"Denaturation of proteins hcl naoh" Essays and Research Papers

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    Materials: 0.20M HCl (±0.02moldm-3) 1.1g solid NaOH (±0.1g) A solution of vinegar of unknown concentration (density= 1.05gcm-3) Phenolphthalein Method: * A solution of NaOH was prepared by dissolving 1.1g in 250cm3 of water * 1.1g of solid NaOH was weighed and then dissolved in the stirred until dissolved in 250cm3 of water * 25cm3 of this solution was placed into a conical flask and 3 drops of phenolphthalein indicator were added * A burette was filled with 0.2±0.02moldm-3 HCl * The

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    Dna Melting Curve

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    Q1. State the experimental aim and describe the term “hyperchromic effect”. (2 mark) The experimental aims are to study the basic structural characteristic of DNA molecules through UV spectrum and thermal denaturation and to understand how pH and ionic strength affect the stability of DNA which lead to the shift of melting temperature Tm. Hyerchromic effect is the increase of absorbance at 260nm during the double strand DNA is unwounded into two single strands

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    react with the solution of NaOH according to this equation: HCl (aq)+ NaOH(aq) → NaCl(aq) + H2O(l) HCl is a strong acid‚ NaOH is also a strong base‚ therefore they will dissolve completely. By adding NaOH into HCl‚ the hydrogen ions neutralized with hydroxide ions‚ the theoretical endpoint and then the solution will be weak basic solution. The endpoint

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    PreLab assesment

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    bluish hues 4.How many grams of NaOH is needed to prepare 250.0 mL of 0.50 M NaOH. Show work.: 22.99+16.00+1.008=39.998 g/mol 62.5 moles (39.998/1mol = 249.875 grams 5. Calculate the molarity of 25 mL of HCL that is neutralized by 30.5 mL of 0.50 M NaOH:  H30+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) --> Na+ (aq) + Cl (aq) + H20 (l))  0.50 M means 0.50 mol/L --> 0.50m mol/mL you added 30.5 mL so:  0.50 *30.5 = 15.25 mmol NaOH.  This means there is also 15.25 mmol HCl in your original solution. 

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    Titration Research Paper

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    Quantitative Chemistry –Titration Determination of the Molarity of an Unknown Solution through Acid-Base Titration Technique 1. Introduction 1.1 Aim The aim of this investigation was to determine the precise molarity of two (NaOH(aq)) sodium hydroxide solutions produced at the beginning of the experiment through the acid-base titration technique. 1.2 Theoretical Background Titration is a method commonly used in laboratory investigations to carry out chemical analysis. The most

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    Lab report

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    Acid Base Titration Purpose: The purpose is to calculate the molarity of a NaOH solution by titrating the base with 5mL of standard HCl solution in each trial. By adding the base with unknown molarity to the acid with 0.10M the molarity of NaOH can be calculated. The base‚ NaOH‚ helps bring the pH of the acid‚ HCl‚ closer to seven‚ which neutralizes it. When using the buret the amount of NaOH used is able to be determined. Then by writing a balanced chemical equation and using the titration

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    Record the colour of this solution in note book FIRST FLASK: add a single drop of 0.1M HCl. Swirl the contents of the flask and continue drop-by-drop addition until a definite colour change is observed. SECOND FLASK: will serve as a control. COMPARE the solution colours Record the new colour and the number of drops required for the change to take place in Table 1. Continue the drop-by-drop addition of 0.1M HCl to the first flask until a second colour change occurs. COMPARE with the control (Flask

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    and advanced chemical equilibrium. In solutions that used in the experiment were two species are present reaction between CuSO4 and NaOH. In this method of experiment‚ the total molar concentration of the two solutions are held constant‚ but their mole fractions are varied. In the second experiment‚ it was observed the reaction of acid-base in solution of NaOH with HCl in different condition of volume for each. An observable that is proportional to complex information is plotted against the mole fractions

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    thermometer‚ two 50.00cm3 measuring cylinders‚ stopwatch‚ three 80cm3 beakers‚ dropper Variables: Manipulated variable: Type of acids used In this experiment‚ type of acids used would be manipulating variable. Different acids such as HCl or CH3COOH are added to NaOH respectively and measure the increase in temperature respectively. Responding variable: Temperature‚ T Responding variable will be the temperature. First‚ we have to measure and record the initial temperature of the sodium hydroxide

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    electrolytes. In this experiment‚ a weak electrolyte‚ phenol is used‚ and the change in enthalpy is endothermic‚ ΔH=q=25.3kJ/mol. The first part of this experiment involves the neutralization of strong electrolytes. 40 mL of 2.0 M HCL was used to neutralized 50 mL of 2.0 M NaOH. Both the solutions were mixed in the calorimeter and the temperature was recorded after 2 seconds until it reaches maximum‚ then every 10 seconds for one minutes‚ and finally every 30 seconds

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