group will devise two syntheses of the compound‚ and compare them for cost effectiveness‚ safety and potential yield of the compound. So to be more specific about the tests that will be conducted on the compound; we will do the following: -Know the solubility of our compound‚ different types of compounds will dissolve in certain amounts
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Extraction and analysis of two compounds from unknown solution C ABSTRACT Liquid C and solid C were both extracted from unknown solution by first using chemically active liquid-liquid extract‚ followed by vacuum filtration. Liquid C and solid C were then purified with the use of simple distillation and recrystallization respectively. Through the process of recrystallization‚ the percentage purity of solid C was found to be 6.01%. The melting point range of purified solid C was 117.0 – 119
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8/14/2013 Complex Ion Equilibria • The dissociation of complex ions can be represented similarly to equilibria. Simultaneous Equilibria: Solubility Product Constants and Complex Ion Equilibria → Ag + + 2 NH Ag(NH3 ) +2 ← 3 [Ag ][NH ] = [Ag(NH ) ] 2 + Kd 3 + 3 2 → Cu 2+ + 4 NH Cu(NH 3 ) 24+ ← 3 [Cu ][NH ] = [Cu(NH ) ] 4 2+ Kd Complex Ion Equilibria 3 2+ 3 4 3 Complex Ion Equilibria • A metal ion coordinated to several neutral molecules or anions forms compounds called complex ions
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which contain replaceable hydrogen. This is because when the hydrogen is replaced they are acting like an acid. E.g. sodium hydrogen sulphate (NaHSO4). •The method chosen to prepare a salt depends on its solubility. •Solubility depends on the combination of positive and negative ions. SOLUBILITY RULES SOLUBLE INSOLUBLE 1 All nitrates All carbonates except for sodium‚ potassium and ammonium carbonates 2 All chlorides except for silver chloride and lead (ii) chloride All sulphides except for
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Experimental The chloride ion (Cl-) is an important anion found in solids and solutions. In this experiment‚ the amount of chloride ion in an unknown sample J (NaCl + KCl mixture) of water using the Mohr method is determined‚ which relies on the solubility differences of two anions and the titration endpoint of a precipitate. The net ionic reaction during the titration is as follows: Ag+ + Cl- → AgCl(s) The Ksp for AgCl is 1.8 x 10-10 and that for Ag2CrO4 is 1.2 x 10-12. Thus‚ as
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dyes (pigments) that when place on porous paper are dissolved in a solvent by capillary action. When the pigments (solutes) are dissolved in the solvent (water & alcohol mixture) they move through the paper at different rates depending on their solubility. The more soluble the pigments are the faster they will move through the paper. Paper chromatography is most commonly used to separate pigments‚ dyes and inks. In forensics‚ paper chromatography is used in crime scene investigation and in sequencing
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(making up the precipitate) should be considered. In gravimetric analysis‚ relatively few crystals are preferred over many small ones. Experimentally‚ it is found that particle size is affected by experimental variables such as precipitate solubility‚ reactant concentrations in the precipitating solution‚ the rate of addition and mixing of reactants‚ and the temperature. Mathematically‚ it is reflected in the Von- Weimarn equation‚ which states that R=Q-SS
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concepts of solubility and conductivity of a substance were studied as well in this experiment helping us understand how the various molecules and its ions react in solvents. An example of this process is when calcium chloride dissolved in water. The polar water molecules attracted the oppositely charged Ca2+ and Cl- ions as calcium chloride is a polar molecule as well. The ions brake apart as the water attraction is greater than the ionic bond within the calcium chloride molecule. Solubility is also
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Step 1: Pb (NO3) 2 (aq) + CaCl2 (aq) PbCl2 (aq) + Ca (NO3) 2 (aq) (double displacement reaction) According to the solubility guidelines lead (II) chloride (PbCl2) is a possible precipitate. This is because even though most chlorides are soluble‚ lead chloride is considered insoluble (p.2‚ Lesson 17). Step 2: PbCl2 ↔ Pb2+ + 2Cl- Q = [Pb2+] [Cl-] 2 Step 3: Ksp = 1.2 x 10-5 (from table 17.1‚ p.5) Step 4: V2 = 20.0 mL (volume of Pb (NO3)2) + 45.0 mL (volume of CaCl2) = 65.0
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I. DISCUSSION: Differences between organic and inorganic compounds based on structure‚ type of bond and some common physical and chemical properties. II. PROCEDURE: A. SOLUBILITY 1. To 2 mL. each of the following solvents: water‚ ethyl alcohol and ether‚ add a pinch of benzoic acid and shake.
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