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Chemistry Test - 17

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Chemistry Test - 17
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  • Question 1
    1 / -0
    If the ionization enthalpy and electron gain enthalpy of an element are \(275\) and \(86\) \(\mathrm{kcal}\) \(\mathrm{mol}^{-1}\) respectively, then the electronegativity of the element on the Pauling scale is:
    Solution

    By this relation:

    \(\mathrm{I.E.+E . A}={275+86=361}\)\(\mathrm{kcal ~mol}^{-1}\)

    \(=361 \times 4.184=1510.42 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

    Where \(\mathrm{I.E.}\) = Ionization enthalpy

    \(\mathrm{E.A.}\) = Electron affinity (electron gain enthalpy)

    Electronegativity \(=\frac{\mathrm{I.E.}+\mathrm{E} . \mathrm{A} .}{540}\)

    So, electronegativity \(=\frac{1510.42}{540}\)

    \(\Rightarrow 2.797=2.8\)

  • Question 2
    1 / -0

    In ___________, a reaction product is itself a catalyst for that reaction leading to positive feedback.

    Solution

    In autocatalysis, a reaction product is itself a catalyst for that reaction leading to positive feedback.

    A single chemical reaction is said to have undergone autocatalysis, or be autocatalytic, if one of the reaction products is also a reactant and therefore a catalyst in the same or a coupled reaction. The reaction is called an autocatalytic reaction. A catalyst is a substance that accelerates the rate of a chemical reaction but remains chemically unchanged afterwards.

  • Question 3
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    The reduction in atomic size with increase in atomic number is a characteristic of elements of:

    Solution

    The reduction in atomic size with increase in atomic number is a characteristic of elements of f- block. It is known as lanthanoid contraction and actinoid contraction. This is due to poor shielding of electrons present in f subshell.

  • Question 4
    1 / -0

    _______ group elements are known as chalcogens.

    Solution

    VIA group or Group 16 elements belong to the oxygen family are also called chalcogens.

    • The name is derived from the Greek word 'chalcos', which means 'ore' and 'gen', which means 'formation'.
    • They are called so because most of the copper ores have copper in the form of oxides and sulphides.
    • It consists of the elements oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and the radioactive element polonium (Po).
  • Question 5
    1 / -0

    Among the following the state function(s) is (are):

    (i) Internal energy

    (ii) Irreversible expansion work

    (iii) Reversible expansion work

    (iv) Molar enthalpy

    Solution

    Among the following, the state functions areInternal energy andMolar enthalpy.

    The sum of all energies of a chemical system is called the internal energy \( (U)\) of the system. The internal energy may change under the following conditions:

    • Heat passes into or out of the system
    • Work done on or by the system
    • Matter enters or leaves the system

    Molar enthalpy is also a state function. Enthalpy is the sum of internal energy and pressure-volume or work. It is defined in terms of state functions.Enthalpy is a state function because it only depends on the initial and final conditions, and not on the path taken to establish these conditions. Therefore, the integral of state functions can be taken using only two values, the final and initial values.

  • Question 6
    1 / -0

    The IUPAC name of \(CH _{3} COCH \left( CH _{3}\right)_{2}\) is:

    Solution

    The IUPAC name of this structure is \(3\)-methyl-\(2\)-butanone.

    The longest chain is of four carbon. The functional group here is keto group. The numbering will start from left carbon as the functional group should also have the lowest numbering. The substituent methyl group will be at third carbon.

    \(1\). Numbering of carbon atoms.

    \(2\). As per rule assign count number to the special group (ketone).

    \(3\). After assigning number to carbon atom we found that it is a chain of four carbon atoms in which ketone group is at second carbon and methyl in connected to third carbon hence IUPAC name of this compound will be \(3\)-methyl-\(2\)-butanone.

  • Question 7
    1 / -0

    The bond length between hybridised carbon atom and other carbon atom is minimum in:

    Solution

    The C – C bond length = 1.54 Å, C = C bond length = 1.34 Å and C ≡ C bond length = 1.20 Å.

    Since propyne has a triple bond, therefore it has minimum bond length.

  • Question 8
    1 / -0

    Which of the following is a tridentate ligand?

    Solution

    Dien (Diethylenetriamine) has following structure:

    InDien,the number of donor atoms is three so it is a tridentate ligand. Each nitrogen atom is a donor atom.

    \(EDT A ^{4-}\) is a hexadentate ligand. \(( COO )_{2}^{2-}\) is a bidentate ligand. \(NO _{2}^{-}\)is a monodentate ligand.

  • Question 9
    1 / -0

    \(R-C H=C R O^{-}+C H_{2}-N^{+} R_{2} \rightarrow N R_{2}-C H_{2}-C H R-C R O\)

    The given reaction in an example of:

    Solution

    The given reaction in an example ofMannich reaction.

    The reaction is named after chemist Carl Mannich. The Mannich reaction is an example of nucleophilic addition of an amine to a carbonyl group followed by dehydration to the Schiff base.

    The Mannich reaction is an organic reaction which consists of an amino alkylation of an acidic proton placed next to a carbonyl functional group by formaldehyde and a primary or secondary amine or ammonia. The final product is a β-amino-carbonyl compound also known as a Mannich base. Reactions between aldimines and α-methylene carbonyls are also considered Mannich reactions because these imines form between amines and aldehydes. The reaction is named after chemist Carl Mannich.

  • Question 10
    1 / -0

    A first order reaction has the rate constant, \(\mathrm{k}=4.6 \times 10^{-3} \mathrm{~s}^{-1}\). The number of correct statement/s from the following is/are____________.

    Given : \(\log 3=0.48\)

    A. Reaction completes in \(1000 \mathrm{~s}\).

    B. The reaction has a half-life of \(500 \mathrm{~s}\).

    C. The time required for \(10 \%\) completion is 25 times the time required for \(90 \%\) completion.

    D. The degree of dissociation is equal to \(\left(1-\mathrm{e}^{-\mathrm{kt}}\right)\).

    E. The rate and the rate constant have the same unit.

    Solution

    \(\begin{aligned} & t_{10 \%}=\frac{1}{K} \ln \left(\frac{a}{a-x}\right)=\frac{1}{K} \ln \left(\frac{100}{90}\right) \\ & t_{10 \%}=\frac{2.303}{K}(\log 10-\log 9) \\ & t_{10 \%}=\frac{2.093}{K} \times(0.04) \\ & \text { Similarly } \\ & t_{90 \%}=\frac{1}{K} \ln \left(\frac{100}{10}\right) \\ & t_{90 \%}=\frac{2.303}{K} \\ & \frac{t_{90 \%}}{t_{10 \%}}=\frac{1}{0.04}=25 \\ & e^{k t}=\frac{a}{a-x} \\ & \frac{a-x}{a}=e^{-k t} \\ & 1-\frac{x}{a}=e^{-k t} \\ & x=a\left(1-e^{-k t}\right) \\ & a=\frac{x}{a}=\left(1-e^{-k t}\right)\end{aligned}\)

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