The Sequence of Chemical Reactions Drew Selfridge Dave Allen‚ Lab partner Instructor Yang February 11‚ 1997 INTRODUCTION This experiment was to recover the most amount of copper after it is subjected to a sequence of reactions. The copper is originally in solid form‚ but the reactions will turn it into free Cu+2 ions floating in solution. The ions will then be regrouped to form solid copper once again. During this process‚ however‚ some of the Cu+2 ions may be lost. The copper will subjected
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complete sentences‚ giving detailed explanations and support for each of your answers. 1. Explain in your own words what it means for a chemical system to be in the state of dynamic equilibrium. After a reaction has occurred for awhile at a given temperature‚ the forward and reverse reaction rate will eventually be equal. Although you may get this confused‚ the concentration may not be equal‚ but the rate will. This occurs in a closed system. In other terms‚ dynamic equilibrium is when no change
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Erin Bolton Chemistry Lab Report April 29‚ 2015 Lab: Reaction Rates Introduction: In this experiment we studied the reaction of potassium persulfate‚ K2S2O8‚ with potassium iodide‚ KI. All chemical reactions have an energy barrier to overcome before the reaction will proceed. We will record data based on the concentration‚ temperature and catalyst for each experiment. Once this has been completed it will be graphed. Procedure: Due to the chemicals being used having hazard gloves are used
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There are two types of chemical reaction that occur one goes only in one direction and the other one is reversible. A reversible reaction is when a products starts to form the backward reaction starts where the products turns back to reactant molecules. When the rate of forward and backward reaction is already equal and the concentrations of the reactants and products no longer change with time we can say that chemical equilibrium is already achieved. A reaction is said to be at equilibrium when
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AP Chem Exam - ‘98 1. Solve the following problem related to the solubility equilibria of some metal hydroxides in aqueous solution. (a) The solubility of Cu(OH)2(s) is 1.72 x10–6 g/100. mL of solution at 25° C. (i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution. Cu(OH)2 Cu 2+ + 2 OH – (ii) Calculate the solubility (in mol/L) of Cu(OH)2 at 25 °C. (1.72 x10–6 g/0.100 L)(1 mol/97.5 g) = 1.76 x10–7 mol/L (iii) Calculate
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II. LEARNING OBJECTIVES - To perform different types of chemical reactions including acid-base‚ precipitation‚ gas forming‚ complex compound forming and oxidation-reduction reactions. - To identify some of the products in these reactions and describe the chemical changes. - To write and balance the chemical equations for the reactions observed. III. EQUIPMENT AND REAGENTS 1. EQUIPMENTThirty test tubes One test tube rack Two test tube holders Two spatulas Three 250 mL beakers One stirring rodOne
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Change of Reaction Purpose of Experiment The purpose of this experiment is to determine the enthalapy change for the displacement reaction: Zn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq) Hypothesis With this experiment I can also not make a hypothesis‚ because we did actually not do the experiment‚ but we were told that the temperature would make a sudden drop ‚ but we can measure the ΔT of the surrounding. The reaction is endothermic because the system will take in energy. During the reaction bonds
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more clearly displays the relationship‚ view Figure 2. The rate at which the reaction occurred for the 100% concentration is 1.45 mg/dL per minute. For the 50%‚ the rate was 3.05 mg/dL per minute‚ and for the 25%‚ the rate was 2.76 mg/dL per minute. As seen‚ the rate fluctuated from the lowest rate at 100% and the greatest rate occurring at 50%. As mentioned earlier‚ the Collision Theory states the rate of a chemical reaction is proportional to the number of collisions that occur. One way to increase
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The effect of time on enzyme reaction. Abstract: In this lab investigation we will observe how the amount of hydrogen peroxide is affected by catalase over time. The enzyme was added to 10 mL’s of hydrogen peroxide and observed over time to determine the relation between time and enzyme activity. The hypothesis stated that as time increased substrate would decrease. Therefore I predicted that at 60 seconds‚ there would be the least amount of H2O2. The enzyme activity mirrored my predictions
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r Lab Report 5 Introduction to the Classes of Chemical Reactions Course: Chem. 1151L‚ Tuesday & Thursday June 23‚ 2011 Mr. Nasir Uddin Pre Lab Questions: 1. CaBr2 (aq) + K3PO4 (aq) → CA(PO4)2(S) + KBr (aq) = Ca3(PO4)2 + 6 KBr Double Replacement 2. Li(s) + O2(g) = Li2O(s) =2 Li2O Decomposition 3. CH4 + O2 = CO2 + H2O = CO2 + 2 H2O Combination 4. AgBr(s) = Ag (s) + Br2(l) = 2 Ag + Br2 Combination 5. Mg(s) + H2SO4 (aq) = MgSO4 + H2
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