Question

When x molecules are removed from 200 mg of n2o 2.89*10^-3 moles of n20 are left x will be?

Answer 1

Answer:

molecular mass of N₂O = 44

weight of N₂O = 200mg = 0.2g

moles of N₂O present = 0.2/44 = 4.55×10^(-3)

Let moles of N₂O removed = y

moles of N₂O remained = 2.89×10^(-3)

Thus 4.55×10^(-3) - y = 2.89×10^(-3)

⇒ -y = 2.89×10^(-3) - 4.55×10^(-3)  

⇒ y = 4.55×10^(-3) - 2.89×10^(-3)

⇒ y = 1.66×10^(-3) moles

No. of molecules in 1 mole = 6.022×10^(23)

No. of molecules in y moles = y×6.022×10^(23) = 1.66×10^(-3)×6.022×10^(23)                                                =9.97×10^(20)  

Thus  x = 9.97×10^(20) molecules

Explanation:

Answer 2

Answer:

olecular mass of N₂O = 44

weight of N₂O = 200mg = 0.2g

moles of N₂O present = 0.2/44 = 4.55×10^(-3)

Let moles of N₂O removed = y

moles of N₂O remained = 2.89×10^(-3)

Thus 4.55×10^(-3) - y = 2.89×10^(-3)

⇒ -y = 2.89×10^(-3) - 4.55×10^(-3)  

⇒ y = 4.55×10^(-3) - 2.89×10^(-3)

⇒ y = 1.66×10^(-3) moles

No. of molecules in 1 mole = 6.022×10^(23)

No. of molecules in y moles = y×6.022×10^(23) = 1.66×10^(-3)×6.022×10^(23)                                                =9.97×10^(20)  

Explanation: