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Chemistry Test - 11

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Chemistry Test - 11
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  • Question 1
    1 / -0

    Soap removes grease by:

    Solution

    Soap removes grease by emulsification.

    Soap is an excellent cleanser because of its ability to act as an emulsifying agent. An emulsifier is capable of dispersing one liquid into another immiscible liquid. This means that while oil (which attracts dirt) doesn't naturally mix with water, soap can suspend oil/dirt in such a way that it can be removed.

    Grease and oil are nonpolar and insoluble in water. When soap and soiling oils are mixed, the nonpolar hydrocarbon portion of the micelles breaks up the nonpolar oil molecules. A different type of micelle then forms, with nonpolar soiling molecules in the centre. Thus, grease and oil and the 'dirt' attached to them are caught inside the micelle and can be rinsed away.

  • Question 2
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    What is an example for a double salt?

    Solution

    Mohr's salt is a compound known to contain two primary cations, namely the ammonium cation (denoted by \(\left.NH _{4}^{+}\right)\)and the ferrous cation (denoted by \(F e ^{2+}\)). Therefore, Mohr's salt can be categorized as a double salt of ammonium sulphate and ferrous sulphate.

    All others are complexes.

  • Question 3
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    Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in this reaction is:

    Solution

    Sodium ethoxide reacts with ethanoyl chloride. This is according to the SN2 mechanism shown in the reaction below.

    Ultimately the compound produced in this reaction is ethyl ethanoate.

  • Question 4
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    Heterolysis of carbon-chlorine bond produces:

    Solution

    ​​Homolysis and Heterolysis:

    The process ofbreakageofcovalent bondswhere the electron density is distributedequallybetween the two atoms. The process ofbreakage of covalent bondswhere the electron density is distributedunequallybetween the two atoms.
    Producesfree radicalswhere each fragment hasone unpairedelectron. Themore electronegativeatomtakes awaythe electron density.
    This occurs in presence offree radical initiatorslike heat, light, peroxides. Acationand ananionare produced.
    Occurs fast. Occurs with a moderate rate.
    Indicated byfish hook arrowsor half arrows Indicated byfull arrows.

    So when the Carbon - Chlorine bond is broken heterolytically, a cation and an anion are formed.

    An example is shown below:

  • Question 5
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    Benzene can't be prepared in the laboratory by which of the following methods?

    Solution

    Vulcanisation is a chemical system in which the rubber is heated with sulfur, accelerator, and activator 140–160°C.

    The method involves the formation of cross-links between long rubber molecules so as to achieve improved elasticity, resilience, tensile strength, viscosity, hardness, and weather resistance.

    Benzene is commercially isolated from coal tar. However, it may be prepared in the laboratory by the following methods:

    Cyclic polymerization of ethyne- here, alkyne molecules add to each other resulting in the formation of a cyclic compound when ethyne is passed through a red hot iron tube at 873 K it undergoes cyclic polymerization. In this process, 3 molecules of ethyne polymerize to form benzene

    Decarboxylation of aromatic acids- It is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO2).

    Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.

    The reverse process, which is the first chemical step in photosynthesis, is called carboxylation, the addition of CO2 to a compound.

    Enzymes that catalyze decarboxylations are called decarboxylases or, the more formal term, carboxy-lyases.

    The term "decarboxylation" usually means replacement of a carboxyl group (-COOH) with a hydrogen atom

    RCO2H → RH + COO Sodium salt of benzoic acid on heating with soda-lime gives benzene.

    Reduction of phenol:

    • Phenol is reduced to benzene by, passing its vapors over, heated zinc dust.
    • Phenol can be converted to benzene by using strong reducing agents like Zn dust with strong heating.
    • When strongly heated, the phenol gets converted into phenoxide ion and proton thus released accepts an electron from Zn forming H radical.
  • Question 6
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    A compound contains three elements \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), if the oxidation of \(A=+2, B=+5\) and \(\mathrm{C}=-2,\) the possible formula of the compound is:
    Solution

    The formula of a compound with elements A, B, \(C\) will be of the form \(\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}} \mathrm{C}_{\mathrm{z}}\), where \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) are the atomicities of individual atoms respectively.

    The oxidation numbers of the elements, A, B, C are \(+2,+5\) and \(-2\) respectively. A compound is always neutral, so the oxidation number of the compound is zero. Hence, the net oxidation number of the individual atoms is zero. \((+2)(x)+(+5)(y)+(-2)(z)=0 ; 2 x+5 y-2 z=0\).

    Restricting it to conditions that maximum oxidation state could be \(+8\) or \(-3\), by Hit and Trial method, the possible solutions of \((\mathrm{x}, \mathrm{y}, \mathrm{z})\) could be \((1,2,6)\) or ( \(3,2,8)\).

    Both the sets are possible as the elements are unknown. Hence the compounds are \(\mathrm{A}\left(\mathrm{BC}_{3}\right)_{2}\) and \(\mathrm{A}_{3} \mathrm{~B}_{2} \mathrm{C}_{8}\)

  • Question 7
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    Who among the following laid the foundation of chemical science by establishing two important laws of chemical combination?

    Solution

    Antoine Laurent Lavoisier was a French Chemist who developed the two important laws of chemical combinations and co-authored the modern system for naming chemical substances.

    He is known as ‘The Father of Modern Chemistry’

    Ernest Rutherford was a New Zealand born physicist who gave the famous ‘Rutherford Model of the atom’ which explains that the atom contains very small charges nucleus (contains most of the mass of atom) and orbited by loss mass electrons.

    Democritus was a Greek philosopher and scientist who observed that a cone or a pyramid has one-third the volume of a cylinder or a prism with the same base and height.

    Joseph Louis Proust was a French chemist who gave the law of constant composition.

  • Question 8
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    Starch is converted to ethanol by fermentation. What is the sequence of enzymes used?

    Solution

    Starch is converted to ethanol by fermentation. The sequence of enzymes used is Diastase, maltase, zymase.

    The manufacture of ethyI alcohol from starch contains the following steps:

    (i) Hydroiysis of starch

    \(2\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{5}\right) n+n \mathrm{H}_{2} \mathrm{O} \stackrel{50-60^{\circ}}{\longrightarrow}\) Diatease \(n \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) This step is called saccharification.

    (ii) Fermentation

    \(\underset{\substack{\text { maltase }}}{n \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\mathrm{H}_{2} \mathrm{O} \underset{\text { teast }}{\stackrel{\text { Maltase }}{\longrightarrow}} 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \stackrel{\text { Zymase yeast }}{\longrightarrow} 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} H+2 \mathrm{CO}_{2}\)

  • Question 9
    1 / -0

    Which of the following is/ are correct statement(s)?

    Solution

    All of the following statements are correct.

    • Portland cement is the most commonly used cement for construction work.
    • Portland cement is made from raw material which consists of 60% lime (calcium oxide), 25% silica (silicon dioxide), 5% alumina. The rest is iron oxide and gypsum (calcium sulphate).
    • Portland cement gets its name from the stone quarried from the Isle of Portland in England which has a similar texture.
  • Question 10
    1 / -0

    Oxidation number of carbon in \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) is:

    Solution

    Consider the oxidation state of carbon in dichloromethane\((\mathrm{CH}_{2} \mathrm{Cl}_{2})\) as \({x}\).

    We know that,

    The charges on \(H\) and \(C l\) are \(+1\) and \(-1\) respectively.

    Therefore,

    \(\mathrm{CH}_{2} \mathrm{Cl}_{2} \rightarrow x+2(+1)+2(-1)=0\)

    \(\Rightarrow x=0\)

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