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701.

Ravi was performing some experiements related to the laws of chemical combination in the science laboratory under the guidance of his chemistry teacher Mr. John. Ravi found that when he burned 1 gram of hydrogen gas in 8 grams of oxygen gas in a closed vessel, he obtained 9 grams of water. He repeated this experiment many times but obtained the same results every time (a) Write a balanced chemical equation for the reaction between hydrogen and oxygen to form water. Also write the names of all the substances involved below their formulae in the equation (b) What are the reactants and products in the above reaction ? (c) Which law of chemical combination is illustrated by the fact that when Ravi burned 1g of hydrogen in 8g of oxygen, he obtained 9g of water ? (d) What mass of water will be obtained if 1g of hydrogen is burned in 10g of oxygen ? Which law of chemical combination will govern your answer ? (e) What values are displayed by Ravi in this episode ?

Answer» (a) The balanced chemical equation for the reaction between hydrogen and oxygen to form water can be written as :
`underset("Hydrogen")(2H_(2))+underset("Oxygen")(O_(2))rarrunderset("Water")(2H_(2)O)`
(b) (i) The reactants in this reaction are : Hydrogen and Oxygen
(ii) The product in this reaction is : Water
(c) Here, Mass of reactants = Mass of hydrogen + Mass of oxygen
`= 1g +8g`
`9g`
And, Mass of product = Mass of water
`= 9g`
Now, since the mass of product (9g) is equal to the mass of reactants (9g), therefore, the given experiemental data illustrates the law of conservation of mass in chemical reactions
(d) Out answer will be governed by the law of constant proportions. It has been given in this question that when 1g of hydrogen is burned in 8g of oxygen of oxygen, then 9g of water is always obtained. Now, since hydrogen and oxygen combine in the fixed proportion of 1 : 8 by mass to produce 9g of water, therefore, the same mass of water (9g water) will be obtained even if we burn 1g of hydrogen in 10g of oxygen. The extra oxygen (10 - 8 = 2g oxygen) will remain unreacted in this case
(e) The various values displayed by Ravi in this episode are (i) Knowledge of the law of conservation of mass and law of constant proportions in chemical reactions and (ii) Application of knowledge in solving problems.
Discussion
702.

Experiements on the emission spectra of______atoms gave rise to a fourth quantum number describing an electron in an atom. (i) `H` (ii) `Na` (iii) `Ag` (iv) `I`A. `(i),(iv)`B. `(ii),(iii)`C. `(i),(ii),(iii),(iv)`D. `(ii),(iii),(iv)`

Answer» Correct Answer - C
Each has one unpaired electron. Experiments revealed that lines in the emission spectra could be split by the application of an external magnetic field. The only ways scientists could explain these results was to assue that electrons behave like tiny bar magnets with north and south poles. If electrons are thorught of as spinning on their own axes, as the earth does, their magnetic properties can be accounted for. According to the electromagnetic theory, a spinnig charge generates a magnetic field, and it is this motion that causes the electron to behave like a magnet. There are two possible spinning motions of an electron-one clockwise and the other counterclockwise. to take the electron spin into account, we use a fourth quantum number called the electron spin quantum number `(m_(s))`, which has a value of `+1//2` or `-1//2`. These values corresponds to the two possible spinning motions of the electron.
Discussion
703.

The electronic configuration of valence shell of Cu is `3d^(10)4s^(1)` and not `3d^(9)4s^(2)`. How is this configuration explained?

Answer» Configuration either exactly half-filled or fully filled orbitals are more stable due to symmetrical distribution of electrons and maximum exchange energy in `3d^(10)4s^(1)` d-orbitals are completely filled and s-orbital is half-filled. Hence it is more stable configuration.
Discussion
704.

Assertion : The outer electronic configuration of Cr and Cu are `3d^(5) 4s^(1) ` and `3d^(10) 4s^(1)` respectively . Reason : Electrons are filled in orbitals in order of increasing energies given by (n+l) rule.A. if both assertion and reason are correct and reason is correct explanation for assertion .B. If both assertion and reason are correct but reason in not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect.

Answer» Correct Answer - B
Correct explanation. This is because completley filled and half-filled electronic configurations impart extra stability due to symmetrical distribution of electrons and more exchange energy.
Discussion
705.

An atom having atomic mass number 13 has 7 neutrons. What is the atomic number of the atom.

Answer» An atom having atomic mass number 13 and number of neutrons 7.
i.e. `A=13,n=7`
As we know that A=n+p
`therefore` p=A-n=13-7=6
Hence, z= p =6
Discussion
706.

Wavelength of different ra diations are given below. `lambda` (A) ` = 300nmlambda` (B) =300 `mumlambda` (C) `=3nmlambda` (D)=`30overset(@)A` Arrange these radiations in the increasing order of their energies.

Answer» `lambda`(A)`=300nm=300xx10^(-9)m,lambda` (B) `=300mum=300xx10^(-6)m`
`lambda` (C)=`3nm=3xx10^(-9) m lambda` (D) `= 30overset(@)A= 30xx10^(-10)m=3xx10^(-9)m`
Energy `E=(hc)/(lambda)`
Therefore `E=(1)/(lambda)`
increasing order of energy is `BltAltC=D`
Discussion
707.

Out of 3d and 4s orbitals, which is filled first ?

Answer» 4s orbital is filled first because it has lower energy. The energies of the orbitals can be compared by their n + l values. For 4s orbital, n +l(4 + 0) value is 4 while for 3d orbital, n + l (3 + 3) value is 5. Therefore 4s orbital is filled before 3d orbital. However, this is true only for multielectron atom. In the excited states of one electron species `(H, He^(+), Li^(2+))`, 3d orbital will be filled before 4s orbital.
Discussion
708.

Table-tennis ball has mass 10g and s peed of 90m/s. if speed can be meausred within an accuracy of 4%. What will be the uncertainly in speed and position?

Answer» Uncertainty in speed of ball `(Deltav)=(90xx4)/(100)=(360)/(100)=3.6 ms^(-1)`
Uncertainty in position `(Delta x)=(h)/(4pi m Delta v)`
`=((6.626xx10^(-34)"kg" m^(2)s^(-1)))/(4xx3.143xx(10^(-3)"kg")xx(3.6 m s^(-1)))`
` =1.46xx10^(-33)m`
Discussion