1. Hsc Chemistry 9 Crack
2. Hsc Chemistry 5.1 Serial Number

Contents.Industrial Chemistry – Contextual outlineIndustry uses chemical reactions to produce chemicals for use by society. This module develops the ideas that some chemicals have been produced to replace naturally occurring chemicals that are longer available or are not economically viable. The concepts of qualitative and quantitative equilibrium are further developed.Industrial chemical processes cover the full range of reactions but concentration on some case studies is sufficient to illustrate the range of reactions and the role of chemists and chemical engineers involved in these processes. This allows some insight into the qualitative and quantitative aspects of the chemicals industry and allows a consideration of the analytical processes and monitoring that are necessary for efficient production.This module increases students’ understanding of the history, application and uses of chemistry, and current issues, research and developments in chemistry. Industrial chemistry processes have enable scientists to develop replacements for natural products.

Discuss the issues associated with shrinking world resources with regard to one identified natural product that is not a fossil fuel, identifying the replacement materials used and/or current research in place to find a replacement for the named material.Rubber is an important resource. It is used for many applications, including tyres, belting, hoses, tubing, insulators, valves and footwear. This is because rubber is elastic, tough, impermeable, adhesive, easily mouldable and an electrical insulator.Natural rubber is made from the sap of rubber trees. It takes 6-7 years before a rubber tree can be suited for harvest of such sap. Natural rubber must be processed thoroughly so that its properties of flexibility and solidity are maintained in all conditions.This involves:. softening by mastication (passing rubber between rollers or rotating blades). grinding and dissolving in a suitable substance for compounding with other ingredients e.g.

Fillers and pigments, antioxidants, plasticisers. sheeting and extrusion into various shapes. vulcanisation (discovered in 1839 by US inventor Charles Goodyear) – heating of rubber with sulfur (and usually other accelerators and improvers) so that cross-links between polymers are formed to improve the properties of rubber. cross-links cause rubber to spring back into shape when stretchedUntil the 1940s, rubber trees were the primary source of rubber, especially trees in tropical areas such as Malaya and Burma. However, the conflict of World War II interrupted natural rubber supplies, and also caused an increase in demand (for military vehicle tyres). German and US scientists developed synthetic polymers to replace rubber.After WWII, the demand for rubber could not be met by natural rubber tree plantations, and thus synthetic rubbers dominated the market instead.

Approximately 80% of the world’s rubber production today is from synthetic polymers, the most common one being SBR (styrene-butadiene rubber).Natural rubber is a polymer of isoprene. The polymer is polyisoprene.The first synthetic rubber was neoprene, or poly(chloroprene). Today’s most common synthetic rubber is SBR, made from two monomers: butadiene (B) and styrene (S) in the pattern: BBBSBBBSBBBSBBBS It is used for its low cost and good properties.

Identify data, gather and process information to identify and discuss the issues associated with the increased need for a natural resource that is not a fossil fuel and evaluate the progress currently being made to solve the problems identified.The increasing population of the world is leading to greater demand on all natural resources, and as they are being depleted, synthetic alternatives much be found. One such resource is rubber.

The massive demand for rubber in WWII could not be met by rubber trees in tropical areas and therefore synthetic substitutes were required, such as poly(chloroprene), and more recently, styrene-butadiene-rubber (SBR). Today, synthetic rubber has become cheaper to produce than natural rubber.Even after WWII, traditional sources of natural rubber could not meet the demand and synthetic rubber began to dominate the market. SBR was particularly useful. Both styrene and butadiene are by-products of petroleum refining, and SBR is less likely to deteriorate than natural rubber. The efficiency of cost and quality of synthetic rubber meant that it quickly overtook natural rubber. SBR used mainly for tyres.In general, synthetic rubber has the following advantages over natural rubber:. Better aging and weathering.

Greater resistance to oil, solvents, oxygen, ozone and certain chemicals. Resilience over a wider temperature rangeHowever, synthetic rubber has a greater build-up of heat from flexing and not as much resistance to tearing when hot, compared to natural rubber.Synthetic rubbers also include specialty elastomers which have been developed for specific purposes. NBR rubbers have good oil resistance and are used in flexible couplings, hoses and washing machine parts.

Neoprene is useful at elevated temperatures and is used for heavy-duty applications such as wetsuits. Many industrial processes involve the manipulation of equilibrium reactions.

Explain the effect of changing the following factors on identified equilibrium reactions: pressure, volume, concentration, temperature. FactorChangeEffectPressureIncreaseFavours side of equation with less moles of gasDecreaseFavours side of equation with more moles of gasVolumeIncreaseFavours side of equation with more moles of gasDecreaseFavours side of equation with less moles of gasConcentrationIncreaseReaction using up the added species is favouredDecreaseReaction creating the removed species is favouredTemperatureIncreaseFavours endothermic processDecreaseFavours exothermic process4.

Interpret the equilibrium constant expression from the chemical equation of equilibrium reactions.aA + bB ⇌ cC + dD 5. Identify that temperature is the only factor that changes the value of the equilibrium constant (K) for a given equation.Temperature is the only factor that changes the value of the equilibrium constant K. Identify data, plan and perform a first-hand investigation to gather information and qualitatively analyse an equilibrium reaction.Two pipettes (10mL and 5mL) were used to transfer water between two identical measuring cylinders.First equilibrium was achieved with 40.8 mL in cylinder A and 56.0 mL in cylinder B.Second equilibrium, after 15 mL was added to cylinder A to disturb the “system”, was achieved with 46.5 mL of water in cylinder A and 64.2 mL in cylinder B.The equilibrium constant K remained around the same. K 1 = 1.40 and K 2 = 1.38.Some water was lost in the process. AdvantagesLimitationsDemonstrates quantitatively the progress of a system reaching equilibriumUses two measuring cylinders instead of one closed containerClearly separates reactants and products in different cylindersA loss of water through cohesion to pipettes or spillage disturbs the processAllows the analysis of the effects of adding or removing speciesCannot show the effect of temperature, volume or pressureAllows analysis of the equilibrium constantWorks with volume rather than concentration or partial pressures7. Choose equipment and perform a first-hand investigation to gather information and qualitatively analyse an equilibrium reaction.Fe 3+ + SCN – ⇌ FeSCN 2+ ΔH.

TranscriptEx 8.1, 1Expand the expression (1– 2x)5(1 – 2x)5We know that(a + b)n = nC0 an b0 + nC1 an – 1 b1 + nC2 an – 2 b2 +. + nCn – 1 a1 bn – 1 + nCn a0 bnHence(a + b)5 5!/0!( 5 − 0)! A5 × 1 + 5!/1!( 5 − 1)! A4 b1 + 5!/2!( 5 − 2)! A3 b2+ 5!/3!( 5 − 3)!

A2b3 + 5!/4!( 5 − 4)! A b4 + 5!/5!( 5 −5)! B5 × 1Ex 8.1, 1Expand the expression (1– 2x)5(1 – 2x)5We know that(a + b)n = nC0 an b0 + nC1 an – 1 b1 + nC2 an – 2 b2 +. + nCn – 1 a1 bn – 1 + nCn a0 bnHence(a + b)5 5!/0!( 5 − 0)! A5 × 1 + 5!/1!( 5 − 1)!

A4 b1 + 5!/2!( 5 − 2)! A3 b2+ 5!/3!( 5 − 3)!

Hsc Chemistry 9 Crack

A2b3 + 5!/4!( 5 − 4)! A b4 + 5!/5!( 5 −5)! B5 × 1= 5!/(0! × 5!) a5 + 5!/(1!

Hsc Chemistry 5.1 Serial Number

× 4!) a4 b + 5!/(2! 3!) a3 b2 + 5!/(3! 2!) a2b3 + 5!/(4! 1!) a b4 + 5!/(5!

0!) b5= 5!/5! A5 + (5 × 4!)/4! A4 b + (5 × 4 × 3!)/(2! 3!) a3 b2 + (5 × 4 × 3!)/(2 × 1 ×3!) a3b2 + (5 × 4 × 3!)/(3! ×1 ×3!) a2b3 + (5 × 4!)/4!

) b5= a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5Thus, (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5Putting a = 1 & b = (– 2x)(1 – 2x)5 = (1)5 + 5(1)4 (– 2x) + 10 (1)3 (– 2x)3 + 10 (1)2 (– 2x)3 + 5 (1) (– 2x)4 + (– 2x)5= 1 – 10x + 10(4x2) + 10 (– 8x3) + 5 (16x4) + (– 32x5)= 1 – 10x + 40x2 – 80x3 + 80x4 – 32x5Show More.