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Chemistry Test - 15

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Chemistry Test - 15
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  • Question 1
    4 / -1

    Hess's law of constant heat summation is based on:

    Solution

    Hess's law of constant heat summation is based on the first law of thermodynamics.

    Hess's law of constant heat summation states that " Regardless of multiple stages, total enthalpy changes is the sum of all the changes."

    Mathematically Hess law can be represented as;

    \(\Delta \mathrm{H}_{\text {net }}=\sum \Delta \mathrm{H}_{\mathrm{Y}}\)

    Where, \(\Delta \mathrm{H}_{\text {net }}=\) Net enthalpy change

    \(\Delta \mathrm{H}_{\mathrm{r}}=\) Sum of enthalpy change of reactions

    So, Enthalpy change is a state function as it depends on the final and initial stage and not the path followed by the system. So, It resembles internal energy as internal energy is also a state function.

    The first law of thermodynamics states that " Some part of internal energy of the system is used to heat the system while another part is used for work done."

    So, Hess's law of constant heat summation is based on the first law of thermodynamics as both suggest the conversation of energy.

  • Question 2
    4 / -1

    Which of the following energy state is filled by an electron after the completion of \(4 p\) orbital?

    Solution

    When \(4 p\) orbital in any atom are filled completely, the next electron goes in \(5 \mathrm{~s}(\mathrm{n}+1=5+0=5)\) not in \(4 \mathrm{~d}(\mathrm{n}+1=4+2=6)\), electron first enters into the subshell which is having less \((n+\) 1) value (energy). So, the \(5 \mathrm{~s}\) is filled first than \(4 \mathrm{~d}\).

  • Question 3
    4 / -1

    The dry ice is _____________.

    Solution

    The dry ice is solid carbon dioxide.

    Carbon dioxide is linear gas at ordinary temperature but at low temperatures, it is in solid form and also known as dry ice or drikold. it is used for storing frozen substances at a lower temperature than normal water and it is used for cooling.

  • Question 4
    4 / -1

    Treatment of ammonia with excess of ethyl chloride will yield:

    Solution

    Treatment of ammonia with ethyl chloride yields Ethylamine.

    When an excess of ethyl chloride is present, the product formed reacts further with alkyl halides to form a secondary amine.

    The reaction proceeds further and forms tertiary amine ultimately.

    Thus, when an excess of ethyl chloride is present,the reaction leads to the formation of tetraethyl amine.

    Finally, the tertiary amine reacts with the alkyl halide to form a quaternary ammonium salt.

    This reaction is called Hofmann ammonolysis of alkyl halides.

    The reaction is shown as follows.

    Therefore, treatment of ammonia with excess ethyl chloride gives Tetraethyl ammonium chloride.

  • Question 5
    4 / -1

    Which of the following can possibly be used as an analgesic, without causing addiction and modification?

    Solution

    N-acetyl-para-aminophenol(or paracetamol) is an antipyretic which can also be used as Analgesic to relieve pain without addiction and mood modification.

  • Question 6
    4 / -1

    The value of gas constant per degree per mole is approximately:

    Solution

    The value of gas constant per degree per mole is approximately\(2 \mathrm{cal}\).

    In the equation of state of an ideal gas \(\mathrm{PV}=\mathrm{nRT}\), the value of universal gas constant would depend only on the units of measurement. It is independent of the nature of the gas, the pressure of the gas and the temperature of the gas.

    For example,

    \(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K}\)

    \(\mathrm{R}=1.99 \mathrm{cal} / \mathrm{mol} / \mathrm{K}\)

    \(\mathrm{R}=0.08206 \mathrm{Latm} / \mathrm{mol} / \mathrm{K}\)

  • Question 7
    4 / -1

    IUPAC name of the following compound is:

    Solution

    The priority of the functional groups is decided first.

    The sequence is given as follows:

    After deciding the priority of the group, substituents are numbered alphabetically.

    In Cyclic compounds, the ring is the parent chain unless it is attached to a larger chain.

    The alkane has a higher priority than ether.

    Here, the parent chain is cyclohexane as it is a six-membered ring.

    While numbering in the substituents in hexane compound 1-position goes to ethoxy as 'e' comes first in alphabetical letter and 2- position to nitro.

    IUPAC name of this compound will be: 1-ethoxy-2-nitrocyclohexane

  • Question 8
    4 / -1

    Phenol on reduction with H2 in the presence of Ni catalyst gives:

    Solution

    Phenol is hydrogenated to cyclohexanol, 

  • Question 9
    4 / -1

    To determine the health of a water body, what is measured?

    Solution

    Dissolved oxygen \((\mathrm{DO})\) is an indicator of the quantity of free oxygen molecules in water.The concentration ofDissolved oxygenis an important indicator of the health of the aquatic environment, as oxygen is essential for almost all forms of life.

    Oxygen is essential for respiration and chemical reactions. Dissolved oxygen in water comes from two primary sources: the atmosphere and photosynthesis.The concentration of Dissolved oxygenis influenced by a variety of factors, including water temperature (cold water retains more oxygen than warm water), salinity (freshwater holds more oxygen than salt water), and atmospheric pressure (the amount of Dissolved oxygenconsumed in water decreases as the altitude increases).The parameter is capable of calculating dissolved oxygen in water within a range of \(0 – 500\%\) or \(0 – 50\) \(mg/l\).

  • Question 10
    4 / -1
    The solubility product of \(\mathrm{A}_{2} \mathrm{X}_{3}\) is \(1.081 .1 \times 10^{-23} \mathrm{M}^{5}\) Its solubility is
    Solution

    \({A}_{2} {X}_{3} \rightarrow 2 {~A}_{3}++3 {X}^{-2}\)

    \({Ksp}=\left[{A}^{3+}\right]^{2}\left[{X}^{2-}\right]^{3}\)

    Let solubility of \({A}_{2} {X}^{3}\) be \({S}\)

    \(\therefore {Ksp}=[2 {~S}]^{2}[3 {~S}]^{3}=108 {~S}^{5}\)

    \(\therefore1.1 \times 10^{-23}=108 \mathrm{~S}^{5}\)

    \(\therefore S 5=1.1 \times 10^{-23} / 108=1 \times 10^{-25}\)

    \(\therefore S=1.0 \times 10^{-5} {~mol} / {L}\)

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