Sample Chapter - Chapter 4
... Water separates ions in a process that greatly reduces the electrostatic force of attraction between them. To see how it does this, let’s examine the water molecule closely. Water’s power as an ionizing solvent results from two features of the water molecule: the distribution of its bonding electron ...
... Water separates ions in a process that greatly reduces the electrostatic force of attraction between them. To see how it does this, let’s examine the water molecule closely. Water’s power as an ionizing solvent results from two features of the water molecule: the distribution of its bonding electron ...
Definitions of terms relating to the structure and
... Examples of colloidal sols are protein sols, gold sols, emulsions, and surfactant solutions above their critical micelle concentrations. ...
... Examples of colloidal sols are protein sols, gold sols, emulsions, and surfactant solutions above their critical micelle concentrations. ...
Concept based notes Chemistry Lab Manual
... Q. 15. What is primary standard substance? Ans. A substance available in high degree of purity, stable & unaffected by air (does not gain & lose moisture in air). Q. 16. What is secondary standard substance? Ans. A substance available in low degree of purity, unstable & affected by air (gains & lose ...
... Q. 15. What is primary standard substance? Ans. A substance available in high degree of purity, stable & unaffected by air (does not gain & lose moisture in air). Q. 16. What is secondary standard substance? Ans. A substance available in low degree of purity, unstable & affected by air (gains & lose ...
QUESTION BANK CHEMISTRY-XII THE SOLID STATE CHAPTER
... 29. Define conductivity and molar conductivity for the solution of an electrolyte .Discuss their variation with concentration. 30. State Faraday’s second law of electrolysis? SHORT ANSWER TYPE QUESTION : (3MARKS EACH) 31. How much Cu is deposited on the cathode of an elect ...
... 29. Define conductivity and molar conductivity for the solution of an electrolyte .Discuss their variation with concentration. 30. State Faraday’s second law of electrolysis? SHORT ANSWER TYPE QUESTION : (3MARKS EACH) 31. How much Cu is deposited on the cathode of an elect ...
TDB-5: Standards and conventions for TDB publications
... The same chemical formula may refer to different chemical species and must often be specified more clearly in order to avoid ambiguities. For example, UF 4 occurs as a gas, a solid, and an aqueous complex. The distinction between the different phases is made by phase designators that immediately fol ...
... The same chemical formula may refer to different chemical species and must often be specified more clearly in order to avoid ambiguities. For example, UF 4 occurs as a gas, a solid, and an aqueous complex. The distinction between the different phases is made by phase designators that immediately fol ...
Chemical Reactions
... 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 200 atoms of sodium react with 200 molecules of water to produce 200 molecules of sodium hydroxide and 100 molecules of hydrogen. ...
... 2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g) The balanced chemical equation can be interpreted in a variety of ways. It could say that 200 atoms of sodium react with 200 molecules of water to produce 200 molecules of sodium hydroxide and 100 molecules of hydrogen. ...
The Major Classes of Chemical Reactions
... Covalent Compounds in Water Water dissolves many covalent compounds also. Table sugar (sucrose, C12H22O11), beverage (grain) alcohol (ethanol, CH3CH2OH), and automobile antifreeze (ethylene glycol, HOCH2CH2OH) are some familiar examples. All contain their own polar O±H bonds, which interact with tho ...
... Covalent Compounds in Water Water dissolves many covalent compounds also. Table sugar (sucrose, C12H22O11), beverage (grain) alcohol (ethanol, CH3CH2OH), and automobile antifreeze (ethylene glycol, HOCH2CH2OH) are some familiar examples. All contain their own polar O±H bonds, which interact with tho ...
Acid + Base Class # 1
... the pH (hydrogen ion concentration) of the solution. Acid Base Indicators are (mostly weak acids) that work by dynamic equilibrium. The molecules are one color, the ions another. The “stress” of adding more acid, or more base, causes a LeChatlier shift forward or reverse. ...
... the pH (hydrogen ion concentration) of the solution. Acid Base Indicators are (mostly weak acids) that work by dynamic equilibrium. The molecules are one color, the ions another. The “stress” of adding more acid, or more base, causes a LeChatlier shift forward or reverse. ...
View Full Text
... Using the model, speciation predictions were made in two solutions: 0.6 m PZ (Figure 4) and 3.6 m K+/0.6 m PZ (Figure 5). In the absence of potassium carbonate, piperazine is much more susceptible to protonation from the addition of CO2 to the system. Model predictions show that protonated piperazin ...
... Using the model, speciation predictions were made in two solutions: 0.6 m PZ (Figure 4) and 3.6 m K+/0.6 m PZ (Figure 5). In the absence of potassium carbonate, piperazine is much more susceptible to protonation from the addition of CO2 to the system. Model predictions show that protonated piperazin ...
precipitate - UniMAP Portal
... precipitation of calcium oxalate and ignition of the oxalate ion to calcium oxide. ...
... precipitation of calcium oxalate and ignition of the oxalate ion to calcium oxide. ...
Precision, accuracy and significant figures
... remaining gases, when bubbled through a NaOH solution, increased its mass by 27.83 g. In a separate experiment, a 6.21 g sample of the compound was vaporised. The vapour occupied 2.17 L at 200°C and 1.25 × 105 Pa. Calculate the molecular formula of the compound. ...
... remaining gases, when bubbled through a NaOH solution, increased its mass by 27.83 g. In a separate experiment, a 6.21 g sample of the compound was vaporised. The vapour occupied 2.17 L at 200°C and 1.25 × 105 Pa. Calculate the molecular formula of the compound. ...
Liquid–liquid extraction
Liquid–liquid extraction (LLE) consists in transferring one (or more) solute(s) contained in a feed solution to another immiscible liquid (solvent). The solvent that is enriched in solute(s) is called extract. The feed solution that is depleted in solute(s) is called raffinate.Liquid–liquid extraction also known as solvent extraction and partitioning, is a method to separate compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent. It is an extraction of a substance from one liquid into another liquid phase. Liquid–liquid extraction is a basic technique in chemical laboratories, where it is performed using a variety of apparatus, from separatory funnels to countercurrent distribution equipment. This type of process is commonly performed after a chemical reaction as part of the work-up.The term partitioning is commonly used to refer to the underlying chemical and physical processes involved in liquid–liquid extraction, but on another reading may be fully synonymous with it. The term solvent extraction can also refer to the separation of a substance from a mixture by preferentially dissolving that substance in a suitable solvent. In that case, a soluble compound is separated from an insoluble compound or a complex matrix.Solvent extraction is used in nuclear reprocessing, ore processing, the production of fine organic compounds, the processing of perfumes, the production of vegetable oils and biodiesel, and other industries.Liquid–liquid extraction is possible in non-aqueous systems: In a system consisting of a molten metal in contact with molten salts, metals can be extracted from one phase to the other. This is related to a mercury electrode where a metal can be reduced, the metal will often then dissolve in the mercury to form an amalgam that modifies its electrochemistry greatly. For example, it is possible for sodium cations to be reduced at a mercury cathode to form sodium amalgam, while at an inert electrode (such as platinum) the sodium cations are not reduced. Instead, water is reduced to hydrogen. A detergent or fine solid can be used to stabilize an emulsion, or third phase.