] / [HC2H3O2] 3.6382/2 = common factor of 1.8197. Final answer needs to be a ratio. Professor will try to trick you on the exam with this. If your answer is 1.8197/1, it can ALSO be 3.6382/2. ● Titrations​ ○ Acid + Base → water + salt + ionic compound. ○ Solids and liquids ARE part of this equation. ○ pH = pKa + log [anion]/[acid]. ○ pKa = +/­ 1. Example​ : Given that 125 mL of H2SO4 is neutralized by 68.05 mL of 0.25 M Al(OH)3. What is the molarity of H2SO4 Acid + Base → Water + Salt + Ionic Compound. A B W S 3H2SO4 + 2Al(OH)3 → 6H2O + Al(SO4)3. 68.05 mL x 0.25 moles Al(OH)3/L x 3 mole H2SO4/2 mole Al(OH)3 x 1 L/125 mL H2SO4 = 0.204 molarity of H2SO4. Example​ : How many mL of a 0.009 NaOH solution would be needed to fully titrate 50 mL of 0.083 M H3PO4 3NaOH + H3PO4 → 3H2O + Na3PO4. 50 mL H3PO4 x 0.083 mole H3PO4/1 L x 3 mole NaOH/1 mole H3PO4 x 1 L/0.009 mol NaOH = ​ 1383 mL NaOH. ● Equivalence Point: the amount of base added was equal to the amount of acid started with meaning reaction was fully titrated/neutralized. * Polyatomics counted as one *

Step 2 of 3

Chapter 2, Problem 2.5 is Solved
Step 3 of 3

Textbook: Chemistry
Edition: 12
Author: Raymond Chang; Kenneth Goldsby
ISBN: 9780078021510

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Answer: Describe the contributions of the following