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Correct option(b)

Explanation:

In potentiometer experiment, galvanometer shows null deflection at balance point due to same potential at jockey and galvanometer circuit. 

Two bulbs marked 200 watt-250 volts and 100 watt-250 volts are joined in series to 250 volts supply. Power consumed in circuit is 67 watt.

In the cuprous oxide (Cu₂O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu⁺ ions is slightly less than twice the number of O²⁻ ions. This is because some Cu⁺ ions have been replaced by Cu²⁺ ions. Every Cu²⁺ ion replaces two Cu⁺ ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor. 

(i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.

 (ii) B (a group 13 element) is doped with Si (a group 14 element). So, there will be an extra electron and the semiconductor generated will be an n-type semiconductor. 

To prevent goitre. Deficiency of iodine may cause Goitre, a disorder affects on thyroid gland.

Cell constant = Conductivity x Resistance

 = 0.146 x 10^-3 S cm-¹ x 1500 = 0.219 cm^-1 

Because it forms a dense fog of sulphuric acid which does not condense easily.

Symbol of the outermost shell of magnesium (3^rd shell) = M 

No. of electrons in outermost shell of magnesium = 2. 

(I) An element X contains 35 electrons and 45 neutrons 

1. The number of protons must be equal to the number of electrons. So the number of protons = 35. 

2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.

3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. 

n = 4, 1 = 1, m= +1, s = -\(\frac{1}{2}\)

(II) Fe → Fe²⁺ + 3e^-

Fe (Z = 26) Fe³⁺ = number of electrons = 23

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom. 

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion. 

So, it contain 5 unpaired electrons. Mn (Z = 25). Electronic configuration is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Mn → Mn²⁺ + 2e^-

Number of unpaired electrons in Mn²⁺ = 5

Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ .

All orbitals are completely filled. So, no unpaired electrons in it.

(a) 1 mole of NH₃ = {1(14) + 3(1)} g of NH₃
= 17 g of NH₃
= 6.022× 1023 molecules of NH₃
Total number of protons present in 1 molecule of NH₃
= {1(7) + 3(1)}
= 10
Number of protons in 6.023 × 10²³ molecules of NH₃
= (6.023 × 10²³) (10)
= 6.023 × 10²⁴
⇒ 17 g of NH³ contains (6.023 × 10²⁴⁾ protons. Number of protons in 34 mg of NH³

6.022*10²⁴*34 mg/17000 mg

= 1.2046 × 10²²
(b) Mass of one proton = 1.67493 × 10^–27 kg
Total mass of protons in 34 mg of NH³
= (1.67493 × 10^–27 kg) (1.2046 × 10²²)
= 2.0176 × 10^–5 kg

¹³₆C:
Atomic mass = 13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7

^16₈ O

Atomic mass = 16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8

^24₁₂ MG

Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24 – 12 = 12

^56 ₂₆ Fe

Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30

⁸⁸₃₈ Sr

Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50

Pure calcium oxide or lime, is an amorphous white solid, having a high melting point (2273K). When lime is added to water, a hissing sound is produced and a large amount of heat is generated.

Cao + H₂O------> Ca(OH)₂ + Heat
Therefore, the correct answer is (a).

Metals are good conductors of electricity because they contain free electrons. These free electrons move easily through the metal and conduct electric current'

Copper is a much better conductor of heat than steel. Also, copper does not react with water at all but steel (an alloy of iron) reacts with steam. So, copper is used to make hot water tanks but not steel (an alloy of iron).

(a) When sulphur is heated in air, sulphur dioxide gas is formed.
(i) Sulphur dioxide has no action on dry litmus paper.
(ii) Sulphur dioxide turns blue litmus paper to red.
(b) S(s) +O₂(g) +SO₂(g)
Sulphur     Orygen       Sulphur dioxide

Non-metals react With oxygen to form acidic oxides or neutral oxides. Carbon forms an acidic oxide CO₂ , sulphur forms an acidic oxide SO₂ and hydrogen forms a neutral oxide H₂O Carbon monoxide (CO). nitrous oxide (N₂O) and nitric oxide (NO) are also the examples of neutral oxides.

Platinum, gold and silver are used to make jewellery because of their bright shiny surface and high resistance to corrosion. Also, they have high malleability and ductility.

Magnlium contains  Mg+Al

NaNO₃ on heating gives  O₂

Concept:

The mode is the value that appears most often in a set of data values.

Calculation:

32 occurred 8 times

14 occurred 4 times

59 occurred 12 times

41 occurred 8 times 

28 occurred 10 times

7 occurred 16 times 

34 occurred 15 times

20 occurred 9 times

∴ Mode will be 7

Given: 

Data is 

21, 22, 24, 25, 25, 19, 26

Concept: 

The mode is the value that appears most often in a set of data.

Calculation: 

Here, we see for the given data 25 comes two times 

∴ The mode of the given data is 25

Given:

The given data,

31, 33, 37, 31, 43, 33, 45, 34, and 33

Concept used:

Range of the data is the difference of the largest value and the smallest value

Mode is the most repeated value in a data

Calculations:

Arranging the values in ascending order

31, 31, 33, 33, 33, 34, 37, 43, 45

Smallest value = 31        

Largest value = 45

Range = 45 – 31

⇒ Range = 14

The most repeated term,

Mode = 33

∴ The range and the mode of the given data is 14 and 33 respectively.

The reaction takes place as: 

Ca + 1/2​O₂​→CaO

CaO + H₂​O→Ca(OH)_2​

Atomic weight of Ca = 40

molecular weight of CaO = 56

molecular weight of Ca(OH)₂​ = 74

40 g of Ca produces 74grm of Ca(OH)₂​

1 g of Ca is equivalent to 74/40 ​gm of  Ca(OH)₂​ = 1.85gm.

So 20 g of Ca = 20×1.85 = 37

Or 37/37 = 1 gm equivalent of  Ca(OH)₂​ (Equivalent weight of Ca(OH)₂​ is 37)

Therefore 1 gm is dissolved in 1 litre

Hence in two litre it will be 

So the normality of alkaline solution is 2.0 N