solvents will be measured in order to explore the rate of reaction on different solvents’ effects Materials: Apparatus: 50.00 mL burette Chemicals: 0.04M NaOH 10.00 mL pipette t-butyl chloride 100 cm3 volumetric flask Acetone Conical flask Isopropanol Measuring cylinder Bromothymol blue Procedures: 1. 150 cm3 of 0.04M NaOH(aq) is placed in a beaker. 2. 100 cm3 of a 50/50 acetone/watermixture (by volume) is put into a stoppered flask or bottle and is mixed well. 3. 1.00 cm3 (±0
Free Chemical kinetics Chemical reaction Solvent
CHEM161 Experiment 7: Calorimetry Introduction Calorimetry is the science of measuring heat flow‚ and heat is defined as thermal energy flowing from an object at a higher temperature to one at a lower temperature. For example‚ if a chunk of metal at room temperature is placed in a beaker of boiling water‚ the metal will absorb heat from the water until it is at the same temperature as the boiling water. Scientists also often study the heat associated with different physical and chemical changes
Premium Sodium hydroxide Chemical reaction Thermodynamics
weighed to 0.12 g mass. Initial Volume of NaOH | Final Volume of NaOH | Volume of NaOH needed | 46.6 mL | 34.8 mL | 46.6-34.8=11.8 mL 46.6-34.8=0.0118 L | 1) Number of moles of NaOH used=0.0118 L of NaOH0.1 M of NaOH 1) Number of moles of NaOH used=0.00118 mol 2) The number of moles of NaOH used is equal to the number of moles of H+ions present in the solution‚ which is equal to the number of moles of the unknown salt. Moles of NaOH=Moles of H+=Moles of Unknown Salt=0.00118 mol
Premium Sodium chloride Chemistry
have greenish and 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
Premium Green PH indicator Water
| |HNO3 + NaOH |28.00 |28.25 |34.50 | |HNO3 + KOH |28.25 |28.25 |34.00 | |HCl + NaOH |28.25 |28.00 |34.75 | |HCl + KOH |28.25 |28.25 |34.00 | |H2SO4 + NaOH |28.00
Premium Sodium hydroxide Thermodynamics Acid
INTRODUCTION Differences between acids and bases An acid-base reaction is based on the reaction involving the ionization of water H2O -> H+ + OH- This means that water can break apart into a hydrogen ion and a hydroxide ion. These two ions can also join together to form a water molecule. When a strong acid is placed in water‚ it will ionize completely‚ and break down into its constituent ions in which one of it a hydrogen ion. When a strong base is placed in water‚ it will ionize
Free PH PH indicator Acid
REAGENT ADDED STRESS (ION ADDED) COLOR OBSERVATION DIRECTION OF EQUILIBRIUM SHIFT HCl (Step 3) HCl (Step 4) NaOH (Step 5) NaOH (Step 6) Table 2 Equilibrium Involving Thiocyanatoiron (III) Ion REAGENT ADDED STRESS (ION ADDED) OBERSERVATION DIRECRTION OF EQUILIBRIUM SHIFT KCl (test tube B) Fe(NO3)3 (test tube C) KSCN (test tube D) NaOH (test tube E) Table 3 Equilibrium Involving Cobalt (II) Complexes STRESS OBSERVATION DIRECTION OF EQUILIBRIUM
Premium Chemistry Experiment Hypothesis
Student: Krystianna Platz Date: November 5‚ 2014 Section: 113 R- 8AM Experiment 8 Determination of Percent Composition of Pennies Using Redox and Double Displacement (Precipitation) Reactions Objectives: The lab experiment will consist of oxidation-reduction and double displacement reactions as well as titration techniques. All these components will be used in order to determine the percent composition of pennies. In
Premium Chemistry Chemical reaction Concentration
[pic] Drops |Water(HcL) |Water(NaOH) |Liver(HcL) |Liver(NaOH) |Egg White(HcL) |Egg White(NaOH0) |Potato(HCl) |Potato(NaOH) |Buffer(HCl) |Buffer(NaOH) | |0 |7 |4 |7.4 |5 |8.2 |7 |6.9 |4 |10.7 |10 | |5 |4.5 |7 |6.9 |6 |7.5 |8 |6.2 |5 |10.5 |10 | |10 |2.7 |9 |6.3 |6 |7 |9 |5.7 |5 |10.4 |11 | |15 |2.6 |12 |5.8 |6 |6.4 |9 |5.3 |6 |10.3 |12 | |20 |2.5 |12 |5.4 |6 |4.5 |10 |4.9 |7 |10.2 |12 | |25 |2.4 |13 |5.1 |6 |3.5 |10 |4.6 |8 |10.1 |13 | |30 |2.3 |13 |4.8 |6 |3.3 |11 |4.2 |8 |10 |13 | | 1.
Free PH Acid dissociation constant
Lab S: Acids and Bases: Titrations | Determining Equivalence Point and Molarity Susheel Palakurthi 11661148 Partners: Justin Mwakule‚ Andrew Farrell July 29‚ 2015 CHEM 123 L02 TA: Amelia Fitzsimmons Experiment conducted: July 22‚ 2015 Introduction Acids and bases is an important aspect in chemistry. A specific example of such is in the body‚ acids and bases have to be balanced in order to provide the optimal pH in the system for normal physiological processes. Anything digested by the body has
Premium Acid Base Chemistry