Of the factors that influence diffusion of respiratory gases, the most variable and, therefore, important factor to consider is the
a) membrane thickness.
b) diffusion distance.
c) membrane surface area.
d) electrical charge.
e) concentration gradient.

a. 2 m

b. 0.15 m

Explanation:

(a) By what deck does the ship sink in fresh.

water, when it loads a cargo of 4000 tonnes

We know that the upward force , U on the ship equal the weight of fresh water displaced, W = ρVg where ρ = density of fresh water = 1000 kg/m³, V = volume of water displaced and g = acceleration due to gravity.

So, U = W = ρVg

This upward force must equal the weight of the ship, W' = mg where m = mass of ship = 4000 tonnes = 4000 × 1000 kg = 4 × 10⁶ kg

So. U = W'

ρVg = mg

V = m/ρ = 4 × 10⁶ kg/10³ kg/m³ = 4 × 10³ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in water h.

So, V = Ah

Thus h = V/A = 4 × 10³ m³/2000 m³ = 2 m

So, the ship sinks to a depth of 2 m in fresh water.

(b) if the ship + Cargo has a displacement tonnage

of 12300 tonnes; by what amount will the ship

rise in the water when it sails from fresh water

into Seawater (density of Sea water - 1025kgm⁻³​

We know that the upward force , U' on the ship in fresh water equals the weight of fresh water displaced, W" = ρV'g where ρ = density of fresh water = 1000 kg/m³, V' = volume of water displaced and g = acceleration due to gravity.

So, U = W" = ρV'g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U' = W₁

ρV'g = m₁g

V' = m₁/ρ = 1.23 × 10⁷ kg/10³ kg/m³ = 1.23 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in fresh water h'.

So, V' = Ah'

Thus h = V'/A = 1.23 × 10⁴ m³/2000 m³ = 6.15 m

So, the ship sinks to a depth of 0.6 m in fresh water.

Also,

We know that the upward force , U" on the ship in sea water equals the weight of sea water displaced, W₂ = ρ'V"g where ρ' = density of sea water = 1025 kg/m³, V"= volume of water displaced and g = acceleration due to gravity.

So, U" = W₂ = ρ'V"g

This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg

So. U" = W₁

ρ'V"g = m₁g

V" = m₁/ρ' = 1.23 × 10⁷ kg/1.025 × 10³ kg/m³ = 1.2 × 10⁴ m³

Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in sea water h".

So, V" = Ah"

Thus h" = V'/A = 1.2 × 10⁴ m³/2000 m³ = 6 m

So, the ship sinks to a depth of 6 m in fresh water.

So, the rise in height from fresh water to sea water is Δh = h' - h" = 6.15 m - 6 m = 0.15 m