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Chemistry Test - 2

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Chemistry Test - 2
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  • Question 1
    1 / -0

    The oxidative rusting of iron under Earth's atmosphere is __________.

    Solution

    The oxidative rusting of iron under Earth's atmosphere is slow reaction.

    The reaction rate (rate of reaction) or speed of reaction for a reactant or product in a particular reaction is intuitively defined as how fast or slow a reaction takes place. For example, the oxidative rusting of iron under Earth's atmosphere is a slow reaction that can take many years.

  • Question 2
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    What is the coordination number of chromium in \(\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) ?

    Solution

    The coordination number of chromium in \(\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) is 6 . There are three oxalate groups attached to the central ion in the given compound, but since oxalate is a didentate ligand, there are a total of 6 donor atoms to which the metal is directly bonded. Thus, the \(\mathrm{CN}\) of \(\mathrm{Cr}\) in \(\mathrm{K}_{3}\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) is 6 .

    Coordination number is also known as the secondary valence of a central metal ion in a complex and is defined as the number of donor atoms it is directly bonded to. Hence, the coordination number is a quantity associated with the metal ion.

  • Question 3
    1 / -0

    Consider the following reaction-

    \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \stackrel{\mathrm{Br}_{2} / \mathrm{NaCl}}{\longrightarrow}\)

    Product of the reaction will be:

    Solution

    In this reaction Mixture of \(1,2-\) dibromopropane and \(1\) -bromo\(-2\) chloropropane is formed as follows:

  • Question 4
    1 / -0

    The basic principle of a hydrogen economy is:

    Solution

    The usage of hydrogen as a low-carbon energy source, such as gasoline as a transport fuel, it is referred to as "Hydrogen economy". The basic principles of Hydrogen economy are:

    (i) Storage of energy in the form of liquid or gaseous dihydrogen.

    (ii) transportation of energy in the form of liquid or gaseous dihydrogen.

  • Question 5
    1 / -0

    The standard Gibbs free energy change and enthalpy change at \(25^{\circ} C\) for the liquid phase reaction

    \(CH _{3} COOH ( l )+ C _{2} H _{5} OH ( l ) \rightarrow CH _{3} COOC _{2} H _{5}( l )+ H _{2} O\) (l) are given as \(AG ^{0}\) ags \(=-4650 J / mol\) and \(\Delta H ^{0}{ }_{298}=-3640J / mol\)

    J/mol. If the solution is ideal and enthalpy change is assumed to be constant, then the equilibrium constant at \(95^{\circ} C\) is:

    Solution

    Given, \( \Delta G_{298}^{\circ}=-4650 J / mol\)

    We know that,

    \(\Delta G^{\circ}=-R T \ln K\)

    \(\ln K=-\frac{\Delta G^{\circ}}{R T}\)

    \(\ln K=\frac{4650}{8.314 \times 298}=1.877\)

    Here, \(K_{i}=6.58, \Delta H^{\circ}=-3640 J / mol\)

    \(T_{1}=298 K , T_{2}=95^{\circ} C =368 K\)

    Putting values,

    \(\ln \left(\frac{K_{2}}{6.58}\right)=\frac{3640}{8.314} \times\left(\frac{1}{368}-\frac{1}{298}\right)\)

    \(\ln \left(\frac{K_{2}}{6.58}\right)=-0.279\)

    \(K_{2}=4.94\)

  • Question 6
    1 / -0

    Which compound amongst the following has the highest oxidation number of Mn?

    Solution

    Let the oxidation number of \(M n\) be \(x\) in all the compounds.

    We know that:

    Oxidation state of \(K=+1, O=-2\).

    (A) \(K M n O_{4}: 1+x+4 \times(-2)=0\)

    \(x=+7\)

    (B) \(K_{2} M n O_{4}: 2 \times 1+x+4 \times(-2)=0\)

    \(x=+6\)

    (C) \(M n O_{2}: x+2 \times(-2)=0\)

    \(x=+4\)

    (D) \(M n_{2} O_{3}: 2 x+3 \times(-2)=0\)

    \(x=+3\)

    Therefore, highest oxidation number of \(\mathrm{Mn}\) is \(+7\) in \(\mathrm{KMnO}_{4}\).

  • Question 7
    1 / -0

    For an ideal solution, the correct option is:

    Solution

    The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. 

    The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,

    For ideal solution, 

    \(\Delta_{\operatorname{mix}} \mathrm{H}=0\)

    \(\Delta_{\operatorname{mix}} \mathrm{V}=0\)

  • Question 8
    1 / -0

    Which of the following is the correct match?

    (a) Eka silicon - Ge

    (b) Eka aluminum - Ga

    (c) Eka manganese - Tc

    (d) Eka boron - Sc

    Solution

    Eka is a prefix used to designate the first element of the same family in the periodic table beyond the one to whose name it is prefixed, as Eka manganese.

     

    • Eka silicon - Ge (germanium)
    • Eka aluminium - Ga (gallium )
    • Eka manganese - Tc (technetium) 
    • Eka boron – Sc (scandium)

     

  • Question 9
    1 / -0

    The electronegativity of Caesium is 0.7 and that of Fluorine is 4.0. The bond formed between the two is:

    Solution

    Such a large difference in electronegativity shows that Caesium is electropositive while Fluorine is electronegative in nature. So, Caesium can donate an electron to Caesium which can accept it. So, they form ionic bond or electrovalent bond.

  • Question 10
    1 / -0

    The main cause of acidity in the stomach is:

    Solution

    The main cause of acidity in the stomach is release of extra gastric acids which decrease the pH level. 

     

    • Gastric acid is a digestive fluid formed in the stomach and is composed of hydrochloric acid \((\mathrm{HCl})\), potassium chloride \((\mathrm{KCl})\) and sodium chloride \((\mathrm{NaCl})\).
    • If gastric acid is released in the stomach, the \(\mathrm{pH}\) level decreases and thereby increasing acidity in the stomach.

     

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