Category: Applied Chemistry Objective Questions

Applied Chemistry Multiple Choice Questions on “Mechanism of Lubrication”.

1. How many mechanisms are there for lubrication?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: There are three mechanisms are there for lubrication. They are thin film lubrication, thick film lubrication and extreme pressure lubrication. The thick pressure lubrication is also called as fluid film or hydro dynamic lubrication. The thin film lubrication is also called as boundary lubrication.

2. The thickness of the lubricant film in the hydro dynamic lubrication is about __________
a) 1000^oA
b) 5000^oA
c) 500^oA
d) 100^oA
Answer: a
Clarification: In hydro dynamic lubrication, the sliding surfaces are separated completely by applying a thin uniform film of the lubricants between the two surfaces. The thickness of the lubricant film in the hydro dynamic lubrication is about 1000^oA.

3. The liquid lubricants stick due to the _________ property.
a) Surface tension
b) Viscosity
c) Temperature coefficient
d) Mass of the liquid
Answer: b
Clarification: The liquid lubricants stick due to their physical property called viscosity or stickiness. The liquid lubricant will cover all the irregularities in the slidings and thus it prevents the formation of welded junctions.

4. The coefficient of friction of unlubricated surface ranges about _________
a) 0.1-1.5
b) 0.1-0.5
c) 0.1-1.0
d) 1.5-2.0
Answer: a
Clarification: The coefficient of the unlubricated surface is about 0.1 to 1.5. If the lubricant energy is higher, then the energy must be required in large amount to circulate and maintain the viscous lubricant film.

5. The effectiveness of the hydro dynamic lubrication does not depends upon the ________
a) Design of bearing
b) Load
c) Rate of rotation of shaft
d) Area of shaft
Answer: d
Clarification: The effectiveness of the hydro dynamic lubrication does not depends upon the area of the shaft. The effectiveness mainly depends upon the design of bearing, load and speed of the shaft rotating.

6. The continuous fluid film cannot be maintained between the rubbing surfaces under the condition such as high load in boundary lubrication.
a) True
b) False
Answer: a
Clarification: The continuous fluid film cannot be maintained between the rubbing surfaces under the condition such as high load, slow rate of rotation and very low viscosity oil, etc. These conditions are seen in boundary lubrication.

7. In boundary lubrication, the coefficient of friction ranges about ______
a) 0.05-0.1
b) 0.5-0.15
c) 0.05-0.15
d) 1-1.5
Answer: c
Clarification: In boundary lubrication, the thickness of the fluid film must be 1000^oA. The coefficient of friction is about 0.05 to 0.15. This thin film contains the one or two layers. Thus it is called as thin lubrication or boundary lubrication.

8. The wear and tear of metal surfaces are involved and _________ are formed in thin lubrication.
a) P-n junctions
b) Welded junctions
c) Layers
d) Thin layers
Answer: b
Clarification: The wear and tear of the metal surfaces is involved and the welded junctions are formed. This also results in friction. These all are formed due to the contact of the surfaces with other leads.

9. Generation of heat takes place in __________ lubrication.
a) Thin lubrication
b) Thick lubrication
c) Extreme pressure lubrication
d) Boundary lubrication
Answer: c
Clarification: Generation heat and friction are also takes place in extreme pressure lubrication. The thin lubrication is also called as thin lubrication. At high load and pressures, the contact between the surfaces produces the more heat.

10. The liquid lubricant may be _______
a) Boiled
b) Evaporated
c) Sublimated
d) Heated
Answer: b
Clarification: The liquid lubricant may be evaporated or decomposed. Due to this, the lubrication becomes ineffective, and the effectiveness can be achieved by adding the extreme pressure additives along with lubricant.

Applied Chemistry Multiple Choice Questions on “Characteristics of Good Fuel and its Advantages”.

1. What is the ignition temperature of a good fuel?
a) High
b) Low
c) Moderate
d) 100⁰C
Answer: c
Clarification: The ignition temperature of a good fuel must be moderate because it is really hard to produce more heat and It requires large amount energy to produce high temperature and to achieve too low temperatures is also a difficult task. Moderate temperature can be produced easily.

2. What is the calorific value of an ideal fuel?
a) 30.8k.cal
b) High
c) 20.2k.cal
d) Low
Answer: c
Clarification: An ideal fuel must contain high calorific value. A calorific value is the amount of heat produced by complete combustion of a good fuel. So, more the fuel which produces more heat on complete burning will be the ideal one.

3. The rate of combustion of a good fuel must be ________
a) Controllable
b) High
c) Low
d) Very high
Answer: a
Clarification: The rate of the combustion of a good fuel must be controllable because if it is high or very high we cannot calculate the time taken by the fuel to burn and if it is low also we cannot observe the time taken.

4. The products of combustion of a good fuel _________
a) CO₂
b) CO
c) Must not be harmful
d) Must be harmful
Answer: c
Clarification: The good fuel must not produce harmful products that affect the surroundings and nature. Both CO and CO₂ are harmful gases.

5. The moisture content of the good fuel must be _________
a) 50%
b) High in amount
c) Low in amount
d) 70%
Answer: c
Clarification: We cannot say exact moisture content of a good fuel but it must be low in amount because the fuel with high moisture content will not burn properly and results in less heat and less calorific value. For a good fuel calorific value is more.

6. Artificial fuels are mainly prepared from ________
a) petrol
b) peat
c) coal
d) natural gas
Answer: a
Clarification: Artificial fuels are produced by man. Peat, coal and natural gas are directly available from nature. Petrol is produced by a man using a fractional distillation process.

7. On combusting a solid fuel, the final product obtained in high amount is ________
a) oxygen
b) nitrogen
c) ash
d) ozone
Answer: c
Clarification: After combusting a fuel the product left in more amount is ash. The oxygen and ozone cannot be produced in most of the cases. Nitrogen may be produced but in small amount than that of ash.

8. The major advantage of a good fuel is _______
a) ease of transportation
b) production of more oxygen
c) production of more nitrogen
d) production of more vapor
Answer: a
Clarification: The good fuel can be transported easily. It is the main advantage of good fuel. It reduces the cost of transportation and makes it ideal.

9. Which of the following is a synthetic fuel in gas form?
a) Natural gas
b) Producer gas
c) LPG
d) Petroleum
Answer: b
Clarification: The producer gas is a synthetic fuel in gas form whereas, natural gas is a natural fuel in gas form. LPG and petroleum are liquid fuels.

10. A fuel is a combustible substance containing ______ as major constitute.
a) nitrogen
b) oxygen
c) carbon
d) hydrogen
Answer: c
Clarification: In every fuel the major constitute is carbon. The other components like oxygen, nitrogen, hydrogen and sulphur are also present but in very small amounts when compared to carbon.

Applied Chemistry Multiple Choice Questions on “Lime Washing Soda Method”.

1. The cost of the lime soda process is _________
a) Less
b) High
c) Very high
d) Moderate
Answer: a
Clarification: The cost of the lime soda process is less and the operational cost of the lime soda process is high.

2. ___________ can be used for the turbid water.
a) Soap titration process
b) Zeolite process
c) Ion exchange process
d) Lime soda process
Answer: d
Clarification: Lime soda process can be used for the turbid water. The lime soda process removes the hardness of the water steps wise by treating with different chemicals.

3. Hardness of the water in the lime soda process is reduced to __________
a) 10-15ppm
b) 15-30ppm
c) 30-35ppm
d) 35-40ppm
Answer: b
Clarification: Hardness of the water in the lime soda process is reduced to 15-30ppm. The lime soda process involves the conversion of the soluble complexes into the insoluble precipitates.

4. Total dissolved solids in the lime soda process is _________
a) Reduced
b) Not reduced
c) Remains same
d) Slightly reduced
Answer: a
Clarification: Total dissolved solids in the lime soda process is reduced. It really works well for eliminating every complex element that cause the hardness of the water.

5. For avoiding the reduction of the residual hardness, incomplete precipitation and slow reaction, The reaction is carried out at __________ temperature.
a) 30-40^oC
b) 40-50^oC
c) 50-60^oC
d) 60-70^oC
Answer: c
Clarification: For avoiding the reduction of the residual hardness, incomplete precipitation and slow reaction, The reaction is carried out at 50-60^oC temperature.

6. Lime soda process removes _________ from the water.
a) Minerals
b) Mineral acids
c) Mineral alkaline
d) Mineral complexes
Answer: b
Clarification: Lime soda process removes the mineral acids from the water. The cold soda lime, hot soda lime and continuous soda lime process are other processes to remove the hardness.

7. Batch process is only used for removing the hardness of the water on _________
a) Small scale
b) Large scale
c) Industrial purposes
d) Irrigation purposes
Answer: a
Clarification: Batch process is only used for removing the hardness of the small scale. The Continuous soda lime process is used for the industrial purposes.

8. The amount of coagulate if added in the hot lime soda process is __________
a) High
b) Low
c) Very high
d) Very low
Answer: d
Clarification: The amount of coagulate if added in the hot lime soda process is very low. The chemical reaction takes place faster and precipitation takes place faster.

9. The final hardness of the water after treating by the hot lime soda process is about __________
a) 10-15ppm
b) 15-20ppm
c) 20-25ppm
d) 25-30ppm
Answer: c
Clarification: The final hardness of the water after treating by the hot lime soda process is about the 20-25ppm. In very less time the water is treated in this method.

10. Due to heating at high temperature in the hot lime soda process reduces the __________
a) Time for treating
b) Energy
c) Corrosion of boiler
d) Cost
Answer: c
Clarification: Due to heating at high temperature in the hot lime soda process reduces the corrosion of boilers. The heating is done to remove the dissolved gases in the water.

11. Lime soda process does not involve steps.
a) True
b) False
Answer: b
Clarification: Lime soda process involves steps like filtration, settling of precipitates, coagulation and removal of sludge.

12. In hot lime soda process, the bacteria in water is reduced to the minimum due to the _________
a) High temperature
b) Faster reactions
c) Slow reactions
d) Low temperature
Answer: a
Clarification: In hot lime soda process the bacteria count in water is reduced to the minimum due to the high temperatures that are applied to the water during the process.

13. In batch process, when the tank gets filled then it indicates that ____________
a) To stir more
b) Softening is completed
c) To add the catalyst
d) To add the lime
Answer: b
Clarification: In batch process, when the tank gets filled then it indicates that the softening of the water is completed. Two tanks are used for construction in the batch process.

14. Reagent used in the lime soda process is _______
a) Regenerated
b) Not regenerated
c) Regenerated depending on the hardness
d) Regenerated some times
Answer: b
Clarification: Reagent used in the lime soda process is can not be regenerated. The high temperatures are applied in this process.

15. In lime soda process, the change in the hardness of the water needs ________
a) Change in dose of lime
b) Change in dose of soda
c) Change in dose of lime and soda
d) Change in dose either lime or soda
Answer: c
Clarification: In lime soda process, the change in the hardness of the water needs the change in the dose of the lime and soda.

Applied Chemistry Multiple Choice Questions on “Types of Polymers”.

1. The polymers are classified into _________ types based on the nature of polymerization.
a) 5
b) 2
c) 3
d) 4
Answer: b
Clarification: The polymers are classified into two types based upon their nature of polymerization. They are homopolymers and co polymers.

2. A straight chain polymer comes under the _________ type of the polymers.
a) Homo polymers
b) Co-polymers
c) Regular chain copolymers
d) Irregular straight chain copolymers
Answer: a
Clarification: The straight chain polymers come under the homo polymers. The regular straight chain copolymers and irregular straight chain copolymer comes under the straight chain copolymers.

3. The different monomers are arranged alternatively in __________
a) Alternate straight chain polymers
b) Regular straight chain copolymers
c) Straight chain copolymer
d) Cross linked polymers
Answer: a
Clarification: The different monomers are arranged alternatively in alternate straight chain polymers. The regular straight chain polymers the monomeric units arranged regularly.

4. The polymers whose backbone is made of same type of atoms called ___________
a) Homo polymers
b) Heterochain polymers
c) Isotactic polymers
d) Atactic polymers
Answer: a
Clarification: The polymers whose back bone is made of some type of atoms is called as homopolymers. The heteropolymers backbone is constructed by the two or more than two different types of atoms.

5. Tacticity of the polymers is the arrangement of the ___________ on carbon backbone.
a) Hydrogen atoms
b) Nitrogen atoms
c) Functional groups
d) Carbons
Answer: c
Clarification: The tacticity is defined as the arrangement of the function groups on the carbon back bone of the polymer is called tacticity of the polymer.

6. Based on tacticity, the polymers are divided into __________ types.
a) Two
b) Three
c) Four
d) Five
Answer: b
Clarification: Based on tacticity, the polymers are divided into three types. They are Isotactic polymers, Atactic polymers and syndiotactic polymer.

7. The functional group are arranged on the same side of the carbon back bone are said to be _________
a) Syndiotactic polymers
b) Atactic polymers
c) Isotactic polymers
d) Chain polymers
Answer: c
Clarification: The functional groups arranged on the same side of the carbon back bone then it is isotactic polymer, if they no regular arrangement then they are said to be Atactic polymers.

8. Which of the following is the example of the syndiotactic polymers?
a) Poly propylene
b) Poly vinyl chloride
c) Gutta percha
d) Poly lactic acids
Answer: c
Clarification: The gutta percha is the natural form of rubber. It is the example of the syndiotactic polymers in which the functional groups are alternately arranged. Poly propylene and poly lactic acids are the examples of the isotactic polymers.

9. Alternate straight chain polymers are ___________
a) Soft and flexible
b) Hard and flexible
c) Soft and brittle
d) Hard and brittle
Answer: a
Clarification: Alternate straight chain polymers are soft and brittle nature. All the straight chain polymers are soft and brittle.

10. Combination of the organic and inorganic polymers are called _________
a) Element organic polymers
b) Inorganic polymers
c) Fibres
d) Thermoplastics
Answer: a
Clarification: Combination of the organic and the inorganic polymers are called as the element of organic polymers. It comes under the synthetic polymers because it is produced by the synthesis.

Applied Chemistry Multiple Choice Questions on “Carbon Nano Tubes”.

1. Carbon nano tubes are the sheets of graphite about ________
a) 0.1nm
b) 0.2nm
c) 0.3nm
d) 0.4nm
Answer: d
Clarification: Carbon nano tubes are the sheets of graphite about 0.4nm in diameter. They are rolled up to make nm in diameter. They are very small in size.

2. Carbon nano tubes are also called as ________
a) Bucky tubes
b) Bulky tubes
c) Bulk tubes
d) Buck balls
Answer: a
Clarification: Carbon nano tubes are also called as bucky tubes because they are also made up of graphite and they are in the form of tubes instead of balls. So, they are also called as bucky tubes.

3. Carbon nano tubes are first observed in _________
a) 1992
b) 1991
c) 1990
d) 1993
Answer: b
Clarification: Carbon nano tubes are observed in an year 1991 in the soot of the graphite during the arc discharge. First production of CNT was in 1992 by arc discharge method.

4. In how many methods the CNT can be prepared?
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: The carbon nano tubes can be prepared in 4 types. They are: 1. arc discharge method, 2. laser ablation method, 3. plasma torch method, 4. chemical vapour deposition method.

5. The current used in the arc discharge method is about _________
a) 100mA
b) 200mA
c) 300mA
d) 400mA
Answer: a
Clarification: The current used in the arc discharge method is about 100mA. By arc discharge of graphite electrodes in presence of ionised gas to reach high temperatures.

6. The carbon contained in the negative electrode sublimates.
a) True
b) False
Answer: a
Clarification: The carbon contained in the negative electrode sublimates because of high discharge temperature. This method is widely used method.

7. The yield of the laser ablation is ______________
a) 30%
b) 70%
c) 80%
d) 90%
Answer: b
Clarification: The yield of the laser ablation is 70%. The reaction temperature determines the diameter of the single walled CNT.

8. Laser ablation method is developed by _______
a) Kelvin
b) Richard Smalley
c) Watson and crick
d) Neils bohr
Answer: b
Clarification: Laser ablation method is developed by Dr. Richard Smalley at rice university, USA. In this process a pulsed laser vaporises a graphite target.

9. A water cooled surface is used in the process to collect __________
a) Nano particles
b) Nano tubes
c) Nano spheres
d) Nano sheets
Answer: b
Clarification: A water cooled surface is used in the process to collect the nano tubes. While an inert gas is blend into chamber, the nano tubes settles on cooler surfaces.

10. To improve the composite of graphite _______ is used as a catalyst.
a) CO
b) NI
c) CO and NI
d) TIO
Answer: c
Clarification: To improve the composite of graphite CO and NI mixture is used as catalyst. They are used to synthesise the single walled CNT.

11. Plasma torch method is used in the year ________
a) 2005
b) 2004
c) 2002
d) 2001
Answer: a
Clarification: Plasma torch method is used in the year 2005 at national research council of Canada. It is used to make the single walled CNT.

12. Plasma torch method is similar to the __________
a) Laser ablation
b) Arc discharge
c) Chemical vapour decomposition
d) Electrolysis
Answer: b
Clarification: plasma torch method is similar to the arc discharge method. It is first started to make the single walled CNT in 2005.

13. Plasma torch method produces ______ of CNT per minute.
a) 1Gms
b) 2Gms
c) 3Gms
d) 4Gms
Answer: b
Clarification: Plasma torch method is to produce 2gms of CNT per minute. A metal crystal is used. In a thermal plasma torch method is a high frequency oscillating current in a coil.

14. Chemical vapour decomposition is developed in an year ________
a) 2001
b) 2002
c) 2006
d) 2007
Answer: d
Clarification: Chemical vapour decomposition is developed in the year 2007 at university of Cincinnati, USA. In this method the substrate is prepared with a layer of metal catalyst nano particles.

15. The substrate is heated to ________
a) 200C
b) 300C
c) 400C
d) 700C
Answer: d
Clarification: The substrate is heated to 700C and a mixture of nitrogen and carbon containing acetylene or ethylene or ethanol or methane was passed.

Applied Chemistry Multiple Choice Questions on “Types of Additives”.

1. The most common reason for the degradation of petroleum oil is their _________
a) Volatility
b) Oxidation
c) Reduction
d) Temperature coefficient
Answer: b
Clarification: Oxidation inhibitors are one of the types of the additives. The oxidation is the main reason for the petroleum oil degradation. At high temperatures, the hydro carbons of the petroleum oil undergo homolytic fission to generate free radicals.

2. The oxidation inhibitors terminate the hydro peroxide chain by combining with ___________
a) Hydrogen peroxide
b) Oxides of metals
c) Hydro carbons
d) Ozone
Answer: a
Clarification: After the homolytic fission of the petroleum oil to produce free radicals and they react with oxides to form the hydrogen peroxides. Some hydro peroxides decompose to form the aldehydes, ketones etc. If they are reacted with hydrogen peroxide then the chain will be terminated.

3. ___________ are surface active additives.
a) Oxide inhibitors
b) Rust inhibitors
c) Dispersants
d) Friction modifiers
Answer: b
Clarification: The rust inhibitors are the surface active additives. They mainly acts on the surface and makes the metal corrosion free. Mainly from the water corrosion. They form thin film on the surface. It is the oxide film that protects the metal.

4. Tiadiazole and triazole derivatives are used for __________
a) Aluminium alloy corrosion
b) Silver corrosion
c) Copper corrosion
d) Manganese corrosion
Answer: c
Clarification: The tiadiazole and triazole derivatives are used to protect the non ferrous metals from corrosion. They also protect the copper derivatives from corrosion. Organic phosphates and sodium and sodium and calcium sulphonates are used as the rust inhibitors in machinery.

5. To reduce the wear and tear of the gears ____________ are used.
a) Anti wear agents
b) Rust inhibitors
c) Oxide inhibitors
d) Dispersants
Answer: a
Clarification: To reduce thee wear and tear of the gears anti wear agents or the extreme pressure agents are used. In high pressure hydraulic components are zinc dialkyl dithiosulphates are used.

6. In steel and steel lubrication, the zinc dialkyl dithiosulphates forms __________
a) ZnSO₄
b) ZnSO₂
c) ZnSO₃
d) ZnS
Answer: d
Clarification: In steel and steel lubrication, the zinc dithiosulphates forms the zinc oxide, zinc sulphide and ferrous oxide and some other zinc organo phosphates. They prevent the wear and tear of steel.

7. In automotive applications __________ are used.
a) Friction modifiers
b) Dispersants
c) Anti wear agents
d) Oxide inhibitors
Answer: a
Clarification: The friction modifiers are used in the auto mobile applications. Mild extreme pressure agent in boundary lubrication conditions are, they prevent stick slip oscillations and control noise in the automatic transmissions.

8. Friction modifiers have 12-18 carbons.
a) False
b) True
Answer: b
Clarification: The friction modifiers have 12 to 18 carbons. The fatty alcohols and esters of fatty acids also act as friction modifiers. Glyceride of the rapeseed and lard oil is used as the friction modifiers in machineries.

9. The oil insoluble sludge can be reduced by using ____________
a) Pour point depressants
b) Friction modifiers
c) Oxide inhibitors
d) detergents
Answer: d
Clarification: The oil insoluble sludge can be reduced by using detergents. Despersants are similar to detergents. Vanish and carbon from fuel combustion in internal combustion engines, there will be depositions of oil insoluble materials. Those are reduced by detergents.

10. The commonly used dispersant is _____________
a) Carboxylic acid
b) Alkyl phosphoric acid
c) Poly butyl succinic acid
d) Sodium salts of alkyl benzene sulfonic acid
Answer: c
Clarification: The calcium, sodium and magnesium salts of the alkyl benzene sulfonic acids are used as detergents. Carboxylic aids and the alkyl phosphoric acid are the other commonly used detergents. The commonly used despersants is poly butyl succinic acid.

11. On cooling below the pour point, the wax will be __________
a) Evaporated
b) Condensed
c) Crystallized
d) Sublimated
Answer: c
Clarification: On cooling below the pour point, the wax will be crystallized and the additive molecules will protect them from crystallization as they will be adsorbed on the surfaces of the crystals.

12. The viscosity index can be improved by adding ___________
a) Linear polymers
b) Branched polymers
c) Cyclic polymers
d) Inorganic materials
Answer: a
Clarification: The viscosity index can be improved by adding linear polymers like poly isobutyls and poly methacrylates and poly alkyl styrene. They range from 10000 to 100000 molecular weight. They increase the viscosity and retains the original viscosity coefficient.

13. Gear oils on adding viscosity index improves can be used in ___________
a) Motor vehicles
b) Machinery
c) Air crafts
d) Engines
Answer: c
Clarification: Gear oils on improving the viscosity, they can be used in air crafts. There are some other materials that can be used in air crafts by increasing their viscosity are multi grade automotive engine oils and automatic transmission oils.

14. Foaming of oils can be prevented by using ___________
a) Despersants
b) Foam inhibitors
c) Rust inhibitors
d) Oxide inhibitors
Answer: b
Clarification: Foam inhibitors are used for preventing the foaming of oil in internal combustion engines. Methyl silicone polymers are used as effective foam inhibitors.they are not soluble in oil they help in breaking form bubbles.

15. vegetable oils are added to the lubricants to improve their ___________
a) Lubrication
b) Oiliness
c) Viscosity
d) Efficiency
Answer: b
Clarification: Oiliness can be improved by adding vegetable oils, castor oils and olive oil to the lubricants. The compounds having strong polar groups like dibenzyl dilsulphide, amyl phenyl phosphate are used as oiliness improvers.