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What are 5 examples of lithium?
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. Sodium, for example, reduces elemental chlorine to chloride anion (sodium is oxidized to its cation), as do the other metals under varying conditions. In a similar fashion these same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to halide anion, and the carbon bonds to the metal (the carbon has carbanionic character). Halide reactivity increases in the order: Cl < Br < I. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination). The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them. The other metals mentioned above react in a similar manner, but the two shown here are the most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.R3C-X + 2Li --> R3C-Li + LiX An Alkyl Lithium ReagentR3C-X + Mg --> R3C-MgX A Grignard ReagentThe metals referred to here are insoluble in most organic solvents, hence these reactions are clearly heterogeneous, i.e. take place on the metal surface. The conditions necessary to achieve a successful reaction are critical.First, the metal must be clean and finely divided so as to provide the largest possible surface area for reaction.Second, a suitable solvent must be used. For alkyl lithium formation pentane, hexane or ethyl ether may be used; but ethyl ether or THF are essential for Grignard reagent formation.Third, since these organometallic compounds are very reactive, contaminants such as water, alcohols and oxygen must be avoided.These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents unique and useful reactants in synthesis.
What is heat of reaction and enthalpy of reaction?
Heat of reaction and enthalpy of reaction are the same thing. Enthalpy, or the heat transfer, cannot be measured, however we can measure the CHANGE of enthalpy which is shown by a value of ∆H. This measured in kilojoules per mole of reactant. (KJ/mol)This value may be positive or negative. For endothermic reactions (which absorb heat), the ∆H value is always positive. For exothermic, where heat is released, the value is negative.
What signs or indicators of chemical changes are shown in these reactions?
Chemical reactions can be obvious. First of all, listen: if you hear any sounds, such as a fizzing sound, that may be one indicator. Another is a change in color, or even state. If a solution is aqueous, and it precipitates (becomes a solid), that's also another indicator of a chemical reaction.
How are titrations carried out in industries and how are they different to titrations in schools and colleges?
The process, operation, or method of determining the concentration of a substance in solution by adding to it a standard reagent of known concentration in carefully measured amounts until a reaction of definite and known proportion is completed, as shown by a color change or by electrical measurement, and then calculating the unknown concentration.
Conversions of stoichiometry?
Stoichiometry Conversions Using Balanced EquationsMol A --> Mol B (Mol -->Mol)("How many moles of B are needed to react with X mol A?") __A + __B --> __AB1. Multiply the number of moles (A) by the number of moles (# molecules)1 for B.2. Divide the result by the number of molecules for A.- FormulaMol A * # molecules B/# molecules A- Conversion factorMol A * # molecules (mol) B = Mol B----- # molecules (mol) A*****1 the number of moles in this case refers to however many molecules of each substance is within the balanced equation (# molecules). For clarity sake I put # molecules in place of "moles" where appropriate; however, on practice problems or demonstrations shown in textbooks, it's likely it will say moles instead of # molecules.*****Mol A --> Mass B (Mol --> Mass)("How many grams B are needed to produce X mol A?") __A + __B --> __AB1. Multiply mol A by # molecules B; divide by # molecules A. (Mol A --> Mol B)2. Multiply result from 1 by molar mass B (mol B --> Mass B).- FormulaMol A x # molecules B/# molecules A- Conversion FactorMol A x # molecules (mol) B x molar mass B = mass B--------- # molecules (mol) A ----- 1 mol BMass A --> Mol B (Mass --> Mol)("How many mol of B are needed to react with X g A?") __A + __B --> __AB1. Multiply Mass A by # molecules B.2. Divide by molar mass multiplied by # molecules A.- FormulaMass A x # molecules B/(Molar Mass A x # molecules A)- Conversion FactorMass A x 1 mol A x # molecules (mol) B = mol B--- molar mass A - # molecules (mol) AMass A --> Mass B (Mass --> Mol --> Mass)("How many grams of A can be produced from X grams B?") __A + __B --> __AB1. Convert from grams (g) to mol for substance A (mass A --> mol A).2. Divide mol A by mol B (mol A --> mol B [Molar Ratio of Substances]).3. Multiply mol B by molar mass B (mol --> mass)In summary, you are converting from grams A to mol A, then mol A to mol B, then mol B to grams B.- FormulaMass A x # molecules B x Molar Mass B/(Molar Mass A x # molecules A)- Conversion FactorMass (g) A * 1 mol A -- x -- # molecules (mol) B -- x -- molar mass B = mass B----- molar mass (g) A --x-- # molecules (mol) A --- x --- 1 mol BLimiting Reagent and Reagent in ExcessThe limiting reagent or limiting reactant is the substance that limits the reaction. ("What substance has the least amount produced from a reaction?"). The reagent in excess or reactant in excess is the product left over or reagent that is leftover from the reaction; in other words, the reagent that has the most product that did not react.To identify the limiting reagent or limiting reactant, identify which substance produced the least amount of product (which reactant yields the least amount of product).Steps (given masses of products):A. Identify amount of product created per reactant (reactant --> product yield).1. Balance the equation if it has not been done already.2. Convert the given masses of reactant (A, B, etc.) to mass product (C) (see "Mass A --> Mass B [Mass --> Mass]" above). (mass reagent --> mass product)B. Identify the limiting reagent. The limiting reagent (reactant) will be the reactant (A, B, etc.) that yields the least amount of C (product).C. Identify the reagent in excess. The reagent in excess (reactant) will be the reactant (A, B, etc.) that yields the most amount of C (product).D. Give how much reagent in excess remain unreacted. (How much reactant is leftover). For simplicity sakes, the limiting reagent will be A and the reagent in excess will be B. "For every X grams of the limiting reagent, there is Y grams of the reagent in excess".Mass Limiting Reactant --> Mass Reagent in Excess:First convert the mass of the limiting reagent to the mass of the reagent in excess (mass limiting reagent : mass reagent in excess [reacting] ratio); then subtract the mass of the limiting reagent from the mass of the reagent in excess (that reacted)1. Convert the mass of the limiting reagent to mass of the reagent in excess (ratio mass limiting reagent: mass reagent in excess). Refer to "Mass A --> Mass B" above.2. Subtract the original amount of B (reagent in excess) from the amount of B needed to react with A (limiting reagent).Reagent in Excess leftover = Starting Mass A - Reacting Mass B (step 1 answer)E. Find % yield.% yield = actual yield (given)------- theoretical yield (must be found)** the theoretical yield is the amount of product theoretically produced by the limiting reagent; the actual yield is the amount of product actually produced by the reactants; the theoretical yield will have been found in step A. The actual yield will be given within the worded problem.Ex. 4Na2CO3 + Fe3Br8 --> 8NaBr + 4CO2 + Fe3O4 A. How many grams of Fe3O4 can be produced from 100.0g Na2CO3 and 300.0g Fe3Br8?100.0g Na2CO3 x 1 mol Na2CO3 x 1 mol Fe3O4 x 231.6g Fe3O4------------------ 106.0g Na2CO3 -- 4 mol Na2CO3 -- 1 mol Fe3O4= 54.62g Fe3O42(2 this is known as the theoretical yield, which will be needed when calculating percentage yield later on).300.0g Fe3Br8 x 1 mol Na2CO3 x 1 molecules Fe3O4 x 231.6g Fe3O4----------------- 106.0g Fe3Br8 ---- 4 molecules Fe3Br8 -- 1 mol Fe3O4= 86.12g Fe3O4B. What is the limiting reagent?The limiting reagent in this case is Na2CO3 because it has the lowest theoretical yield for producing Fe3O4 (54.62g Fe3O4 vs 86.12g Fe3O4).C. What is the reagent in excess?The reagent in excess is Fe3Br8 because it has the highest theoretical yield for producing Fe3O4.D. How many grams of the reagent in excess(Fe3Br8)remain unreacted?100.0g Na2CO3 x 1 mol Na2CO3 x 1 molecules Fe3Br8 x 806.8g Fe3Br8------------------- 106.0g Na2CO3 -- 4 molecules Na2CO3 - 1 mol Fe3Br8= 190.3g Fe3Br8300.0g Fe3Br8 - 190.3g Fe3Br8 = 109.7g Fe3Br8 leftover (unreacted)E. If 42.75g of Fe3O4 were isolated, what is the % yield?% yield = 42.75g----------- 54.62g2 x 100% = 78.27%
What are 5 examples of lithium?
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. Sodium, for example, reduces elemental chlorine to chloride anion (sodium is oxidized to its cation), as do the other metals under varying conditions. In a similar fashion these same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to halide anion, and the carbon bonds to the metal (the carbon has carbanionic character). Halide reactivity increases in the order: Cl < Br < I. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination). The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them. The other metals mentioned above react in a similar manner, but the two shown here are the most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.R3C-X + 2Li --> R3C-Li + LiX An Alkyl Lithium ReagentR3C-X + Mg --> R3C-MgX A Grignard ReagentThe metals referred to here are insoluble in most organic solvents, hence these reactions are clearly heterogeneous, i.e. take place on the metal surface. The conditions necessary to achieve a successful reaction are critical.First, the metal must be clean and finely divided so as to provide the largest possible surface area for reaction.Second, a suitable solvent must be used. For alkyl lithium formation pentane, hexane or ethyl ether may be used; but ethyl ether or THF are essential for Grignard reagent formation.Third, since these organometallic compounds are very reactive, contaminants such as water, alcohols and oxygen must be avoided.These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents unique and useful reactants in synthesis.
Hydrogen and chlorine combine to produce hydrochloric acid as shown in the equation below H2 C12 - 2HC1 is this reaction exothermic or endothermic?
This reaction is exothermic because there is a net RELEASE energy.
What is heat of reaction and enthalpy of reaction?
Heat of reaction and enthalpy of reaction are the same thing. Enthalpy, or the heat transfer, cannot be measured, however we can measure the CHANGE of enthalpy which is shown by a value of ∆H. This measured in kilojoules per mole of reactant. (KJ/mol)This value may be positive or negative. For endothermic reactions (which absorb heat), the ∆H value is always positive. For exothermic, where heat is released, the value is negative.
How do you know if an enthalpy change diagram is endothermic or exothermic?
If you plot the reaction coordinate (what I think you mean by "enthalpy change diagram"), the reaction will be exothermic if the products are lower on the graph than the reactants. If they are higher than it is endothermic. For instance, if you go to the linked Wikipedia page (link to the left of this answer), the graph shown is of an exothermic reaction.
What chemical reactions shown from H2o O2 H2O2?
The simplified chemical reaction is:H2 + O2 = H2O2
What happens to redox reaction when it shown as half reactions?
The redox reaction is split into its oxidation part and its reduction part.
What happens to redox reaction when it is shown as half-reactions?
The redox reaction is split into its oxidation part and its reduction part.
A chemist synthesize tin iodide according to the chemical reactions shown below the chemist starts with 10.0g of tin and excess iodine after the reaction 36.8g of tin iodide are obtained what percent?
69.7
What signs or indicators of chemical changes are shown in these reactions?
Chemical reactions can be obvious. First of all, listen: if you hear any sounds, such as a fizzing sound, that may be one indicator. Another is a change in color, or even state. If a solution is aqueous, and it precipitates (becomes a solid), that's also another indicator of a chemical reaction.
What are the 5 characteristics of chemical change?
During a chemical change the structure of molecules (as chemical composition) is changed. Signs of chemical reactions may be: - Gas release - Formation of a precipitate - Change of color - Change of odor - Change of pH - Change of aspect - Change of viscosity - Change of the temperature - Visible formation of new compounds
How are titrations carried out in industries and how are they different to titrations in schools and colleges?
The process, operation, or method of determining the concentration of a substance in solution by adding to it a standard reagent of known concentration in carefully measured amounts until a reaction of definite and known proportion is completed, as shown by a color change or by electrical measurement, and then calculating the unknown concentration.
Conversions of stoichiometry?
Stoichiometry Conversions Using Balanced EquationsMol A --> Mol B (Mol -->Mol)("How many moles of B are needed to react with X mol A?") __A + __B --> __AB1. Multiply the number of moles (A) by the number of moles (# molecules)1 for B.2. Divide the result by the number of molecules for A.- FormulaMol A * # molecules B/# molecules A- Conversion factorMol A * # molecules (mol) B = Mol B----- # molecules (mol) A*****1 the number of moles in this case refers to however many molecules of each substance is within the balanced equation (# molecules). For clarity sake I put # molecules in place of "moles" where appropriate; however, on practice problems or demonstrations shown in textbooks, it's likely it will say moles instead of # molecules.*****Mol A --> Mass B (Mol --> Mass)("How many grams B are needed to produce X mol A?") __A + __B --> __AB1. Multiply mol A by # molecules B; divide by # molecules A. (Mol A --> Mol B)2. Multiply result from 1 by molar mass B (mol B --> Mass B).- FormulaMol A x # molecules B/# molecules A- Conversion FactorMol A x # molecules (mol) B x molar mass B = mass B--------- # molecules (mol) A ----- 1 mol BMass A --> Mol B (Mass --> Mol)("How many mol of B are needed to react with X g A?") __A + __B --> __AB1. Multiply Mass A by # molecules B.2. Divide by molar mass multiplied by # molecules A.- FormulaMass A x # molecules B/(Molar Mass A x # molecules A)- Conversion FactorMass A x 1 mol A x # molecules (mol) B = mol B--- molar mass A - # molecules (mol) AMass A --> Mass B (Mass --> Mol --> Mass)("How many grams of A can be produced from X grams B?") __A + __B --> __AB1. Convert from grams (g) to mol for substance A (mass A --> mol A).2. Divide mol A by mol B (mol A --> mol B [Molar Ratio of Substances]).3. Multiply mol B by molar mass B (mol --> mass)In summary, you are converting from grams A to mol A, then mol A to mol B, then mol B to grams B.- FormulaMass A x # molecules B x Molar Mass B/(Molar Mass A x # molecules A)- Conversion FactorMass (g) A * 1 mol A -- x -- # molecules (mol) B -- x -- molar mass B = mass B----- molar mass (g) A --x-- # molecules (mol) A --- x --- 1 mol BLimiting Reagent and Reagent in ExcessThe limiting reagent or limiting reactant is the substance that limits the reaction. ("What substance has the least amount produced from a reaction?"). The reagent in excess or reactant in excess is the product left over or reagent that is leftover from the reaction; in other words, the reagent that has the most product that did not react.To identify the limiting reagent or limiting reactant, identify which substance produced the least amount of product (which reactant yields the least amount of product).Steps (given masses of products):A. Identify amount of product created per reactant (reactant --> product yield).1. Balance the equation if it has not been done already.2. Convert the given masses of reactant (A, B, etc.) to mass product (C) (see "Mass A --> Mass B [Mass --> Mass]" above). (mass reagent --> mass product)B. Identify the limiting reagent. The limiting reagent (reactant) will be the reactant (A, B, etc.) that yields the least amount of C (product).C. Identify the reagent in excess. The reagent in excess (reactant) will be the reactant (A, B, etc.) that yields the most amount of C (product).D. Give how much reagent in excess remain unreacted. (How much reactant is leftover). For simplicity sakes, the limiting reagent will be A and the reagent in excess will be B. "For every X grams of the limiting reagent, there is Y grams of the reagent in excess".Mass Limiting Reactant --> Mass Reagent in Excess:First convert the mass of the limiting reagent to the mass of the reagent in excess (mass limiting reagent : mass reagent in excess [reacting] ratio); then subtract the mass of the limiting reagent from the mass of the reagent in excess (that reacted)1. Convert the mass of the limiting reagent to mass of the reagent in excess (ratio mass limiting reagent: mass reagent in excess). Refer to "Mass A --> Mass B" above.2. Subtract the original amount of B (reagent in excess) from the amount of B needed to react with A (limiting reagent).Reagent in Excess leftover = Starting Mass A - Reacting Mass B (step 1 answer)E. Find % yield.% yield = actual yield (given)------- theoretical yield (must be found)** the theoretical yield is the amount of product theoretically produced by the limiting reagent; the actual yield is the amount of product actually produced by the reactants; the theoretical yield will have been found in step A. The actual yield will be given within the worded problem.Ex. 4Na2CO3 + Fe3Br8 --> 8NaBr + 4CO2 + Fe3O4 A. How many grams of Fe3O4 can be produced from 100.0g Na2CO3 and 300.0g Fe3Br8?100.0g Na2CO3 x 1 mol Na2CO3 x 1 mol Fe3O4 x 231.6g Fe3O4------------------ 106.0g Na2CO3 -- 4 mol Na2CO3 -- 1 mol Fe3O4= 54.62g Fe3O42(2 this is known as the theoretical yield, which will be needed when calculating percentage yield later on).300.0g Fe3Br8 x 1 mol Na2CO3 x 1 molecules Fe3O4 x 231.6g Fe3O4----------------- 106.0g Fe3Br8 ---- 4 molecules Fe3Br8 -- 1 mol Fe3O4= 86.12g Fe3O4B. What is the limiting reagent?The limiting reagent in this case is Na2CO3 because it has the lowest theoretical yield for producing Fe3O4 (54.62g Fe3O4 vs 86.12g Fe3O4).C. What is the reagent in excess?The reagent in excess is Fe3Br8 because it has the highest theoretical yield for producing Fe3O4.D. How many grams of the reagent in excess(Fe3Br8)remain unreacted?100.0g Na2CO3 x 1 mol Na2CO3 x 1 molecules Fe3Br8 x 806.8g Fe3Br8------------------- 106.0g Na2CO3 -- 4 molecules Na2CO3 - 1 mol Fe3Br8= 190.3g Fe3Br8300.0g Fe3Br8 - 190.3g Fe3Br8 = 109.7g Fe3Br8 leftover (unreacted)E. If 42.75g of Fe3O4 were isolated, what is the % yield?% yield = 42.75g----------- 54.62g2 x 100% = 78.27%
