called ions. Those ionic substances are broken down by electrolysis. Electrolysis is the breakdown of a substance by electricity‚ and it only happens in liquids. The liquids that can be electrolyzed are called electrolytes. When said electrolyzed‚ means that the compounds in solutions are broken down when they conduct electricity. Its necessary the use of electrodes since it conducts the electricity to the solution. The electrodes are needed to be an anode‚ which receives the negative ions‚ and the
Premium Potassium Sulfuric acid Sodium chloride
The first step in both mechanisms is the protonation of the alcohol to form an oxonium ion‚ converting the OH group into a good leaving group. What happens next depends on the nature of the alkyl group‚ R. If R is a group that readily forms a carbocation‚ then the slow‚ rate-determining step is the loss of a water molecule from the oxonium ion. Once formed‚ the carbocation then reacts rapidly with a halide ion to form the alkyl halide. SN1 Mechanism: The first step is protonation of the alcohol‚
Premium Alcohol Sodium chloride Functional groups
reactions below‚ (i) state and explain what would be observed; and (ii) write an appropriate equation. (a) adding excess sodium sulphite solution to iodine solution (b) adding aqueous chlorine to potassium bromide solution (c) adding excess potassium iodide solution to acidified potassium permanganate solution 1 (d) adding excess iron(II) sulphate solution to acidified potassium dichromate solution (e) adding concentrated nitric acid to magnesium ribbons (f) (g) (h) (i) (j) (k) (l) (m) (n) 5. adding
Premium Oxygen Hydrogen Acetic acid
HKDSE CHEMISTRY – A Modern View (Chemistry) Experiment Workbook 5 Suggested answers Chapter 52 Importance of industrial processes Chapter 53 Rate equation Experiment 53.1 Determining the rate equation of a reaction using method of initial rate (A microscale experiment) 1 Chapter 54 Activation energy Experiment 54.1 Determining the activation energy of a chemical reaction 3 Chapter 55 Catalysis and industrial processes Experiment 55.1 Investigating the action of a catalyst 6 Experiment
Premium Chemical reaction Chemistry Reaction rate
the reducing solution is potassium iodate solution and the oxidizing solution is sodium thiosulphate solution. Potassium iodate solution which is an oxidizing agent is added into an excess solution of acidified potassium iodide. This reaction will release iodine. Potassium iodide is acidified with sulphuric acid and the iodine released quickly titrated with sodium thiosulphate until it become light yellow. The iodine then detected with starch solution and it turn into dark blue solution and titrated
Premium Iodine Titration
Solubility Rules Chart Negative Ions (Anions) + Positive Ions (Cations) = Solubility of Compounds in water Example any anion + Alkali Ions (Li+‚ Na+‚ K+‚ Rb+‚ Cs+‚ Fr+) = soluble Sodium fluoride‚ NaF‚ is soluble any anion + hydrogen ion [H+ (aq)] = soluble hydrogen chloride‚ HCl‚ is soluble any anion + ammonium ion (NH4) = soluble ammonium chloride‚ NH4Cl‚ is soluble nitrate NO3- + any cation = soluble potassium nitrate‚ KNO3‚ is soluble acetate (CH3COO-) + any cation (except
Premium Ion Ammonia Potassium
Top of Form Bottom of Form Chemical Compound Formulas Chemistry is the study of the composition of matter and its transformation. A substance can be considered matter with definite properties that establishes its identity. The tremendous number of chemical compounds has been categorized into numerous categories. A broad classification distinguishes between inorganic and organic compounds. Organic compounds are carbon based. Inorganic compounds exclude compounds exclude compounds based on carbon
Premium Ammonia Chlorine Sulfuric acid
EXPERIMENT TWENTY Qualitative Analysis of Anions Analysis of Solutions Containing the Ions Cl-‚ Br-‚ l-‚ SO42-‚ CO32-‚ and NO3- This experiment continues the qualitative analysis begun in Experiment 19. Here we will be analyzing solutions to determine the presence of anions. The same techniques that were used for the cation analysis must be used for the anions. If you have not carried out Experiment 19‚ read the introductory section before starting this experiment. The major difference
Premium Ion Ammonia Sodium chloride
oxidize potassium iodide as follow: Equation: K S 0 (aq) + 2KI (aq)->2K S0 (aq)+I (aq) 2 2 8 2 4 2 KI(aq) +I (aq) -> KI (aq) 2 3 _________________________________________________ K S 0 (aq) +3KI(aq) -> 2K S0 (aq)+ KI (aq) 2 2 8 2 4 2 The rate law of this reaction can be represented as follow: Rate=k[S208 2-]^a [I-]^b When the concentration of peroxydisulphate ions is fixed‚ the order of reaction with respect to iodide ion is formed. Alos‚ when the concentration of iodide ion is fixed‚ the
Premium Chemistry Chemical reaction
I2. The brown iodine solution can be reduced by vitamin C (ascorbic acid) to form colourless iodide ions. However‚ I2 solution is not normally prepared directly by dissolving iodine in water because iodine is too volatile so it is almost impossible to avoid loss while the solution is being prepared. Therefore iodine is prepared in situ by mixing pure potassium iodate (KIO3 ) and potassium iodide (KI) in acidified medium. IO3- + 5I- + 6H+ → 3I2 + 3H2O The excess iodine
Premium Iodine