Lab chemisty 03/27/2011 Five unlabeled bottles Set 1: A) colorless B) colorless C) blue D) blue E) colorless A: Ba(NO3)2 B: AgNO3 C: CuSO4 D) CuCl2 E) KCl Description how to identify solution: _ We have two blue solution which are CuSO4 and CuCl2 or C and D‚ according to chemical reaction experiment‚ C didn’t have any reaction with other solution like B and D beside A‚ so if we look at the solubility
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Kassidy Caviness- Thames Lab Partner: Dena Jackson Reaction lab “I certify that this lab report is my own work‚ except for properly referenced and cited information. I have adhered to all guidelines published in the student handbook on Academic Integrity‚ as well as all guidelines published for this class in the Syllabus and Academic Integrity Handouts.” Purpose- The purpose of this lab was to display to us a variety of different reactions using an eclectic of things in the chemistry lab: including
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Qualitative Observations of Double Displacement Reactions Lab Table 1.0 Qualitative Observation of Products Formed |Balanced Chemical Equations |Qualitative Observations | |BaCl2 (aq) + 2NaOH (aq)( BaOH2(aq) + 2NaCl(s) |An aqueous solution formed | | |Precipitate
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Calcium Calcium is present in fly ash in multiple modes of occurrence‚ primarily lime‚ anhydrite‚ calcite and within the glassy matrix. It is the most largely released cation (Kim et al.‚ 2003). In the present column experiment‚ leached concentration of calcium was recorded as 50 ppm at 10th day of experiment. At 20th day of experiment‚ calcium concentration in leachate was found as 120 ppm. While at 30th day of experiment‚ calcium concentration in leachate was 280 ppm and at 45th day it was 450
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INTRODUCTION The lab introduced the relationship between reactants and products‚ and sought to discover which ratio of an acid and base reaction produced the most amount of carbon dioxide gas (CO2) without leaving leftover reactants. 5 varying amounts of bases were added to a constant amount of acid (10 ml) to better understand which ratio was the most efficient. RESULTS Data collected from the lab suggests that the ratio of acid to base that produced the most carbon dioxide gas (CO2) was 1:0.5
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balance chemical equations to represent chemical reactions. Based on the similarities in color‚ the unknown solution #1 is predicted to be Sodium Hydroxide. Procedure: Materials: * Well Plates * Dropper Bottles with Solutions(Sodium Carbonate‚ Silver Nitrate‚ Iron(III) Nitrate‚ Strontium Nitrate‚ Potassium Nitrate‚ Sodium Chloride‚ Zinc Sulfate‚ Sodium Hydroxide) Directions: 1. Use the dropper bottles to fill 9 well
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solution would be 0.00935 M. Enter your calculated molarity of the primary standard KIO3 solution. Please use 3 significant figures. Your Answer: 0.01 You Scored 3 points out of 3 Possible 7) Data Entry - No Scoring Standardization of the sodium thiosulfate solution using the potassium iodate primary standard solution. We must examine each of the three acceptable trials. First‚ let’s consider the analyte volume. You were instructed to pipet two 10.00 mL aliquots of potassium iodate into
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What is Soap? Soaps are mixtures of sodium or potassium salts of fatty acids which can be derived from oils or fats by reacting them with an alkali (such as sodium or potassium hydroxide) at 80°–100 °C in a process known as saponification. fat + NaOH ---> glycerol + sodium salt of fatty acid CH2-OOC-R - CH-OOC-R - CH2-OOC-R (fat) + 3 NaOH ( or KOH) both heated ---> CH2-OH -CH-OH - CH2-OH (glycerol) + 3 R-CO2-Na (soap) R=(CH2)14CH3 Soap is a cleansing agent created by the chemical reaction
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How sodium chloride concentration affects the germination‚ growth and development of Brassica rapa. Osumi‚ K.‚ Nihei‚ B.‚ Takanishi‚ K.‚ Nakamura‚ C.‚ Warehime‚ R. Period 2. 03/05/17 Introduction: To have a successful germination‚ the plant seed needs a sufficient amount of water and oxygen present. Other conditions include room temperature‚ and the amount of light intake. Hydration of a seed‚ is an essential step for seed germination (Martínez-Andújar‚ 2014). Plants tend to germinate better in
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Test #2 Volume CaCl2: 5.0 mL = .005 L Volume NaOH: 4.0 mL = .004 L Moles CaCl2 = (.005 L) * (1.0 mol/L) = .005 moles Moles NaOH = (.004 L) * (2.5 mol/L) = .010 moles Limiting Reactant: NaOH‚ because precipitate was formed when Sodium Hydroxide
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