General Chemistry Exam 2

Front 1

Define term: dissolve

Back 1

to suspend in another substance

Front 2

Define solute

Back 2

substance dissolved in a liquid

Front 3

Define solvent

Back 3

the dissolving medium in a solution

Front 4

Define solution

Back 4

a solute dissolved in a solvent

Front 5

Define aqueous solution

Back 5

solution where water is the solvent

Front 6

Define dipole moment

Back 6

electric field generated by fact electrons aren't shared equally

Front 7

Define solubility

Back 7

Amount to which 1 solute will dissolve in a solvent or solution

Hint 7

like dissolved like meaning polar dissolved polar. Oil has no dipole moment and therefore doesn't dissolved in water but electrolytes have dipole moment and do dissolve in water

Front 8

Define strong electrolyte

Back 8

dissolved completely in water (aqueous solution) and ionized completely

Front 9

Define weak electrolyte

Back 9

dissolves in water completely but only partially ionizes

Hint 9

acetic acid

Front 10

Define non electrolyte

Back 10

dissolved completely in water but creates no ions

Hint 10

sugar and ethanol

Front 11

Define acid

Back 11

substance producing H⁺ ions in a solution

Hint 11

HCl(g) →H⁺(aq) + Cl⁻(aq)

Front 12

Define strong acid

Back 12

ionizes entirely in water, strong acid is also a strong electrolyte

Front 13

Define weak acid

Back 13

ionizes partially in water

Hint 13

HC₂H₃O₂(ℓ)→H⁺(aq) + C₂H₃O₂⁻(aq)
Acetic acid

Front 14

Define base

Back 14

Brønsted Lowry- a substance that accepts H⁺ ions

Arrhenius- a substance that produces OH⁻ ions

Front 15

Define precipitate

Back 15

solid created as a result of a reaction

Front 16

rules of solubility in water

Back 16

-Most Nitrate (NO₃⁻) salts are soluble
-Most salts containing Alkali metal ions (row 1A: Li⁺, Na⁺, K⁺, Cs⁺, Rb⁺) and the ammonium ion (NH₄⁺) are soluble
-Most Chloride, Bromide, and Iodide salts are soluble
-most sulfate (SO₄²⁻) salts are soluble

Front 17

Define spectator ion

Back 17

ions that don't participate in a given reaction

Front 18

Define Oxidation reduction reaction AKA redox Reax

Back 18

a reaction where electrons are transferred

Front 19

Define oxidation

Back 19

an increase in oxidation state/ loss of electrons. Whatever is oxidized is also the reducing agent (Electron donor)

Front 20

Define reduction

Back 20

a decrease in oxidation state/gain of electrons. Whatever is reduced is the oxidizing agent (Electron acceptor)

Front 21

Define titration

Back 21

mixing acid and base in a controlled way so pH is as close to 7 as possible

Front 22

Define gas

Back 22

substance with no fixed shape or volume

Front 23

what's Boyle's law

Back 23

assuming constant temp: P•V=b or P₁•V₁ =P₂•V₂

Hint 23

b=constant

Front 24

what's Charle's law

Back 24

assuming constant pressure: V=b•T or V₁/T₁=V₂/T₂

Hint 24

b=constant

Front 25

what's Avogadro's law

Back 25

assuming a constant temperature and pressure: V=a•n or V₁/n₁=V₂/n₂

Hint 25

a=constant
n=moles

Front 26

What's the ideal gas law

Back 26

PV=nRT

Hint 26

R=universal gas constant with two values depending on the units used:
R=0.08206 L•atm/mol•K
R=8.31451 J/mol•K

Front 27

what is STP

Back 27

standard temperature and pressure which is O°C and 1atm

Front 28

what is Dalton's law of partial pressure

Back 28

sum of partial pressure of gases:
P_total= P₁+P+₂+P₃...

assuming ideal behavior of all gases the partial pressure of each gas can be calculated from the ideal gas law:
P₁=(n₁RT)/ V, P₂=(n₂RT)/V, P₃= (n₃RT)/V

Front 29

define diffusion

Back 29

mixing of gases

Front 30

define effusion

Back 30

passage of gas through and orifice into an empty chamber

Front 31

Define energy

Back 31

capacity to do work

Front 32

Law of conservation of energy

Back 32

Can't create or destroy energy in a given proces

Front 33

Define potential energy

Back 33

energy due to position or composition

Hint 33

ball on hill

Front 34

Define kinetic energy

Back 34

energy of motion KE= 1/2m• v²

Hint 34

m=mass of object
v=velocity

Front 35

Define heat

Back 35

transfer of energy between 2 objects, occurring when there is a difference in temperature

Front 36

Define work

Back 36

force acting over a distance F=m• a

Hint 36

m=mass
a=acceleration

Front 37

Define state function

Back 37

property of a system which depends only on the current state of the system, not the path it takes

Front 38

non-state function

Back 38

heat and work which are simply numerical quantities of a process

Front 39

Define system

Back 39

portion of the universe under consideration

Front 40

Define surroundings

Back 40

everything else not of the system

Front 41

Define exothermic process

Back 41

when heat flows out of system

Front 42

Define endothermic process

Back 42

when heat flows into system

Front 43

Define thermodynamics

Back 43

study of energy

Front 44

1st law of thermodynamics

Back 44

energy of the universe is constant

Front 45

Internal energy of a system (E)

Back 45

Σ=sum of all kinetic and potential energy in a system

∆E=q+w (heat+work) can change the internal energy of a system

Front 46

define calorimetry

Back 46

study of measuring heat

Front 47

heat capacity equation

Back 47

C=heat absorbed/ increase in temperature

Front 48

specific heat capacity equation

Back 48

Specific heat capacity= heat capacity(C) /grams

Hint 48

in defining heat capacity of a substance the amount of substance must be identified

Front 49

Enthalpy(H) equations

Back 49

H= E + PV: equally ΔH= ΔE+ΔPV but P is constant so ΔH=ΔE+PΔV

ΔH=q; @ constant pressure the enthalpy ΔH of system is equal to the heat flow

for a chemical reaction the enthalpy change is given by: ΔH+H_products-H_reactants

Front 50

what's 1 calorie

Back 50

heat required to raise temperature of exactly 1 gram of water by exactly 1°C

Front 51

When an object changes temperature the heat capacity can be calculated by:

Back 51

q=s•m•ΔT

Hint 51

where s=specific heat capacity
m=mass
ΔT =change in temp

when ΔT is positive q is also positive

Front 52

things that are a given unless stated otherwise

Back 52

-1 mole of any gas occupys the same volume; 1 mol gas=22.4 L@STP
-Pressure @ sea level is 760mm Hg
-STP=O°C and 1 atm
-1atm=760mm Hg=760 torr=101.325 Pa (pascals)

Front 53

calculate the molarity of a 5.623g sample of NaHCO₃ thats dissolved in enough water to make 250.0mL sample of a solution

Back 53

answer: .2677 M of NaHCO₃

method:
- 5.623gNaHCO₃ (1mol NaHCO₃/ 84.008g NaHCO₃)=6.693x10⁻²
-6.693x10⁻²/250.0mL (1000mL/1L)=.2677 M NaHCO₃

Hint 53

-turn grams into moles
-moles divided by proper unit of volume

Front 54

Calculate the concentration of all ions present in 0.100 mol Ca(NO₃)₂ in 100.0mL of solution

Back 54

ans: 1 M Ca²⁺ & 2 M NO₃⁻

method:
-Ca(NO₃)₂→Ca²⁺ + 2NO₃⁻
-(.1mol Ca(NO₃)/100.0mL)(1000mL/1L)=1 M Ca(NO₃)₂
-M_Ca²⁺= 1(1)=1 M Ca²⁺
-M_NO₃⁻=2(1)=2 M NO₃⁻

Hint 54

-write a balance equation
-find molarity of compound
-using the balanced equation multiply the molarity of solution by number of moles of ions

Front 55

what is the number of moles of Cl⁻ ions in 100.0mL of 0.30 M AlCl₃

Back 55

answer: 9.0 x10⁻² mol Cl⁻

method:
-AlCl₃→Alᶟ⁺ +3 Cl⁻
-100mL (1L/1000mL)=0.1L
-Mol Cl⁻ = 0.1L(0.30 mol AlCl₃/L ) (3 mol Cl⁻/ 1 mol AlCl₃)= 9.0 x10⁻² mol Cl⁻

Hint 55

-write balanced equation
-convert mL to L
-use mole of solute equation; Mol solute=volume (L) x Molarity (mol/L)
-use conversion factor for relation between mole of compound to mole of ion

Front 56

What mass of NaOH is contained in 250.0mL of a 0.400 M NaOH

Back 56

answer: 4 g NaOH

method:
-250.0mL(1L/1000mL)=.25
-.25L (0.4mol/ L)(40g NaOH/1 mol NaOH)=4g NaOH

Hint 56

-convert mL to L
-mole of solute equation: mol solute-volume (L) x Molarity (mol/L)
-calculate molar mass of NaOH and use conversion factor of grams per mole of compound

Front 57

A is prepared by dissolving 10.8 g of ammonium sulfate in enough water to make 100.0mL of stock solution. A 10.00 mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in final solution

Back 57

answer: M_NH₄⁺=.272 & M_SO₄²⁺=.136

method:
-(NH₄)₂SO₄→2NH₄⁺ + SO₄²⁺
-10.8 g (NH₄)₂SO₄ (1 mol(NH₄)₂SO₄/132.154g(NH₄)₂SO₄)=8.17 x 10⁻²
-(8.17x10⁻²/100mL)(1000mL/L)=.817 M (NH₄)₂SO₄
-10.0mL/1000mL=.01L(.817mol (NH₄)₂SO₄/L)=8.17x10⁻ᶟ mol (NH₄)₂SO₄
-(8.17x10⁻ᶟ mol (NH₄)₂SO₄/10+50mL)(1000mL/L)=.136
-M_NH₄⁺=2(.136) & M_SO₄²⁺=1(.136)

Hint 57

-write a balanced equation
-determine the molarity
-calculate the moles of the 10.00mL sample
-calculate the molarity of the final solution
-using the balanced equation determine the concentration of ions

Front 58

Calculate the sodium ion concentration when 70.0mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate

Back 58

answer: 4.5 M Na⁺

method:
-Na₂CO₃→2Na⁺ + CO₃²⁻
-70.0mL-1000.0mL=.07L
-.07L (3 mol Na₂CO₃/L )(2 mol Na⁺/1 mol Na₂CO₃)=.42 mol Na⁺
-30.0mL/1000mL=.03L
-.03L (1mol NaHCO₃/L)(1 mol Na⁺/ NaHCO₃)=.03 mol Na⁺
- (.42mol Na⁺+.03 mol Na⁺)/ (.07L+.03L)=4.5 M Na⁺

Hint 58

-write balanced eq. for 1st solution
-convert mL to L
-calc moles solute and use conversion factor of the relationship between moles of compound and ions
- balance eq. for 2nd compound
-convert mL to L
-calc moles solvent and used onverson factor for the relationship between moles of compound and moles of ions
-M_Na⁺= total mol Na⁺/ total volume=(.42mol Na⁺+.03 mol Na⁺)/ (.07L+.03L)= .45 mol Na⁺ /.1L=4.5 M Na⁺

Front 59

On the basis of the general solubility rules, predict which of the following substances are likely to be soluble in water:
-Aluminum Nitrate
-magnesium chloride
-rubidium sulfate
-nickel (II) hydroxide
-lead (II) sulfide
-magnesium hydroxide
- iron (III)phosphate

Back 59

ans:
-Aluminum Nitrate=soluble
-magnesium chloride=soluble
-rubidium sulfate=soluble
-nickel (II) hydroxide= insoluble
-lead (II) sulfide=insoluble
-magnesium hydroxide=insoluble
- iron (III)phosphate=insoluble

Hint 59

-Aluminum Nitrate=most nitrate ions are soluble
-magnesium chloride=most chloride salts are soluble
-rubidium sulfate=most sulfate salts are soluble
-nickel (II) hydroxide= most hydroxides are only slightly soluble
-lead (II) sulfide=most sulfide salts are only slightly soluble and we consider them insoluble
-magnesium hydroxide=most hydroxides are only slightly soluble
- iron (III)phosphate=most phosphate salts are only slightly soluble

Front 60

what precipitate (if any) will form from FeSO₄(aq) + KCl(aq)

Back 60

possible products=FeCl₂(aq) & K₂SO₄(aq); both salts are soluble so no precipitate forms

Front 61

what precipitate (if any) will form from Al(NO₃)₃(aq) + Ba(OH)₂(aq)

Back 61

Possible products=Al(OH)₃ & Ba(NO₃)₂

Al(OH)₃=hydroxides are only slightly soluble so this is the precipitate. Ba(NO₃)₂=most nitrates are soluble

Front 62

what precipitate (if any) will form from CaCl₂(aq) + Na₂SO₄(aq)

Back 62

possible products=CaSO₄ & NaCl

CaSO₄ is an exception to rule that most sulfate salts are soluble
NaCl=most chloride salts are soluble

Front 63

what precipitate (if any) will form from K₂S(aq)+Ni(NO₃)₂(aq)

Back 63

possible products=K(NO₃)& NiS

K(NO₃)=most nitrates are soluble
NiS=is the precipitate

Front 64

Write the balanced formula equation, the complete ionic equation and the net ionic equation of Al(NO₃)₃(aq) + Ba(OH)₂(aq)

Back 64

balanced=2Al(NO₃)₃(aq) +3Ba(OH)₂(aq)→2Al(OH)₃(s) + 3Ba(NO₃)₂(aq)

complete ionic=2Alᶟ⁺(aq) + 6NO₃⁻(aq) + 3Ba²⁺(aq) +6OH⁻(aq) →2Al(OH)₃(s) + 3Ba²⁺(aq) + 6NO₃⁻(aq)

net ionic= Al(OH)₃(s)→Alᶟ⁺(aq) +3OH⁻(aq)

Front 65

Write the balanced formula equation, the complete ionic equation and the net ionic equation of CaCl₂(aq) + Na₂SO₄(aq)

Back 65

balanced=CaCl₂(aq) + Na₂SO₄(aq)→CaSO₄(s)+ 2NaCl(aq)

complete ionic=Ca²⁺+2Cl⁻ +2 Na⁺ + SO₄²⁻→ CaSO₄(s) + 2Na⁺ + 2Cl⁻

not ionic=CaSO₄→Ca²⁺ + SO₄²⁻

Front 66

Write the balanced formula equation, the complete ionic equation and the net ionic equation of K₂S(aq)+Ni(NO₃)₂(aq)

Back 66

balanced=K₂S(aq)+Ni(NO₃)₂(aq)→2K(NO₃)(aq)+ NiS(s)

complete ionic=2K⁺(aq) +S²⁻(aq) + Ni²⁺(aq) + 2NO₃⁻(aq) → 2K⁺(aq) +2NO₃⁻(aq) + NiS(s)

net ionic=NiS(s)→ Ni²⁺(aq) + S²⁻(aq)

Front 67

What mass of Na₂CrO₄ is required to precipitate all of the silver ions from 75.0mL of a 0.100 M solution of AgNO₃

Back 67

answer=0.607 g Na₂CrO₄

method:
-2AgNO₃ + Na₂CrO₄→ AgCrO₄ + 2NaNO₃
-75mL/1000mL=0.075L
-0.075L(.1mol AgNO₃/ L)(1mol Na₂CrO₄/2mol AgNO₃) (161.98g Na₂CrO₄/1 mol Na₂CrO₄)=0.607 gNa₂CrO₄

Hint 67

-balance the equation
-convert mL to L
-use mole of solute equation Volume (L) x Molarity (mol/L)
-use conversion factor of relationship between Na₂CrO₄ & AgNO₃
-use conversion factor for grams per mole of Na₂CrO₄

Front 68

What volume of 0.100 M Na₃PO₄ is required to precipitate all lead(II)ions from 150.0mL of 0.250 M Pb(NO₃)₂

Back 68

answer=0.250L

method:
-2Na₃PO₄ +3Pb(NO₃)₂→6NaNO₃ + Pb₃(PO₄)₂
-150.0mL of Pb(NO₃)₂(0.250M Pb(NO₃)₂/ L) (1L/ 1000mL)=0.0375M Pb(NO₃)₂
-0.0375M Pb(NO₃)₂ (2 mol Na₃PO₄/ 3 mol Pb(NO₃)₂)=0.0250 mol Na₃PO₄
-0.100 mol Na₃PO₄ (1L/0.0250mol Na₃PO₄)=0.25L

Hint 68

-balance equation
-calc molarity of Pb(NO₃)₂
-use conversion factor between the relationship between Pb(NO₃)₂ & Na₃PO₄ to find mole of Na₃PO₄
-use conversion factor of molarity in way that moles cancel out to leave Liters