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Station 1 - Unit 5 Ag + H2S(aq) → Ag2S + H2 HgO(s) → Hg + O2
Balance the Chemical Equation and identify the type of reaction. Use the solubility rules and your periodic table to identify states of matter of each substance. Ag + H2S(aq) → Ag2S + H2 HgO(s) → Hg + O2 C3H8(l) + O2 → CO2 + H2O(g) CaCl2 + Na2SO4 → CaSO4 + NaCl P + O2→ P2O5(s) Aqueous aluminum sulfate reacts with aqueous copper (II) chloride to produce aluminum chloride and copper (II) sulfate.
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Station 1 - KEY 2Ag(s) + H2S(aq) → Ag2S(s) + H2 (g) SR
2HgO(s) → 2Hg(s) + O2(g) Decomposition C3H8(l) + 5O2(g) → 3CO2(g) + 4H2O(g) Combustion CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq) DR 4P(s) + 5O2(g)→ 2P2O5(s) Synthesis Al2(SO4)3(aq) + 3CuCl2(aq) + 2AlCl3aq) + 3CuSO4(aq) DR
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Station 2 - Unit 5 Na3PO4(aq) + CuCl2(aq) → Ag + MgBr2 →
Predict the products and balance the following chemical equations. Use the activity series to determine if a SR reaction will occur. Use the solubility rules and your periodic table to identify the states of matter of each substance. Na3PO4(aq) + CuCl2(aq) → Ag + MgBr2 → Pb(NO3)2 + FeI3 → Al + CoCl2 → C2H6O(l) + O2 →
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Station 2 KEY 2Na3PO4(aq) + 3CuCl2(aq) → 6NaCl(aq) + Cu3(PO4)2(S)
Ag(s) + MgBr2(aq) → No Reaction 3Pb(NO3)2(aq) + 2FeI3(aq) → 3PbI2(s) + 2Fe(NO3)3(aq) 2Al(s) + 3CoCl2(aq) → 2AlCl3(aq) + 3Co(s) C2H6O(l) + 3O2(g) → 2CO2(g)+ 3H2O(l)
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Station 3 - Unit 6 2KBr + MgCl2 → MgBr2 + 2KCl
How many moles of magnesium chloride are needed to produce 13.1moles of KCl? How many grams of KBr are needed to completely react 142.6g of MgCl2? 2NaCl + CuO → CuCl2 + Na2O 23.5g of sodium chloride are reacted with an excess of copper (II) oxide. How many molecules of copper (II) chloride will be produced? CH4 + 2O2 → CO2 + 2H2O 29g of methane (CH4) reacts with an excess of oxygen gas. How many liters of CO2 will be produced?
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Station 3 KEY = 6.55 mol MgCl2 = 356.5g KBr = 81.00L O2
2KBr + MgCl2 → MgBr2 + 2KCl 2NaCl + CuO → CuCl2 + Na2O CH4 + 2O2 → CO2 + 2H2O 13.1 moles of KCl 1 mol MgCl2 = 6.55 mol MgCl2 2 mol KCl 142.6g MgCl2 1 mol MgCl2 2 mol KBr 119.00g KBr = 356.5g KBr 95.21 gMgCl2 1 mol KBr 23.5g NaCl 1 mole NaCl 1 mole CuCl2 6.02x1023 molecules CuCl2 = 1.21 x 1023molecules of CuCl2 58.44g NaCl 2 mole NaCl 29g CH4 1 mole CH4 2 moles O2 22.4L O2 = 81.00L O2 16.04g CH4 1 mole O2
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Station 4 - Unit 6 2C2H6O(l) + 6O2(g) → 4CO2(g)+ 6H2O(l)
58.6g of C2H6O are reacted with 28.5L of oxygen gas How many liters of carbon dioxide will be produced? What is the limiting reactant in this equation? What is the excess reactant in this equation? How much excess is there?
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C2H6O(l) + 3O2(g) → 2CO2(g)+ 3H2O(l)
Station 4 - KEY C2H6O(l) + 3O2(g) → 2CO2(g)+ 3H2O(l) 19L of CO2 can be Produced Limiting Reactant: oxygen gas Excess Reactant = C2H6O . 58.6 g (what we have) g (needed) = 38.68g C2H6O in excess 58.6g C2H6O 1 mole C2H6O 2 mole CO2 22.4L CO2 = 55.89L CO2 46.97g C2H6O 1 mole CO2 28.5L O2 1 mole O2 2 mole CO2 22.4L CO2 = 19L CO2 22.4L O2 3 mole O2 1 mole CO2 19L CO2 1 mole CO2 1 mole C2H6O 46.97g C2H6O = 19.92g C2H6O needed 22.4LCO2 2 mole CO2
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Station 5 - Unit 7 M= mol/L 1L = 1000mL
How many moles of Na2CO3 are in 10.0mL of a 2.0M solution? What volume (in mL) of 12.0M HCl is needed to contain 3.00moles of HCl? How many liters of 4M solution can be made with 100g of lithium bromide? How many grams of ammonium sulfate are needed to make a 0.35L solution with a concentration of 6.0M? A student boils a sample of water and slowly adds as much NaCl as possible while still having NaCl be dissolved. If the student adds any more NaCl, it will not be able to dissolve. Is this a saturated, supersaturated, or unsaturated solution.
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Station 5 KEY 2.0M = moles/0.01L Na2CO3 = 0.02moles
12.0M HCl = 3.00moles/L 0.25L = 250mL of 12.0M HCl 100gLiBr= 1.15 moles LiBr 4M= 1.15moles/L Volume of LiBr = 0.288L 6.0M= moles/0.35L moles = 2.1moles of (NH4)2SO4 2.1 moles (NH4)2SO4 = g (NH4)2SO4 Supersaturated
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Station 6 - Unit 7 1) If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K. I then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas? 2) In the diagram below, write the equation that represents the gas laws relationship. 3) Explain why water boils at a lower temperature in Colorado vs. places along the coast.
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Station 6 KEY 29.57 Liters P1V1 = P2V2
Since Colorado is at a higher elevation, there is less air pressing down on the water here. Therefore, it is easier for water to boil since there is less atmospheric pressure. With less pressure pushing down, there is less energy required to evaporate the molecules.
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Station 7 - Unit 8 What was the specific heat of the metal?
A colorimeter containing 100.0mL of room temperature water (23.2℃) warms up to 48.1℃ when a 225g sample of an unknown metal is placed inside. The initial temperature of the metal was 100.0℃. 𝚫H = m𝚫TCp Cp(H2O) = 4.18J/g℃ 1mLH2O = 1gH2O Metal Specific Heat (J/g℃) Aluminum 0.91 Iron 0.46 Lead 0.13 Silver 0.23 Tin 0.21 Titanium 0.54 Zinc 0.39 What was the specific heat of the metal? What is the identity of the metal? Perform a percent error calculation for the specific heat of your metal. Was the metal endothermic or exothermic? Was the water endothermic or exothermic?
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Station 7 - KEY 𝚫HH2O = (100.0g)(48.1℃-23.2℃)(4.18J/g℃) 𝚫HH2O = J 𝚫HH2O= -𝚫Hmetal 𝚫Hmetal= J 𝚫Hmetal=m𝚫TCp J = (225g) (48.1℃ ℃) Cp Cp metal= 0.89J/g℃ Metal = Aluminum % error = [( )/(0.91)] x 100 = 2.20% error Metal- Exothermic Water- Endothermic
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NH3(g) + HCl(g) ↔ NH4Cl(g) 𝚫H = -92,048.0 J
Station 8 - Unit 8 Stress Equilibrium Shift [NH3] [HCl] [NH4Cl] 1. Add NH3 Products ____ decreases increases 2. Add HCl 3. Add NH4Cl 4. Remove NH3 5. Remove HCl 6. Remove NH4Cl 7. Increase Temperature 8. Decrease Temperature Complete the following chart to describe the temporary equilibrium shift of the reaction due to stressors (Le Chatelier’s Principle) Is this reaction Endothermic or Exothermic? ______________
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NH3(g) + HCl(g) ↔ NH4Cl(g) 𝚫H = -92,048.0 J
Station 8 - KEY Stress Equilibrium Shift [NH3] [HCl] [NH4Cl] 1. Add NH3 Products ____ decreases increases 2. Add HCl P Decrease Increase 3. Add NH4Cl R 4. Remove NH3 5. Remove HCl 6. Remove NH4Cl 7. Increase Temperature 8. Decrease Temperature Complete the following chart to describe the temporary equilibrium shift of the reaction due to stressors (Le Chatelier’s Principle) Is this reaction Endothermic or Exothermic? ___Exothermic_____
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Station 9 - Unit 8 Draw a graph to represent how energy changes during an exothermic and an endothermic reaction. Label Activation Energy, Products, Reactants, and Enthalpy.
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Station 9 KEY Exothermic Endothermic
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Station 10 - Unit 9 H2SO4 + NH3 ↔ NH4+ + HSO4-
Draw an arrow for the transfer of H+. Label Acid, Base, Conjugate Acid, and Conjugate Base H2SO4 has a [H3O+] of 3.8 x 10-3M, identify the pH of H2SO4 NH3 has a pH of 12.87, calculate the concentration of hydronium ions.
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Station 10 KEY H2SO4 + NH3 ↔ NH4+ + HSO4- 2. pH of H2SO4= 2.42 3. [H3O+] NH3= 1.35 x M
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Station 11 - Unit 9 A student performs a titration of 25.0mL of an unknown concentration of KOH is titrated with 0.6M HBr. The initial reading of the burette was 47.2mL while the final burette reading was 6.3mL. Phenolphthalein was used as the indicator (turns pink when basic, clear when acidic). Calculate the molarity of the KOH using M1V1= M2V2. Describe the titration lab setup and method for performing a titration. Identify which substance is the acid- where will the acid be during the experiment? Identify which substance is the base- where will the base be during the experiment? What color will each of the substances be at the beginning and end of the titration? Describe how the titration is performed and how to know when to stop. Calculate moles of acid and moles of base at the equivalence point using M=mol/L. Check your answer for #3 using your knowledge of what the relationship between moles of acid and moles of base at the equivalence point should be.
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Station 11 KEY M1V1= M2V2 (25.0ml) M1 = (47.2mL - 6.3mL) (0.6M) M1= 0.98M KOH Describe the titration lab setup and method for performing a titration. Acid = HBr, HBr will be in the burette Base = KOH, KOH will be in the erlenmeyer flask at the bottom of the burette, 3-ish drops of phenolphthalein will be placed in the flask The burette (HBr) will be clear at the start and end of the experiment. The flask (KOH) will start out pink and become clear when it reaches the equivalence point. When the base has turned clear, the equivalence point was reached. Moles of Base (0.98M) = moles/(0.025L) KOH= moles Moles of Acid (0.6M) = moles/(0.0409L) HBr = moles Moles of Acid = Moles of Base at the equivalence point
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