step to step explanation and answer neetly
Answers
question 3 will be SAME
Question 4 will be 10^20 MOLECULES
BECAUSE
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)
HOPE THIS HELPS
THE FIRST QUESTION IS LOGIC
2) 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)
1) Moles of ch4 = 0.1
Number of e = 0.1 na
Moles of h20 = 0.1
Number of e = 0.1 Na
Total = 0.2 Na
Explanation: In a neutral atom, atomic number represents the number of protons inside the nucleus and equal number of electrons around it.