opportunities‚ this industry is burgeoning ahead without looking back or halting. Though the growth potential in the laptop market is tremendous‚ so is the competition. In this market their exist at least 6 major players: HP‚ Lenovo‚ Dell‚ Sony VAIO‚ Acer and HCL; as also many smaller but niche players like Toshiba‚ Panasonic and Apple. This tremendous competition has led to a mad scramble for a larger portion of this ever-growing pie‚ i.e.‚ the laptop industry. In this attempt to increase one’s market share
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1 Assignment I - Consumer Behaviour (23 Oct 12) AFP 12/ STUDY GROUP 6 REPORT ON UNDERSTANDING CONSUMER BEHAVIOUR: BUSINESS TO BUSINESS (B2B)- COMPUTER NETWORKING SYSTEMS Objective 1. To understand the concepts of ‘Buyer Behaviour and Segmentation’ in a specific product situation through application. Definition of Product Category and Description of Product 2. The product category is B2B Products in which Computer Networking Services has been selected. A brief description of these is given below:(a)
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c)i) Moles of HCl in 25cm3 of solution D: HCl + NaOH NaCl + H2O 1 : 1 ½ Moles of HCl in 25cm3 = 0.002125 ½ ( Mole ratio) ii) Moles of HCl on 100cm3 of solution D 25 0.002125 100x 0.002125 ½ = 0.0085 moles ½ 25 iii) Moles of HCl in 100cm3 Solution B: 0.5 moles 1000cm3 0.5 x 100cm3 = 0.05moles ½ 1000 iv) Moles reacting with x2CO3 X2CO3 + 2HCl 2xCl + CO2 + H2O 1 : 2 Moles of HCl reacting =
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Department of Pure and Applied Chemistry College of Arts and Sciences Visayas State University Name: Almera P. Padolina Course & Year: BS Chem 2 Date Performed: July 24‚ 2012 Date Submitted: August 6‚ 2012 Experiment No. 4 Solubility Equilibrium- Common Ion Effect INTRODUCTION: The common ion effect is another example of Le Châtelier ’s Principle in action.The common ion effect tells us that the solubility of an
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Equation 4 Measured Molar Enthalpy Change Figure 4. Combination of 100 mL 6 M HCl and approximately 1.0 g MgO Reaction Figure 5. Combination of 120 mL 2 M HCl and approximately 0.5 g Mg Reaction The given theoretical value was 602 kJ/mol and the measured ΔH was -552.05 kJ/mol. The calculated percent error was 8.30% which is very good‚ considering less than
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volumetric flask‚ and fill to the line with distilled water. The alkali will now have a concentration of 0.1 mol dm-3. • In the titration‚ the NaOH will be in the burette‚ and will be titrated into HCl in a conical flask: NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) • The reaction is 1:1 between NaOH and HCl‚ so a solution of 0.2 mol dm-3 should be used. • Rinse a 25cm3 pipette out with some of this solution‚ and then transfer 25cm3 of it to a clean conical flask. • Add 3 drops of phenolphthalein.
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molarity of Hydrochloric Acid Solution Part II Determination of Percentage Composition of Vinegar Data and Observations Part I Determination of Molarity of Hydrochloric Acid Solution Table 1 Volume of NaOH Needed to Neutralize 10.00 ml of Unknown HCl Molarity of NaOH= Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Initial volume of NaOH (ml) Final volume of NaOH (ml) Volume of NaOH used Average volume of NaOH Part II Determination of Percentage Composition of Vinegar Table 2 Volume of
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a) Titration of HCl with Sodium Carbonate: Na2CO3 + 2HCl 2NaCl + H2O + CO2 (2) Concentration of Sodium Carbonate (Na2CO3) Molar mass of Na2CO3 : 22.99*2+12.01+16*3=105.99 Mass of Na2CO3 : 1.3 g Number of moles : n=mM =1.3 105.99=0.012265308 mol Volume Na2CO3 solution : 250 mL : 0.250 L Concentration: Concentration of Sodium Carbonate is 0.049 mol L-1 c=nV = 0.012265308 0.250=0.049061232 mol L-1 (3)Concentration of HCl Volume of Na2CO3 solution
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end-point is greater than 7. 2. Calculate the pH after the following solutions are mixed together: a) 15 cm3 of 0.1 moldm-3 HCl and 10 cm3 of 0.1 moldm-3 NaOH b) 10 cm3 0.1 moldm-3 HCl and 15 cm3 of 0.1 moldm-3 NaOH 3. Sketch pH curves for the following titrations: a) 20 cm3 0.10 moldm-3 NH3 against 0.1 moldm-3 HCl b) 20 cm3 0.10 moldm-3 NaOH against 0.2 moldm-3 HCl c) 20 cm3 0.10 moldm-3 CH3COOH against 0.06 moldm-3 NaOH d) 20 cm3 0.10 moldm-3 CH3COOH against 0.15 moldm-3 NH3
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Experiment 5 COMMON ION EFFECT MALUBAY‚ Justin Damian PADRILLAN‚ Hazel Rose CD2‚ Group 5 Ms. Sarah Sibug 6 April 2013 ------------------------------------------------- ------------------------------------------------- I. ABSTRACT The common ion effect occurs when a given ion is added to an equilibrium mixture that already contains that ion‚ and the position of equilibrium shifts away from forming more of it. This paper is a follow-up of the experiment which aims to determine the common-ion effect
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