An air bubble with a radius of 2.5 cm at the bottom of a lake where the temperature is 5.4 0c and the pressure is 3.2atm rises to the surface, where the temperature is 25.0 0c and the pressure is 1.0 atm. calculate the radius of the bubble when it reaches the surface. assume ideal gas behavior.

3.8 cm

Explanation:

Given data

Initial pressure (P₁): 3.2 atmInitial volume (V₁): 65 cm³

Volume of a sphere: V = 4/3 × π × r³ = 4/3 × π × (2.5 cm)³ = 65 cm³

Initial temperature (T₁): 5.4°C + 273.15 = 278.6 KFinal pressure (P₂): 1.0 atmFinal volume (V₂): ?Final temperature (T₂): 25.0°C + 273.15 = 298.2 K

We can find the final volume using the combined gas law.

The final radius of the bubble is:

V = 4/3 × π × r³

223 cm³ = 4/3 × π × r³

r = 3.8 cm

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na

Ithink that the answer is b