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Structure of Atom Test - 62

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Structure of Atom Test - 62
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Weekly Quiz Competition
  • Question 1
    1 / -0
    General electronic configuration of  outermost and penultimate shell is  $$(n-1)s^{2}(n-1)p^{6}(n-1)d^{x}ns^{2}$$  If n = 4 and x = 5,  then number of protons in  the nucleus will be :- 
    Solution

  • Question 2
    1 / -0
    Write the S.I. unit of activity.
    Solution
    $$S.I.$$ Unit of activity : It is the count rate of radioactivity.
    It's $$S.I$$ Unit is decay per second $$=1\ Bq$$
    where $$Bq=Becquerel  (\because 1\ Bq=1\ decay\ per second)$$
  • Question 3
    1 / -0
    What mass of solid ammonium carbonate $$H_2NCOONH_4$$, when vaporized at $$273^oC$$, will have a volume of $$ 8.96\ 1$$ at $$760\ mm$$ of pressure. Assume that the solid completely decomposes as
    $$H_2NCOONH_4(s)\to CO_2(g)+2NH_3(g)$$  
    Solution
    $$H_2NCOONH_4(s)\longrightarrow CO_2(g)+2NH_3(g)$$
    gases moles $$=a+2a=3a$$
    $$pv=nRT$$
    $$n=\dfrac {Pv}{RT}=\dfrac {760\times 8.96}{760\times 0.082\times 546}$$
    $$3a=0.2$$
    $$a=0.056$$
    mass $$=a\times molar\ mass\ NH_2COONH_4$$
    $$=0.066\times 78$$
    $$=5.14$$ gram
    option $$"b"$$ correct
  • Question 4
    1 / -0
    Twenty molecular of $$SO_3$$ will weight as much as _____ molecular of oxygen.
    Solution
    The weight of one $$SO_3$$ molecule $$ = 32+( 3\times16 )$$ 

                                                             $$= 32+48=80u$$

    Weight of twenty $$SO_3$$ molecule $$= 20\times80=1600u$$

    Weight of each molecule of oxygen$$=2\times16=32u$$

    Therefore, the no. of oxygen molecules $$= \dfrac{1600}{32}$$ $$=50$$

    So, the correct option is $$B$$
  • Question 5
    1 / -0
    How many spectral line of balmer series present in visible region?
    Solution
    Solution:- (B) $$4$$
    Four of the Balmer lines are in the technically visible part of the spectrum, with wavelengths between $$400 \; nm - 700 \; nm$$.
  • Question 6
    1 / -0
    The commonly used pain reliever, aspirin, has the molecular formula $$C_9H_8O_4.$$ If a sample of aspirin contains $$0.968\ g $$ of carbon, what is the mass of hydrogen in the sample?
    Solution

    Molar mass of $$C_9H_8O_4=9 \times 12+8\times 1+4\times 16$$

    $$=180$$ gram/mole 

    $$180$$ gram  $$C_9H_8O_4 \equiv 108$$ gram carbon 

    $$\therefore x$$ gram $$C_9H_8O_4 \equiv 0.968$$ gram carbon 

    $$\therefore x=\dfrac{0.968 \times 180}{108}$$

    $$= 1.613$$ gram $$C_9H_8O_4$$

    Now

    $$180$$ gram $$C_9H_8O_4\equiv 8$$ gram Hydrogen 

    $$\therefore 1.613$$ gram $$C_9H_8O_4 \equiv \dfrac{8\times
    1.613}{180} $$

    $$=0.0717$$ gram '$$H$$'

    Option $$(b)$$ correct.

  • Question 7
    1 / -0
    When a mixture of aluminium powder and iron $$(III)$$ oxide is ignited, it produces molten iron and aluminium oxide. In an experiment, $$5.4\ g$$ of aluminium was mixed with $$18.5\ g$$ of iron $$(III)$$ oxide. At the end of the reaction, the mixture contained $$11.2\ g$$ of iron, $$10.2\ g$$ of aluminium of unreacted iron $$(III)$$ oxide. No aluminium was left. What is the mass of the iron $$(III)$$ oxide left?
    Solution
    $$Al+Fe_{2}O_{3}\rightarrow Fe+Al_{2}O_{3}$$
    Mass of $$Al=5.4\ gm$$
    Mass of $$Fe_{2}O_{3}=18.5\ gm$$
    Mass of $$Fe=11.2\ gm$$
    Mass of $$Al_{2}O_{3}=10.2\ gm$$
    $$\therefore$$ Amount of unreacted $$Fe_{2}O_{3}$$
    $$=(5.4+18.5)-(11.2+10.2)\ gm$$
    $$=(23.9-21.4)\ gm$$
    $$=2.5\ gm$$
    Option $$a$$
  • Question 8
    1 / -0
    Which statement about a $$3$$p orbital is correct?
    Solution
    Solution:- (A) It can hold a maximum of $$6$$ electrons.
    Any $$p$$ orbital can hold a maximum of $$6$$ electrons.
    Hence statement $$\left( A \right)$$ is correct.
  • Question 9
    1 / -0
    What is the nucleos number of an atom?
    Solution
    Solution:- (B) The number of neutrons and protons in the nucleus
    The nucleon number or mass number of an atom is the sum of number of neutrons and protons in the nucleus.
  • Question 10
    1 / -0
    If rocket were fuelled with kerosene and liquid oxygen, what mass of oxygen would be required for every liter of kerosene? Assume kerosene to have the average composition $$C_{14}H_{30}$$ and density, $$0.792\ g/ml$$. 
    Solution
    $$d=\dfrac {m}{v}\quad v=1$$ litre $$=100\ ml$$
    $$m=d\times v=0.792\ g\times 1000\ ml$$
    $$=792$$ gram/mole
    moles of kerosene $$=\dfrac {792}{M}$$
    Molar mass $$=C_{14}H_{30}=14\times 12+30\times 1$$
    $$=198$$
    moles of kerosene $$=\dfrac {792}{198}$$
    $$=4$$ moles
    combusion of kerosene
    $$2C_1H_{30}+43O_2\longrightarrow 28CO_+30H_2O$$
    $$2$$ mole $$C_{14}H_{30}\equiv 43$$ mole $$O_2$$
    $$4$$ mole $$C_{14}H_{30}\equiv =43\times 4=85$$ mo,e $$O_2$$
    mass of $$O_{2}=86\times 32$$
    $$=2752$$ gram 
    $$=2.752\ kg$$
    $$\therefore $$ option $$"b"$$ correct
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