Question

from 10th std Chemistry 2nd chapter
please answer fastly if you know
anyone tell the right answer I choose it the brainliest answer​

Answer 1

Answer:

Acids, Bases And Salts

Understanding The Chemical Properties Of Acids And Bases

What Do All Acids And All Bases Have In Common?

How Strong Are Acid Or Base Solutions?

More About Salts

Answer 2

Answer:

Given, 1 gram atomic weight= 1mol

Q1. 12 g C= 1 mol

⇒   24 g C= 2 mol

Q2. 34 g NH₃ = ?

GMW of NH₃= (1*14) + (3*1) = 17 g

Given, 1 GMW= 1 mol

Therefore 34 g = 2 GMW= 2 MOL

Q3. 6×10²³= 1 mol

Therefore, 12 × 10²³= 2 mol

Q4. 22.4 l of any gas at STP = 1 mol

Therefore, 44.8 l of CO₂= 2 mol

Q5. In 48 g CH₄

a) no. of moles= ?

GMW of CH₄= (1*12) + (4*1)= 16 g= 1 mol

Therefore, 48 g= 3 mol

b) no. of GMW= ?

16 g= 1 GMW

48 g= 3 GMW

c) no. of molecules= ?

1 mol= 6× 10²³ molecules

3 mol= 18 × 10²³ molecules

d) no. of atoms= ?

In 1 mol → 6× 10²³ × 5 atoms

In 3 mol → 90 × 10²³ atoms

( Please note that in order to find out atoms in 1 mol of a molecule, you need to multiply the no. of molecules with atomicity. )

e) 22.4 l at STP= 1 mol

3 mol= 67.2 l at STP

Q6. 1 GMW of CO₂= 12 + 32= 44= 1 mol

In 2 mol, we will have 2 GMWs.

1n 1 mol, we have 6× 10²³ molecules.

In 2 mol, we will have 12 × 10²³ molecules.

No. of atoms= no. of molecules × atomicity= 6 × 10²³ × 3= 18×10²³ atoms

Vol. at STP for 2 mols= 44.8 l

3 mol NH₃= (3 × 17)g= 51 g= 3 GMW= 18× 10²³ molecules= 18 × 10²³ ×4 atoms= 72× 10²³ atoms= 67.2 l at STP