Preparation of Sodium Chloride through titration Abstract: acid-base titration is a technique commonly used to determine the moles of acid in a sample by adding a known volume of strong base of a known concentration. The strong base provides the hydroxide ion‚ to react quantitatively with the acid. The point at which the acid is completely and exactly consumed the known quantity of base is called the equivalence end point and is signalled by a colour change in the solution (end point). This colour
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Acids and Bases Q1.This question is about several Brønsted–Lowry acids and bases. (a) Define the term Brønsted–Lowry acid. ........................................................................................................................ ........................................................................................................................ (1) (b) Three equilibria are shown below. For each reaction‚ indicate whether the substance immediately above the box is acting
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effect of citric and buffered lactic acids on the flavour of hard-boiled sweets and the effect of acids on various flavours in high temperature applications. Introduction Materials and Methods An amount of water‚ sugar and glucose syrup of 30g‚ 100g and 70g were weighed respectively into a stainless steel pot. The mixture was then heated and removed immediately from the induction cooker after reaching the desired temperature of 145˚C. Flavours of 0.51g and acid of 1.20g was added immediately afterwards
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Accounting Equation Paper Student Course Date Instructor Accounting Equation Paper The accounting equation which we know as Assets equals to Liabilities plus Equity for a sole proprietorship and for a corporation we know it as Assets equals to liabilities plus stockholders & equity. Assets are company owned‚ liabilities are what company owes and the difference between the both of them is the owner’s equity‚ these three things are what the companies
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used (ml) | 30.6 | 30.2 | 30.0 | | | | | Acid to Base Ratio | 0.87 | 0.86 | 0.86 | Average acid/base ratio | 0.86 | | | | | Base to Acid Ratio | 1.14 | 1.16 | 1.17 | Average base/acid Ratio | 1.16 | Name: Jared Philip Condez Date Performed: June 28‚ July 2 & 5‚ 2013 Partner: Shiela Mae Molina Date Submitted: July 12‚ 2013 Experiment 3 ACID – BASE TITRATION I. Objectives * Determine the purity of Potassium Acid Phthalate * To titrate effectively
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Studying the pH of Strong Acid‚ Weak Acid‚ Salt‚ and Buffer Solutions The purpose of the current experiment was to determine the pH of various hydrochloric acid and acetic acid solutions‚ to determine the pH of various salt solutions‚ to prepare a buffer solution‚ and determine the effects of adding a strong acid and strong base to the buffer solution versus adding a strong acid and strong base to water. The measured pHs for the hydrochloric acid solutions were 1.6‚ 2.2‚ 2.9‚ and 3.8. The measured
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| An acid is a chemical substance any of various typically water soluble and sour compounds that are capable of reacting with a base to for a salt‚ that redden litmus paper‚ that evolve hydrogen on reaction with various metals‚ that in water solution yield hydrogen ions‚ and that have hydrogen containing molecules or ions able to give up a proton to a base or that are substances able to accept and unshared pair of electrons from a base. An alkaline is a chemical compound that neutralizes or effervesces
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GIVEN: TO DESIGN A 1000TPD CAPACITY H2SO4 ACID PLANT BASIS: 1 HOUR OF OPERATION. PURITY: PRODUCT WHICH IS TO BE MANUFACTURED IS ASSUMED TO HAVE STRENGTH OF 98% ACID. 1000TPD implies that we have Acid 1000 x 10 / 24 = 41666.67 Kg/Hr of 3 With 98% purity‚ the acid that is produced per hour = (98 x 41666.67) / 100 = 40833.34 Kg/Hr Kmoles of Sulfuric acid to be produced = 40833.34 / 98 = 416.667 Kmoles/Hr It’s assumed that overall absorption of the acid is 100 % = 416.667 / 1.0 Then‚ SO3 required
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Kinematics / Projectiles x =?vt ?v = (v + vo)/2 v = vo + at x = vot + ½at2 v2 = vo2 + 2ax y =?vt ?v ’ ½(vo + v) v = vo – gt y = vot – ½gt2 v2= vo2 – 2gy R = (v02/g)sin(2θ) Forces Fnet = ma Fgravity = mg Ffriction ≤ μsN Ffriction = μkN Circular Motion Fnet = mv2/r ac = v2/r v = 2πr/T f = 1/T T = 1/f Gravitation F = GM1M2/R2 g = GM/R2 T2/R3 = 4π2/GM = constant GM = Rv2 Energy W = Fdcosθ KE
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The Vector equation of a plane To find the vector equation of a plane a point on the plane and two different direction vectors are required. The equation is defined as: where a is the point on the plane and b and c are the vectors. This equation can then be written as: The Cartesian equation of a plane The cartesian equation of the plane is easier to use. The equation is defined as: One of the advantages to writing the equation in cartesian form is that we can easily find the normal
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