What mass of oxygen is needed for the complete combustion of 1.60-10^-3 g
of methane?
Express your answer with the appropriate units.

6.4×10¯³ g of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4

= 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂

Thus, 6.4×10¯³ g of O₂ is needed for the reaction.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.