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The p-Block Elements Test - 60

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The p-Block Elements Test - 60
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  • Question 1
    1 / -0
    Which among the following substances is not a component of the mixture used for glazing pottery?
    Solution
    Ceramic glaze is an impervious layer or coating of a vitreous substance which has been fused to a ceramic body through firing. Glaze can serve to colour, decorate or waterproof an item.

    Pottery glaze is made up of five basic components. These components are silica, alumina, flux, colourants and modifiers.

    Hence, option A is correct.
  • Question 2
    1 / -0
    Which of the following does not react with cold water?
    Solution
    As $$SiC$$ is a covalent compound, it requires some amount of energy to react with water. So, it doesn't react with cold water.

    Option A is correct.
  • Question 3
    1 / -0
    $$Me_3SiCl$$ is mixed with $$Me_2SiCl_2$$ to get a polymers called silicones of the type:

  • Question 4
    1 / -0
    Silica reacts with $$Mg$$ to form a magnesium compound $$A$$. $$A$$ reacts with dilute $$HCl$$ and forms $$B$$. The compound $$B$$ is:
    Solution
    It is given that silica reacts with $$Mg$$ to form a magnesium compound $$A$$. $$A$$ reacts with dilute $$HCl$$  to form $$B$$.
    The reactions are as follows:
    $$2Mg + SiO_2 \rightarrow 2MgO + Si$$
    $$MgO + 2HCl \rightarrow MgCl_2 + H_2O$$
    Hence, the compound $$B$$ is $$MgCl_2$$.
  • Question 5
    1 / -0
    Borax in its crystal possess:
    Solution
    Borax $$[Na_2(B_4O_5(OH)_4.8H_2O]$$

    $$Sp^2 \to$$ trigonal planar

    $$Sp^3 \to$$ tetrahedral

    In borax $$2(Sp^2)$$ trigonal planar and $$2(Sp^3)$$ tetrahedral

  • Question 6
    1 / -0
    Microcosmic salt and borax are used in the identification of cations by dry tests. They are respectively :
    Solution
    Microcosmic salt bead test is useful to identify cations like in borax bead test. When the microcosmic salt is heated on platinum wire loop to get sodium meta phosphate,
    $$Na(NH_4)HPO_4.4H_2O \rightarrow Na(NH_$)HPO_4 + 4H_2O$$
    $$Na(NH_4)HPO_4 \rightarrow NaPO_3 +NH_3 +H_2O$$
                                      (Glass mass)
    Now, we have $$NaPO_3$$ which reacts with metallic oxides to give coloured orthophosphates.
    Thus, option (B) is the correct answer.
  • Question 7
    1 / -0
    Read the following write-ups and answer the questions at the end of it. 
    Silicons are synthetic polymers containing repeated $$R_{2}SiO$$ units. Since, the empirical formula is that of a ketone $$(R_{2}CO)$$, the name silicone has been given to these materials. Silicones can be made into oils, rubbery elastomers and resins. They find a variety of applications because of their chemical inertness, water repelling nature, heat- resistance and good electrical insulating property. Commercial silicon polymers are usually methyl derivatives and to a lesser extent phenyl derivatives and  are synthesised by the hydrolysis of $$R_{2}SiCl_{2}$$[R=methyl (Me) or phenyl (Ph)]

    If we mix $$Me_{3}SiCl$$ with $$Me_{2}SiCl_{2}$$, we get silicones of the type:
  • Question 8
    1 / -0
    One natural source of atmospheric $$CO_2$$ is precipitation reaction such as the precipitation of silicates in the oceans:

    $$Mg^{2+}(aq)+SiO_2(dispersed)+2HCO_3(aq)\rightarrow MgSiO_3(s)+2CO_2(g)+H_2O(l)$$

    How many grams of magnesium silicate would be precipitated during the formation of $$100$$ L of $$CO_2$$ at $$30^oC$$ and $$775$$ torr? 
    $$(Mg=24, Si=28, C=12, O=16)$$
    Solution
    Moles of $$CO_2$$ formed$$=n=\displaystyle\frac{PV}{RT}$$

    $$PV=\displaystyle\frac{775}{760} atm\times 100L$$

    $$RT=0.0821\ L \ atm$$ $$mol^{-1}$$ $$K^{-1}\times 303K$$

    $$n = 4.0992$$

    If $$2$$ moles $$CO_2$$ are formed then $$MgSiO_3$$ precipitated is $$1mol$$ and when $$4.0992$$ moles $$CO_2$$ and formed $$MgSiO_3$$ would be $$=\displaystyle\frac{4.0992}{2}$$ $$=2.0496\  mol$$

    (molar mass of $$MgSiO_3=100\ g$$ $$mol^{-1}$$

    $$=2.0496\times 100 \ g$$
    $$=204.96\ g$$ is nothing but $$205\ g$$

    Hence, option $$C$$ is correct.
  • Question 9
    1 / -0
    $$Al + N_2 \xrightarrow {\triangle} (A) \xrightarrow {H_2O} (B) \xrightarrow {K_2HgI_4/OH^-} (D)$$ (brown ppt.)

    What is $$(D)$$?
    Solution
    $$Al + N_2 \xrightarrow {\Delta} AlN \xrightarrow {H_2O}NH_3 \xrightarrow {(K_2HgI_4/OH^-)}\, \, \, (D) : H_2N - Hg - O - Hg - I$$ (brown ppt.) (Iodide of Millon's base)
    Hence option $$B$$ is correct.
  • Question 10
    1 / -0
    Which is incorrect statement about silicons?
    Solution
    Organic polymers containing silicon in them are called as silicones. These are organosilicon polymers containing $$R_2SiO$$ repeating units and empirical formula analogous to ketone $$(R_2CO)$$. They are formed by hydrolysis of $$R_2SiCl_2$$.
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