CHM1022 Tutorial 2 – Semester 2‚ 2012 (Chemical Equilibria) 1. The reaction 2 HCl(g) +I2(s) [pic] 2 HI (g) + Cl2(g) has Kc = 1.0 x 10-34 at 25˚C. If a 1.00 L reaction vessel initially contains 0.100 mol of each HCl and solid I2‚ what are the concentrations of HI and Cl2 at equilibrium? 2. Consider the following gas-phase reaction and equilibrium constant at 25 oC: 4 HCl(g) + O2(g) [pic] 2 Cl2(g) + 2 H2O(g) The concentrations of all species were measured at a particular
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Chapter 6 - Answer Key to Section Review 1-3 Section Review 1 1. What is the main distinction between ionic and covalent bonding? Answer (A): Ionic bonding involves the electrical attraction between large numbers of anions and cations. Covalent bonding involves the sharing of electron pairs between two atoms. Translation: -Ionic bonding happens between a metal and a non-metal (east coast and west coast) -One atom completely donates its valence electrons to another atom -Metals
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Chemistry: A Molecular Approach (Tro) Chapter 10 Chemical Bonding II: Molecular Shapes‚ Valance Bond Theory‚ and Molecular Orbital Theory 1) Determine the electron geometry (eg) and molecular geometry(mg) of BCl3. A) eg=trigonal planar‚ mg=trigonal planar B) eg=tetrahedral‚ mg=trigonal planar C) eg=tetrahedral‚ mg=trigonal pyramidal D) eg=trigonal planar‚ mg=bent E) eg=trigonal bipyramidal‚ mg= trigonal bipyramidal Answer: A 2) Determine
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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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INTRODUCTION Enzyme is a biological catalyst that acts on a molecule called substrate and it also significantly speed up a chemical reaction by lowering the activation energy. In order to learn about the enzyme and its behaviour‚ this lab practical is conducted to examine the kinetic of the enzyme alkaline phosphatase. As illustration‚ when alkaline phosphatase is added to a substrate called p-Nitrophenyl phosphate (colourless in alkaline solution)‚ a series of reaction takes place and eventually
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Analysis of a Chemical Reaction Purpose: To observe a chemical reaction and to use qualitative and quantitative evidence to identify this reaction from among four possibilities. Hypothesis: I think the result is going to produce water. I think this is going to happen because there is hydrogen and oxygen inNaHCO3. Materials: -Test tube clamp - 150 mm test tubes (2) - burner - retort stand - clay triangle - iron ring - crucible Procedure: Part A: 1. Add 0.5 g of NaHCO3
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From the graphs‚ it is evident that an increase in both catalase concentration and substrate concentration resulted in a higher rate of reaction or‚ as observed in the kPa graphs‚ a higher volume of O2(g) formed at the end of the 5 minute trial. Interestingly‚ it should also be noted‚ as it was mentioned in the Figure 2‚ that the trend for the 6mL of 3% H2O2(aq) was more of a linear trend than an exponential decay‚ steadily rising until the end of the 5 minute trial. From this‚ it can be inferred
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The relationship between shape and diffusion rate Aim- To investigate the relationship between the shape of an object and the time taken for a substance to diffuse to its centre. Hypotheses- Part A) The 5x5x2.5mm block will turn clear first as the Surface area to volume ratio is high and the 10x10x10mm block will take the longest to completely diffuse the Acid to its centre. Part B) The flat‚ rectangular block will be the first to completely turn clear due to its high Surface Area/Volume Ratio
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Ligia Ramos 11th Grade HL Biology – Ms. Bartels Due Monday October 21st Limitations on Cell Size Research Question: What is the correlation between surface-to-volume ration and ion exchange and how does this relate to cells? Evaluation of Method and Results: Errors/ Limitations: | Suggestions for improvement: | Impact on Results: | Precision - Human error –cutting cubes freely‚ resulting in very different sized cubes‚ especially for the 1cm x 1cm x 1 cm cubes. All the cubes also had slightly
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Investigating the Limits of Cell Growth Background: To find factors that limit cell growth In multicellular organisms‚ In order for an organism to grow cells must divide through mitosis. Cell division happens after the cell is big enough so the daughter cells will be replicated without any flaws. The regulation and amount of materials that can enter and leave the cell is based on how large the cells surface area is. The rate of movement is also determined by the volume of the cell. Many factors like
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