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Gaseous fuel – Liquified Petrolum Gas (LPG) 

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Gaseous fuel – Liquified Petrolum Gas (LPG) 

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Gaseous fuel – Compressed Natural gas (CNG)

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Gaseous fuel – Compressed Natural gas (CNG)

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CNG 101: An Introduction to Compressed Natural Gas

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Compressed natural gas (CNG) is a readily available alternative to gasoline. It is domestically produced, clean burning and affordable. This video is your introduction to the history, benefits and many uses of compressed natural gas. To learn more visit CNGnow.com

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The Technology of Compressed Natural Gas (CNG)

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The Technology of Compressed Natural Gas (CNG)

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The Technology of Compressed Natural Gas (CNG)

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Natural gas has enormous potential as a versatile energy source. While it's had a history of powering electric generators and heating stove tops, natural gas is growing in use as an efficient fuel that also powers cars and trucks. But what exactly is natural gas?

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Research Octane Number and Motor Octane Number

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Dow Chemical Builds New Ethylene Production Plant

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ENGS102P Explanatory Video on Octane Number.

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Octane Ratings Explained - EricTheCarGuy

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I know it has been a while but better late than never. I recently had a conversation with someone who wanted to know why octane ratings were important and what they meant, I thought this would make a perfect video for ETCG and here we are.

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Motorcraft Cetane Boost Video 6.0 & 6.4 Power Stroke EGR Related

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What is cetane and how does it impact combustion?

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As the piston rises in a diesel engine, air is compressed to very high pressures, which also generates temperatures capable of igniting diesel fuel. Therefore, combustion timing is controlled by when the diesel fuel is injected into the combustion chamber. In reality, the combustion does not start instantaneously.

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Crude Oil Cracking Reforming and Isomerisation

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Crude Oil Cracking Reforming and Isomerisation

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Crude Oil Cracking Reforming and Isomerisation

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Comparision between catalytic cracking, thermal cracking and steam cracking
  

      
    
  

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Comparision between catalytic cracking, thermal cracking and steam cracking
  

      
    
  

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C.2.2 - Compare catalytic cracking, thermal cracking and steam cracking

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fccimp.mov   (FCC=FLUID CATALYTIC CRACKING)

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CRACKING CATALITICO A LETTO FLUIDO (FCC=FLUID CATALYTIC CRACKING) Il cracking catalitico permette di ottenere benzine con alto potere indetonante (N.O.) da gasoli pesanti e residui della distillazione atmosferica, spesso si ricorre anche a cariche ottenute dalla distillazione Vacuum. Si usano catalizzatori a base di silicoalluminati cristallini a carattere acido (zeoliti) sotto forma di microsfere con dimensioni inferiori al millimetro (20-100 mesh).

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Moving Bed (or) Fluid Catalytic Cracking Unit

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Moving Bed (or) Fluid Catalytic Cracking Unit

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Fluid catalytic cracking (FCC) is one of the most important conversion processes used in petroleum refineries. It is widely used to convert the high-boiling, high-molecular weight hydrocarbon fractions of petroleum crude oils to more valuable gasoline, olefinic gases, and other products.

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Cracking and Reforming in petroleum refining

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Cracking and Reforming in petroleum refining

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cracking and reforming in petroleum refining

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Crude Oil Fractions & Their Uses | Organic Chemistry | Chemistry | FuseSchool

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Like our Facebook: https://www.facebook.com/oilvips - Geologists and geophysicists have agreed on the existence of a "prospect", a potential field. In order to find out if hydrocarbons are indeed trapped in the reservoir rock, we must drill to hit them.

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http://www.cteskills.com Crude oil is processed into more useful petroleum products such as gasoline, diesel fuel, and jet fuel. In this video, learn the theory and process behind how crude oil distilation works.

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The Role of Synthetic Fuel in Nazi Germany During World War II

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The Role of Synthetic Fuel in Nazi Germany During World War II

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"The Role of Synthetic Fuel in Nazi Germany During World War II"

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An introduction to the German synthetic fuel requirements during the second world war.

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The Role of Synthetic Fuel in Nazi Germany During World War II

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The Role of Synthetic Fuel in Nazi Germany During World War II

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Fischer Tropsch synthesis - an introduction

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Fischer Tropsch synthesis - an introduction

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Fischer tropsch synthesis - an introduction

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Fischer Tropsch synthesis is crucial in the preparation of fuels and chemicals from gas, coal or biomass. This presentation is a short introduction to Fischer Tropsch synthesis. This synthesis is crucial in for example to preparation of plastics from biomass.

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Animation showing the formation of hydrocarbon chains on surfaces of small ruthenium particles starting from carbonmonoxide and hydrogen gas. The carbon atoms are shown in grey, oxygen atoms in red, hydrogen atoms in white, and ruthenium atoms in green. For more information see: http://dx.doi.org/10.1039/C1CY00118C http://dx.doi.org/10.1039/C1CC11499A Animation: ICMS Animation Studio Voice-over: Matthew F.

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Hydrogenation of Coal (or) Synthesis of gasoline from coal

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Hydrogenation of Coal (or) Synthesis of gasoline from coal

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This video is made available as part of the biofuels education projects funded by the National Science Foundation and the U.S. Department of Agriculture. Converting coal to a liquid fuel, coal liquefaction, allows coal to be utilized as an alternative to oil.

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Analysis of Coal: Proximate and Ultimate analysis

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Analysis of Coal: Proximate and Ultimate analysis

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Uploaded by Shashikant Dhuppe on 2015-03-22.

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Solid Fuel: Coal - The Classification of Coal

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Solid Fuel: Coal - The Classification of Coal

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Defining and Explaining Coal, Analysis of Coal

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COAL -- WHAT IS COAL? Coal is a rich C-containing matter formed due to alternation of plant and trees under high temperature and pressure condition. Following is the list of features the good quality coal have: 1) Moisture content should be low. 2) Ash content should be low.

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Comparision of Solid, Liquid and Gaseous fuel

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Comparision of Solid, Liquid and Gaseous fuel

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MODULE-V  FUELS AND COMBUSTION: Fuels – Classification – Calorific value (Problems using Dulong’s formula) – Coal – proximate and ultimate analysis – Hydrogenation of coal – Petroleum –  processing and fractions – cracking – catalytic cracking and methods – knocking – octane number and cetane number – synthetic petrol – Fischer-Tropsch and Bergius processes – Gaseous fuels – Natural gas, CNG and LPG – Flue gas analysis – Orsat apparatus – Combustion – theoretical calculation of air (Problems on calculation of weight of air only).
Module Overview: This module deals with most common solid fuels, liquid fuels and gaseous fuels. The combustion part explains flue gas analysis.

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Question Bank for Unit -V Fuels and Combustion

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Question Bank for Unit -V Fuels and Combustion

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PART–A (2 Mark Question including Problems)

1. Calculate the net calorific value of a coal sample containing 7% elemental hydrogen,given that its GCV is 8828 kcal/kg.

2. Calculate the gross and net calorific value of coal having following composition: C =84%, S = 1.5%, N = 0.6%, H = 5.5% and O = 8.4%.

3. Calculate the net calorific value, if the gross calorific value is 2550 cal/g and contains 6.5% H.

4. What is flue gas?

5. Define calorific value with its types.

6. Write the characteristics of a good fuel.

7. Distinguish between coal and coke.

8. Why is coke used in metallurgical process than coal? (OR) what are the requisites of good coke for metallurgy?

9. Mention the two advantages of using alcohol-gasoline blends in I.C. engine.

10. Why a good fuel must have low ash content?

11. Name the reagent used for absorbing CO₂, CO and O₂ during flue gas analysis by Orsat apparatus.

12. Define cetane number. How can it be improved in diesel?

13. What are the advantages of catalytic cracking over thermal cracking of gasoline?

14. Define GCV and LCV of a fuel. How they are related?

15. Define octane number. How can it be improved in petrol?. (OR) What is meant by octane number of gasoline? How can it be improved? (OR) Define octane number. What structural features of hydrocarbons in unleaded petrol make its octane rating on par with leaded petrol?

16. Select the compound which possesses highest octane number and highest cetane number out of n-hepatane, n-hexadecane, n-octane and iso-octane. Explain why?

17. Why KOH is preferred to NaOH in the flue gas analysis using Orsat’s apparatus for CO₂?

18. What is the drawback of presence of sulphur in the coal?

19. Define the terms: (a) Cracking. and (b) Knocking with chemical structure..

20. What is the cause of knocking in internal combustion engine? Name a commonly used antiknocking compound.

21. Distinguish between proximate analysis and ultimate analysis of coal.

22. In the catalytic process, catalyst requires regeneration. Give reason.

23. What are the advantages of compressed natural gas?

24. Give the composition and uses of water gas (or) blue gas.

25. Define calorific value. Mention its units.

26. What are the uses of LPG?

PART–B (8 Mark Question)

1. What is flue gas analysis? How is it carried out by Orsat’s process? (OR) With a neat diagram explain the analysis of flue gas by Orsat apparatus and mention the precautions to be followed during the analysis.

2. Describe the manufacture of gasoline by Fischer-Tropsch method?

3. Describe the manufacturing of metallurgical coke by Otto Hoffman’s method. List out by-product.

4. What is synthetic petrol? How is it manufactured by Bergius process?

5. What are LPG and CNG? Discuss the advantages of LPG over gaseous fuel and CNG over LPG.

PART–B (8 Mark Question including Problems)

1. By using Dulong’s formula, calculate the approximate calorific value of a sample of coal in cal g⁻¹. Express the value in joules also. The ultimate analysis of this sample is C = 80%, H = 3.5%, S = 2.8%, N = 1.0%, O = 5.0% and ash = 7.7%.

2. Calculate the gross and net calorific value of coal having the following compositions: C = 85%, H = 8%, S = 1%, N = 2%, ash = 4%.

3. A coal sample has the following composition C = 81%, H₂ = 5%, S = 1%, O₂ = 8.5%, N₂ = 1.5% and ash = 3%. Find the gross and net calorific value of the sample. Given that the caloric values of C, H and S are 8060 Kcal/Kg, 34000 Kcal/Kg and 2200 Kcal/Kg respectively.

4. A coal sample has the following composition by weight: C = 90%, O = 3.0%, S = 0.5%, N = 0.5 and ash = 2.5%. Net calorific value of the coal was found to be 8490.5 kCal / kg. Calculate the percentage of hydrogen and higher calorific value of coal.

5. A sample of coal contains 87% carbon, 2% hydrogen, 1% oxygen, 1% sulphur and  9% ash. Calculate the theoretical weight and volume at NTP, of air required for complete combustion of 1 kg of the sample of coal.

6. Calculate the total air required for the complete combustion of a gaseous fuel, which gives the following composition:  H₂ = 48%, CH₄ = 35%, CO = 15% and N₂ = 2%.

7. A sample of coal was found to have the following percentage composition: C = 75%,H₂ = 5.2%, S = 1.2%, O₂ = 12.8%, N₂ = 3.7% and ash = 2.1%. Find the minimum amount of air required for complete combustion.

8. Calculate the volume of the air required for the complete combustion of 1 dm³ of the gaseous fuel having the following composition by volume: H₂ = 50%, CH₄ = 36%, N2 = 1.5%, CO = 6%, C₂H₄ = 4% & H₂O vapour =2.5 %.

9. A sample of coal was found to contain the following C = 81%, H = 4%, O = 2%, S = 2% and remaining being ash. Estimate the quantity of minimum air required for the complete composition of 3 kg of the sample.

10. Calculate the amount of minimum air required for complete combustion of 2 kg of coal sample with 80% carbon, 5% hydrogen, 1% oxygen, 2% nitrogen and 12% ash.

11. A gas used in an internal combustion engine had the following composition by volume: H₂ = 50%, CH₄ = 30%, CO = 15%, N₂ = 5%. Find the volume of air required for the combustion of 1 dm³ of the gas.

12. A coal sample on analysis gives C = 80%, S = 1%, H = 4.5%, O = 2% and rest ash. Find the theoretical amount of air required per 2 kg of the coal burnt.

13. A gaseous fuel containing H₂ = 10%, CH₄ = 30%, C₂H₂ = 15%, C₄H₁₀ = 25% and  C₂H₆ = 20% by volume was burnt in 10% excess of theoretical air by mass required for complete combustion. Find out the analysis of dry products of combustion.

14. A boiler fired with a coal with composition C = 75%, H = 9%, O = 4%, N = 3% and  ash = 7%. Calculate (a) Gross and net calorific value of 1 kg of coal (latent heat of steam = 587 kcal/kg); (b) minimum theoretical air required for composition of 1 kg of coal by weight and by volume and (c) percentage composition of dry flue gas, if 25% excess air is used?

15. A hydrocarbon fuel on buring give a flue gas of the following volume percentage composition. CO₂ = 9.39, N₂ = 83.23 and O₂ = 7.38. Determine the composition of fuel by weight and also the percentage of air used in excess?

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NanoMaterials: From Synthesis to Applications

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NanoMaterials: From Synthesis to Applications

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NanoMaterials: From Synthesis to Applications

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Speaker: Fernando Patolsky "Tel Aviv University Center for Nanoscience and Nanotechnology Annual workshop - 2013", Dates: 02-04/07/13

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CCVD method of Carbon Nanotube Synthesis

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Hello My name is Sasha Asheghi and this is my Engineering 1166 project for Doug Cooper's class at the University of Connecticut.

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Video from Nova s38e09 showing impressive production of long carbon nanotube strands.

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Nanoseries 1/5: What is a carbon nanotube?

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What are carbon nanotubes? First episode of the nanoseries. Videos sponsored from WomenInNano, nano2hybrids and Vega Science Trust. http://www.nano2hybrids.net http://vega.org.uk Also available in spanish and french

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Magnetic and Optical Properties of Nanomaterials

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Carbon nanotubes (CNTs) are allotropes of carbon with a cylindrical nanostructure. Nanotubes have been constructed with length-to-diameter ratio of up to 132,000,000:1, significantly larger than for any other material. These cylindrical carbon molecules have unusual properties, which are valuable for nanotechnology, electronics, optics and other fields of materials science and technology.

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Nanotechnology - Revolutionizing medicine and healthcare

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Nanotechnology - Revolutionizing medicine and healthcare

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Best Full Documentary Fidel Castro Biography

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Nanotechnology Documentary - Revolutionizing medicine and healthcare

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Nanotechnology Documentary. Nanotechnology and its applications are revolutionizing medicine, healthcare and the philosophy of healthcare in unprecendented ways!

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Nanotechnology for Targeted Cancer Therapy

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Nanotechnology for Targeted Cancer Therapy

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Nanotechnology for Targeted Cancer Therapy

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What is Nanomaterial?, Classification of Nanomaterials

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What is Nanomaterial?, Classification of Nanomaterials

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What is Nanomaterial?, Classification of Nanomaterials ....

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NANOMATERIAL -- WHAT IS NANOMATERIALS? The material with average particle size is less than 100 nanometer is called Nanomaterial. CLASSIFICATION OF NANOMATERIALS 1) Carbon based nanomaterials 2) Nanomaterial of metals and alloys 3) Nano-Polymers 4) Nano-glasses 5) Nano-ceramics 6) Nano composites 7) Biological Nanomaterial

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Nanomaterials Nanotechnology, The Technology of Future

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Nanomaterials Nanotechnology, The Technology of Future

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Structure of Molybdenumdisulphide with Graphite

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Structure of Molybdenumdisulphide with Graphite

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How to lubricate your bike chain with graphite powder for a maintenance free chain all winter

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One of the best in the country for Education, Training for ISEET, CET, NEET, NTSE, etc. Bags several ranks every year and their students get admission to prestigious engineering institutes such as IIT's, NIT's, etc. in large numbers. Run by professionals with a vision. Visit: www.ace-online.co.in

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Viscosity index is an arbitrary measure for the change of viscosity with variations in temperature. The lower the VI, the greater the change of viscosity of the oil with temperature and vice versa. It is used to characterize viscosity changes with relation to temperature in lubricating oil.

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Properties of Lubricants Viscosity Viscosity index Flash point Fire point Cloud point Pour point thankz to A.Angel Ezhilarasi Mam ( chemistry department) thankz to all my Friends...

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The Video describes with the help of demonstrations on how lubricants perform their role in engines.

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Unit-5  Engineering Materials (Mechanism of Lubrication,Liquid Lubricants,Properties) - Chemistry

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http://www.gurug.net Unit-5 Engineering Materials (Mechanism of Lubrication,Liquid Lubricants,Properties,Viscosity Index,Flash & Fire Points, Cloud ar pour points,Oilyness) - Chemistry

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Properties of Lubricats and Lubrication by Mohd. Hanif Dewan, Senior Engg. Lecturer, International Maritime Academy, Bangladesh.

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Refractories Manufacturing and its Properties

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Refractories Manufacturing and its Properties

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Refractories Manufacturing and its Properties

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Refractories Manufacturing and its Properties

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HOW IT'S MADE: Fire Bricks - Full Documentary (720p HD)

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A refractory material is one that retains its strength at high temperatures. ASTM C71 defines refractories as "non-metallic materials having those chemical and physical properties that make them applicable for structures, or as components of systems, that are exposed to environments above 1,000 °F (811 K; 538 °C)".

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My Dear Mech-B and Civil students Wish you all a very happy and Safe Dewali!

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Happy "Diwali" sir

1.what is the purpose of airbath in pensky mortenes apparatus?

2.difference between hydrocarbon lubricants and vegetable oil lubricant's?

3.which is more corrosive to metal,hardwater or softwater?

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Module Syllabus: Refractories – classification – acidic, basic and neutral refractories – properties (refractoriness, refractoriness under load, dimensional stability, porosity, thermal spalling) – manufacture of alumina, magnesite and zirconia bricks. Lubricants – mechanism of lubrication, liquid lubricants – properties (viscosity index, flash and fire points, cloud and pour points, oiliness) – solid lubricants – graphite and molybdenum disulphide – semi solid lubricants and emulsions. Nanomaterials – introduction to nano chemistry – carbon nanotubes and their applications.

Module overview: This module explains the basic concepts of various engineering materials and their industrial applications. Also the synthesis, properties and applications of carbon nanotubes are discussed in detail. After the completion of this module, the students are able to identify the appropriate engineering material.

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Question Bank for Unit -IV Engineering Materials

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Question Bank for Unit -IV Engineering Materials

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PART–A (2 Mark Question including Problems)

1. Explain why silica bricks expand on heating?

2. Under what conditions would you use lubricating oils and solid lubricants? [OR] Under what conditions solid lubricants are used?.,What is meant by thermal

3. spalling with respect to a refractory?. Where do you find thermal spalling?. How it is minimised?

4. Define oiliness of a lubricant. How can this be improved?

5. Mention any four applications of carbon nanotubes….. List any four nanometrials. Define refractoriness of a refractory..

6. An oil sample under test has a Saybolt universal viscosity same as that of standard Gulf oil (low viscosity standard) and Pennsylvanian oil (high viscosity standard) at 210oF. Their Saybolt universal viscosities at 100 oF are 61s, 758s and 420s respectively. Calculate the viscosity index of the oil sample.

7. Define the term (i) fire point and (ii) flash point with their significance..

8. What is meant by dimensional stability? Mention their types.

9. Why is graphite used as a lubricant whereas other form allotropes of carbon are not?

10. What are the types of lubrication? Mention some important characteristics of a lubricant.

11. What is viscosity index? What are viscosity index improvers? Give examples.

12. What are carbon nanotubes? How they are classified? [OR] How CNTs are classified on the basis of chirality or helicity?

13. What are greases? How are they classified? How lithium based greases are prepared?

14. What are refractory? How are they classified?

15. Explain the lubrication action of graphite..

16. What are single-walled carbon nanotubes and multi-walled carbon nanotubes? How they are differ from each other?

17. What should be the flash point of lubrication is involved in delicate machine like watches, clocks, etc.?

18. What is meant by pyrometric cone equivalent? How will you find it?

19. Define the terms ‘Lubricant’ and Lubrication’.

20. What is garnet? Give its uses.

21. Define (i) cloud point and (ii) pour point with their significance….

22. List any four properties of refractory.

23. How is carborundum prepared? Mention some important properties of carborundum. [OR] How will you prepare silicon carbide? What are the uses of silicon carbide?

24. What are nanometrials? Mention their characteristic properties?

25. How lubricants classified? Give examples for each type..

26. Give two examples for neutral and acidic refractory.

27. Under what conditions extreme pressure lubricant is used?

28. Define porosity of refractory.

29. An oil sample under test has a Saybolt universal viscosity same as that of Standard Gulf Oil (low viscosity standard) and Pennsylvanian oil (high viscosity standard) at 210 oF. Their Saybolt universal viscosity at 100 oF. Their Saybolt universal viscosities at 100 oF are 600 s, 800s and 500s respectively. Calculate the viscosity index.

PART–B (Big Questions)

1. How will you measure refractoriness of refractory? Explain with neat diagram.

2. Draw and explain the structure and lubrication action of (i) Graphite and (ii) Molybdenum disulphide. [OR] Give the structure of graphite and molybdenum disulphide. How do they function as solid lubrication? Explain with neat diagrams.

3. Explain the preparation, properties and uses of the following: (a) Zironia bricks (Neutral Refractory), (b) High Alumina bricks (Acidic Refractory) and (c) Magnesite bricks (basic Refractory).

4. What is meant by lubrication? Mention, draw and discuss the three types of mechanism of lubrication with examples. [OR] What are lubricants? Discuss the different types of lubrication.

5. Discuss the different types, their synthesis methods,properties and applications of carbon nanotubes (CNT).

6. Explain any five properties of refractory. [OR] Explain the following with respect to refractory: (a) Refractoriness under load, (b) Thermal spalling,(c) Dimensional stability and (d) porosity.

7. Draw a neat structure of graphite and molybdenum disulphide. Explain how they act as good solid lubricants

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ONLINE ASSIGNMENT:
O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

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1.carbonate hardness=30mg/litre

non carbonate hardness=32.9mg/l

2.carbonate hardness=12mg/l

non carbonate hardness=7mg/l

3.total hardness=150ppm of calcium carbonate equivalent

permanent hardness=80 ppm of calcium carbonate equivalent

temporary hardness=70 ppm of calcium carbonate equivalent

4.amount of hardness=300ppm of calcium carbonate equivalent
5.amount of hardness=500 ppm of calcium carbonate equivalent

# data[].attributes.body
1.Permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness = 30 ppm of CaCO3 equivalent

2.Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

3.Temporary hardness =70 ppm of CaCO3 equivalent

4.Amount of hardness = 300 ppm of CaCO3 equivalent

5.Amount of hardness = 500 ppm of CaCO3 equivalent

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Sir any easy way for mugupping emf and galvonic series

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1.Non-Carbonate Hardness =32.84 ppm of CaCO3 Equivalent Carbonate Hardness =30 ppm of CaCO3 Equivalent 2.Permanent Hardness =12 ppm of CaCO3 Equivalent Temporary Hardness =7 ppm of CaCO3 Equivalent 3.Temporary Hardness =70 ppm of CaCO3 Equivalent 4.Amount of Hardness =300 ppm of CaCO3 Equivalent 5.Amount of Hardness =500 ppm of CaCO3 Equivalent

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sir,please give the answers for the questions from corrosion and its control

Q.NO:4,9,21,35 and which is more corrosive to metal, hard water or soft water? why?

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SIR this is very helpful but for hostel students it is not helping,  so please got permission as soon as possible to visit this site at the college/hostel

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1]non carbonate hardness=32.4ppm of CaCO3equivalent

temporary hardness=30ppm of CaCO3equivalent

2]permanent hardness=12ppm of CaCO3equivalent

temporary hardness=7ppm of CaCO3equivalent

3]temporary hardness=70 ppm of CaCO3equivalent

4]amount of hardness=300 ppm of equivalent

5]amount of hardness=500 ppm of equivalent

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sir,the videos are really helpful to me and give the answers for the following questions
from corrosion and its control.

which is more corrosive to metal,hard water or soft water? why?

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1.Non-Carbonate Hardness  =32.84 ppm of CaCO3 Equivalent

Carbonate Hardness =30 ppm of CaCO3 Equivalent

2.Permanent Hardness  =12  ppm of CaCO3 Equivalent

Temporary Hardness  =7 ppm of CaCO3 Equivalent

3.Temporary Hardness  =70 ppm of CaCO3 Equivalent

4.Amount of Hardness  =300 ppm of CaCO3 Equivalent

5.Amount of Hardness  =500 ppm of CaCO3 Equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

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Sir put the offline question paper for aasignment

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Sir put the offline question paper for aasignment

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1.Non-Carbonate Hardness  =32.84 ppm of CaCO3 Equivalent

Carbonate Hardness =30 ppm of CaCO3 Equivalent

2.Permanent Hardness  =12  ppm of CaCO3 Equivalent

Temporary Hardness  =7 ppm of CaCO3 Equivalent

3.Temporary Hardness  =70 ppm of CaCO3 Equivalent

4.Amount of Hardness  =300 ppm of CaCO3 Equivalent

5.Amount of Hardness  =500 ppm of CaCO3 Equivalent

# data[].attributes.body
1.Non-Carbonate Hardness  =32.84 ppm of CaCO3 Equivalent

Carbonate Hardness =30 ppm of CaCO3 Equivalent

2.Permanent Hardness  =12  ppm of CaCO3 Equivalent

Temporary Hardness  =7 ppm of CaCO3 Equivalent

3.Temporary Hardness  =70 ppm of CaCO3 Equivalent

4.Amount of Hardness  =300 ppm of CaCO3 Equivalent

5.Amount of Hardness  =500 ppm of CaCO3 Equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
ONLINE ASSIGNMENT:

O.N:1  permanent hardness = 32.84 ppm of CaCO3 equivalent

Temporary hardness  = 30 ppm of CaCO3 equivalent

O.N:2  Permanent hardness = 12 ppm of CaCO3 equivalent

Temporary hardness = 7 ppm of CaCO3 equivalent

O.N:3  Temporary hardness = 70 ppm of CaCO3 equivalent

O.N:4  amount of hardness = 300 ppm of CaCO3 equivalent

O.N:5  amount of hardness =  500 ppm of CaCO3 equivalent

# data[].attributes.body
1.carbonate hardness=30mg/litre

non carbonate hardness=32.9mg/l

2.carbonate hardness=12mg/l

non carbonate hardness=7mg/l

3.total hardness=150ppm of calcium carbonate equivalent

permanent hardness=80 ppm of calcium carbonate equivalent

temporary hardness=70 ppm of calcium carbonate equivalent

4.amount of hardness=300ppm of calcium carbonate equivalent
5.amount of hardness=500 ppm of calcium carbonate equivalent

# data[].attributes.body
1.carbonate hardness=30mg/litre

non carbonate hardness=32.9mg/l

2.carbonate hardness=12mg/l

non carbonate hardness=7mg/l

3.total hardness=150ppm of calcium carbonate equivalent

permanent hardness=80 ppm of calcium carbonate equivalent

temporary hardness=70 ppm of calcium carbonate equivalent

4.amount of hardness=300ppm of calcium carbonate equivalent

5.amount of hardness=500 ppm of calcium carbonate equivalent

# data[].attributes.body
1.carbonate hardness=30mg/litre
non carbonate hardness=32.9mg/l
2.carbonate hardness=12mg/lnon carbonate hardness=7mg/l
3.total hardness=150ppm of calcium carbonate equivalentpermanent
hardness=80 ppm of calcium carbonate equivalent
temporary hardness=70 ppm of calcium carbonate equivalent
4.amount of hardness=300ppm of calcium carbonate equivalent5.amount of hardness=500 ppm of calcium carbonate equivalent

# data[].attributes.body
sir where are you put the offline assingment questions

# data[].attributes.body
sir i saw the trick for reactivity series but i have a doubt on that is the series denotes galvanic series or emf series?

# data[].attributes.body
1.Carbonate hardness=30 ppm of Calcium carbonate equivalent

Non-Carbonate Hardness= 32.85 ppm of calcium carbonate equivalent

2.Total Hardness=19 ppm of calcium carbonate equivalent

Temporary Hardness=7 ppm of calcium carbonate equivalent

Permanent Hardness=12 ppm of calcium carbonate equivalent

3.Total amount of hardness=70 ppm of calcium carbonate equivalent

4.Total amount of hardness= 300 ppm of calcium carbonate equivalent

5.Total amount of hardness=500 ppm of calcium carbonate equivalent

# data[].attributes.body
1.Non-Carbonate Hardness=32.84 ppm of CaCO3 Equivalent

Carbonate Hardness=30 ppm of CaCO3 Equivalent.

2.Permanent Hardness=12  ppm of CaCO3 Equivalent.

Temporary Hardness=7 ppm of CaCO3  Equivalent.

3.Temporary HArdness=70 ppm of CaCO3 Equivalent.

4.Amount of hardness=300 ppm of CaCO3 Equivalent.

5.Amount of Hardness=500 ppm of MgSO4 Equivalent.

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Chrome Plating video: 'How to' guide for (real) chrome plating as a Business Opportunity

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Chrome Plating: www.gold-plating-kits.co.uk tutorial. brush plating chrome onto a coin (gold to show the contrast) this is real chrome not substitute (chrome plating) Available from Goldsolutions or visit our site, www.gold-plating-kits.co.uk With a Gold Solutions 'Chrome Plating kit' anyone can start up a small local electroplating business producing very high profits, on a full or part time basis.

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Electroplating - Part 1 | Electrochemistry | Chemistry

# data[].attributes.attachment_link.description
This video is created by www.onlinetuition.com.my. More videos and free notes are available at http://spmchemistry.onlinetuition.com.my/ Electroplating is a process to coat an object with a thin protective layer of metal. A very common use of electrolysis is to form a thin protective coating of a metal on the surface of another which is likely to corrode.