various elements and compounds in the unknown‚ its true identity can be predicted. Chloride‚ ammonia‚ and cobalt are three examples of percent compositions determined to help narrow the selection of possible unknowns. Titrations using Na2S2O3 and HCl to determine percent cobalt and ammonia‚ respectively‚ are used. Silver nitrate is used to precipitate the chloride ions in the unknown‚ which can be measured to determine the percent composition of chloride in the unknown. The results from these three
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concentration affect the rate of a reaction? In this investigation I will be investigating to see whether the concentration affects the rate of a reaction. To see if the concentration does affect the rate of a reaction I will be reacting various HCL strengths with Na2S2O3 to see if the concentration affects the rate the two substances react. Also in this investigation I will be looking into see if there are any states the substance should be in that would affect the rate of reaction. The
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NaOH(aq) and Na2CO3(aq) 2. NaHCO3(aq) and Na2CO3(aq) Introduction Consider a mixture of NaOH(aq) and Na2CO3(aq). Reaction between HCl(aq) and Na2CO3(aq) takes place in two stages: HCl(aq) + Na2CO3(aq) ⎯→ NaHCO3(aq) + H2O(l) …………………. (1) HCl(aq) + NaHCO3(aq) ⎯→ NaCl(aq) + CO2(g) + H2O(l) …………. (2) While that between HCl(aq) and NaOH(aq) completes in only one step: HCl(aq) + NaOH(aq) ⎯→ NaCl(aq) + H2O(l) ……………….………. (3) Solution mixture of reaction (1) at the equivalence point is alkaline‚ that of
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SYSTEM 2.A. Cardiac Drugs Generic name | Brand name | Dopamine HCl | BAXTER DOPAMINE HYDROCHLORIDE & 5% DEXTROSE INJ. | | DOPAMAX | | DOPAQARD | | HOSPIRA DOPAMINE HCL | | MYOCARD | Dopamine HCl in 5 % dextrose sol’n | HOSPIRA DOPAMINE HCL IN 5% DEXTROSE SOLUTION | | MYOCARD- DX | Amiodarone HCl | AMIO | | ANOION | | CORDARONE | | MYODIAL | | RYTHMA | | Dbutamine HCl | BAXTER DOBUTAMINE HCL IN 5% DEXTROSE INJ | | CARDOMIN | | DOBUCARD | | DOBUJECT
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antacids by looking at their weight of HCl and weight of antacid values. The analysis of antacid tablets was highlighted in this experiment. The efficiency of antacid tablets was determined and compared when the number of grams of HCl can be neutralized by 1 gram of the tablet was found. First‚ the two antacid tablets (Kremil-S) were crushed and weighed to the nearest 0.01 g which was 0.5003 g and 0.5014g. Then‚ transferred into a 250 mL flask and added 50.0 ml 0.1M HCl using an acid burette. Then‚ the
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Unknown HCl Molarity of NaOh | Trial 1 | Trial 2 | Trial 3 | Trial 4 | Initial Volume of NaOH(mL) | 0.00 | 11.00 | 20.85 | 30.45 | Final Volume of NaOH(mL) | 11.00 | 20.85 | 30.45 | 39.98 | Volume of NaOH used(mL) | 11.00 (Cancel out) | 9.85 | 9.60 | 9.53 | Average Volume of NaOH = (9.85+9.60+9.53)/3 = 9.66mL Sample Calculations: (9.85+9.60+9.53)/3 = 9.66mL The average volume of NaOH used. Calculations: 1. Moles NaOH = M x V = (0.1M) (0.00966L) = 0.000966 moles 2. Moles HCl = moles
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DEMONSTRATOR: RICHARD TIA DATE: 2ND APRIL‚ 2009. AIM To prepare and standardize HCl solution. INTRODUCTION Standardization is a process of determining the relationship between the measured signal and the amount of analyte. Standardization can be defined also as a titration experiment in which the concentration of a solution becomes
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Hydrolysis of tert-butyl Chloride in different solvents Practical conducted on 5 March‚ 2013 Reported by Pham Vu Hung on 10 March‚ 2013 Introduction: This practical is meant to measure the rate of reaction of the hydrolysis of tertiary-butyl chloride –a colorless‚ liquid organic compound at room temperature that is sparingly soluble in water - in water/acetone and water/isopropanol mixtures. Since there are many influencing factors for the rate of reaction‚ all are kept constant but the
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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. So there is 15.25 mmol / 25 mL --> 15.25/25‚0=0‚ mmol/mL --> 0‚61 mol/L ANSWER: 0.61 M HCl 6. Calculate the molarity of 15 mL of NaOH that is neutralized by 38.8 ML of 0.20 M HCl 0.20 m HCL 0.20
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