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Chemistry Test - 23

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Chemistry Test - 23
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  • Question 1
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    Inhaling _________ in large amounts can cause pulmonary edema.

    Solution

    Inhaling of nitrogen oxides by humans results in pulmonary edema and hemorrhage. Several sensitive plants get adversely affected by NOx even at the concentration of 1 ppm for a day or 0.35 ppm for a few months. The affected plants start shedding their leaves and fruit. Nitrogen oxides cause fading of textile dyes and deterioration of cotton and nylon fabrics.

  • Question 2
    1 / -0

    Roasting of sulphides gives the gas \(\mathrm{X}\)as a by-product. This is a colourless gas with choking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic acts as a reducing agent and its acid has never been isolated. The gas \(\mathrm{X}\) is:

    Solution

    In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal. Some of the reactions involving sulphide ores are:

    \(2 \mathrm{ZnS}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{ZnO}+2 \mathrm{SO}_2\)

    \(2 \mathrm{PbS}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{PbO}+2 \mathrm{SO}_2\)

    \(2 \mathrm{Cu}_2 \mathrm{~S}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{Cu}_2 \mathrm{O}+2 \mathrm{SO}_2\)

    The sulphide ores of copper are heated in a reverberatory furnace.

    \(\mathrm{SO}_2\)has a choking smell and causes acid rain.

    Therefore, the gas \(\mathrm{X}\) is\(\mathrm{SO}_2\).

  • Question 3
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    Data given for the following reaction is as follows.

    \(\mathrm{FeO}(\mathrm{s})+\mathrm{C}_{\text {(graphite) }} \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}(\mathrm{g})\)

    Substance  \(\Delta \mathbf{H}^{\circ}\left(\mathbf{k J ~ m o l}^{-1}\right)\)  \(\Delta \mathbf{S}^{\circ}\left(\mathbf{J ~ m o l}^{-1} \mathbf{K}^{-\mathbf{1}}\right)\)
    FeO(s)  -266.3 57.49 
    \(\mathrm{C}_{\text {(graphite) }}\) 0 5.74 
    Fe(s)  0 27.28 
    CO(g)  -110.5 197.6 
    The minimum temperature in K at which the reaction becomes spontaneous is......... (Integer answer)
    Solution

    \(\mathrm{FeO}(\mathrm{s})+\mathrm{C}_{\text {(graphite) }} \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \)

    \(\Delta \mathrm{H}_{\text {reaction }}^{\circ}=\Delta \mathrm{H}_{\text {ffrpoduac) }}^{{ }^{\circ}}-\Delta \mathrm{H}_{\text {ffrectants) }}^s \)

    \( =\left[\Delta \mathrm{H}_{\mathrm{f}(\mathrm{Fe})}+\Delta \mathrm{H}_{\mathrm{f}(\mathrm{CO})}\right]-\left[\Delta \mathrm{H}_{\mathrm{f}(\mathrm{Fe})}^{\circ}-\Delta \mathrm{H}_{\mathrm{f}(\mathrm{C})}\right] \)

    \(=[0+(-110.5)]-[-266.3-0]\)

    \(=156 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1} \)

    \(\Delta \mathrm{S}_{\text {reaction }}^{\circ}=\Delta \mathrm{S}_{\text {product }}^{\circ}-\Delta \mathrm{S}_{\text {reactant }}^{\circ}\)

    \(=\left[\Delta \mathrm{S}_{(\mathrm{Fe})}^{\circ}+\Delta \mathrm{S}_{(\mathrm{CO})}^{\circ}\right]-\left[\Delta \mathrm{S}_{(\mathrm{Fe} 0)}^{\circ}-\Delta \mathrm{S}^{\circ}{ }_{(0)}\right] \)

    \(=[27.28+197.6]-[57.49+5.79]\)

    \(=161 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

    According to Gibb's equation,

    \(\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}\)

    The reaction becomes spontaneous when \(\Delta \mathrm{G}^{\circ}\) is atleast zero or negative.

    \(0=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \)

    \(\mathrm{T} \Delta \mathrm{S}^{\circ}=\Delta \mathrm{H}^{\circ} \)

    \(\Rightarrow \mathrm{T}=\frac{\Delta \mathrm{H}^{\circ}}{\Delta \mathrm{S}^{\circ}}=\frac{156 \mathrm{~kJ} \mathrm{~mol}^{-1}}{161 \mathrm{~J}^{-1} \mathrm{~mol}^{-1}} \)

    \(=\frac{156000 \mathrm{~mol}^{-1}}{161 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}=964 \mathrm{~K}\)

    The temperature at which reaction becomes spontaneous is 964 K.

  • Question 4
    1 / -0

    Amorphous solids are:

    Solution

    Amorphous solids have an irregular arrangement of constituent particles. They have the same physical properties in all directions due to irregular arrangement. Hence the amorphous solids are isotropic. Here the constituent particles (atoms, molecules or ions) are scattered and disordered.

    Due to irregular arrangement, they process fluidity like liquids. The structure of amorphous solids is similar to that of liquids. This type of solid melts over a range of temperatures. Hence these are also called pseudo solids or supercooled liquids.

  • Question 5
    1 / -0

    If we have an element with atomic number 115, then which of the following statements is correct about this element?

    Solution

    If we place an element with atomic number 115 in the periodic table, it will come under the 15th group. So, it will be a p-block element.

  • Question 6
    1 / -0

    The structure given below is known as:

    Solution

    The structure given below is known aspenicillin G.

    Penicillin G is an antibiotic used to treat a number of bacterial infections. This includes pneumonia, strep throat, syphilis, cellulitis, tetanus, etc.

    Penicillin G has a narrow spectrum. Those effective mainly against Gram-positive or Gram-negative bacteria are narrow-spectrum antibiotics.

  • Question 7
    1 / -0

    For which of the following mixture of gases, Dalton's law of partial pressure is not applicable at room temperature?

    Solution

    Dalton's law of partial pressure is not applicable at room temperature for a mixture of reacting gases.

    Thus, Dalton's law of partial pressure is not applicable at room temperature for a mixture of hydrogen chloride and ammonia.

  • Question 8
    1 / -0

    Which of the following antibiotics is used to cure typhoid?

    Solution

    (A). Novalgin: Also called Metamizole, is an analgesic and antipyretic.

    (B). Quinine: Used to treat Malaria.

    (C). Chloromycetin: Also called Chloramphenicol, is an antibiotic used to treat typhoid.

    (D). Paracetamol: Also called Acetaminophen, is an analgesic and antipyretic.

  • Question 9
    1 / -0

    Sodium burns in dry air to ordinarily give:

    Solution

    Sodium monoxide \(\mathrm {(Na_2 O)}\) is ordinarily formed upon oxidation of sodium in dry air. The superoxide \(\left(\mathrm{NaO_2}\right)\) can be prepared by heating metallic sodium to \(300^{\circ} \mathrm{C} \space \left(570^{\circ} \mathrm{F}\right)\) in an autoclave (a heated pressure vessel) containing oxygen at high pressure.

  • Question 10
    1 / -0

    The isotope of hydrogen, \(D _{2}\) forms heavy water, \(D _{2} O\). It is qualified as \(D _{2} O\) because it is:

    Solution

    D2O or heavy water is an oxide of Deuterium, an isotope of hydrogen with a mass double that of hydrogen. Due to this, heavy water is 10% heavier than ordinary water. The physical and chemical properties of heavy water are similar to that of ordinary water but due to an increase in mass by 10%, it has a higher boiling point than ordinary water.

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