Question

★ Science problem !!

1: What are $\sf{\alpha -particles }$ ?

2: What is the Valency of Magnesium ?

★ Maths Problem !!

1: A room's height is double it's breadth , and it's Length is three times it's breadth . Let the breadth be 11 m . Find the area of whitewashing the room excluding its ceiling.

§ Don't Spam dear users
§ I will appreciate good answerers​

Answer 1

1) Alpha particles, also called alpha rays or alpha radiation, consist of two protons and two neutrons bound together into a particle identical to a helium-4 nucleus. They are generally produced in the process of alpha decay, but may also be produced in other ways.

2) Magnesium has valency equal to 2 + because the electronic configuration of Mg is [2,8,2]. The nearest noble gas to magnesium is neon with electronic configuration of [2,8], to achieve this stable electronic configuration Mg can lose 2 valence electrons, hence its valency is 2 + .

That's all i know .

I hope it's helpful for you.

Answer 2

(1.) Science Questions :-

1: What are $\sf{\alpha -particles }$ ?

2: What is the Valency of Magnesium ?

____________________

Answers :-

Answer 1 :-

Alpha particles, also alpha radiation are the particles that consist of two protons and two neutrons bound together into a particle identical to the nucleus of a helium-4 atom.

• Production :- Generally, alpha particles are produced by the 'alpha decay method'.

• Symbols :-  $\sf{ \alpha , \: {\alpha }^{2 + } , \: { He}^{2 + } }$

• Composition :- 2 protons and 2 electrons

• Discovered by :- Earnest Rutherford

• Velocity :- 2,00,00,000 m/sec

__________

Answer 2 :-

In the periodic table, Magnesium is 12th element, therefore it's electronic configuration is [ 2, 8, 2 ]. Since the valence electrons or the electrons in the valence shell of Magnesium is 2, the valence of Magnesium also turns out to be 2 +. While making a bond, Magnesium tries to lose its 2 valence electrons, thus again, making it's valency 2 +.

__________

___________________

(2.) Maths Question :-

A room's height is double it's breadth , and it's Length is three times it's breadth . Let the breadth be 11 m . Find the area of whitewashing the room excluding its ceiling.

____________________

Solution :-

Given Information :-

• Breadth of room = 11 m
• Length of room = Three times the breadth of the room.
• Height of the room = Two times the breadth of the room.

To Find :-

• Area of whitewashing the room excluding the ceiling.

Calculation :-

• Calculation of Length :-

Since it is told that the length of the room is three times the breadth of the room,

⇒ Breadth = 11 m ( given information )

⇒ Length = Three times Breadth

⇒ Length = 3 × 11 m

⇒ Length = 33 m

• Calculation of Height :-

Since it is told that the height of the room is two times the breadth of the room,

⇒ Breadth = 11 m ( given information )

⇒ Height = Two times the Breadth

⇒ Height = 2 × 11 m

⇒ Length = 22 m

• Calculation the area of the room ( Including the ceiling ) :-

Formula Used :-

Lateral Surface Area of cuboid = 2 × h ( l + b )

Substituting the values calculated above,

⇒ Lateral Surface Area of cuboid = 2 × 22 ( 33 + 11 ) m²

⇒ Lateral Surface Area of cuboid = 2 × 22 ( 44 ) m²

⇒ Lateral Surface Area of cuboid = 44 × 44 m²

⇒ Lateral Surface Area of cuboid = 1936 m²

∵ We know, that while whitewashing, the floor of a room isn't engaged.

∴ Area of room if the ceiling is excluded is calculated my the formula of 'Lateral Surface Area of cuboid' which is mentioned above.

Answer :- Area of room whitewashed is 1936 m².

____________________

Final Answer :-

Area of room whitewashed is 1936 m².

____________________