Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.
T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
Δn = 0,589 mol - 0,2356 mol = 0,3534 mol.
The mass percentage of water (H₂O) in cadmium chloride tetra hydratesolution is equal to the 28.2%.
<h3>How do we calculate mass percentage?</h3>
Mass percentage of any substabce present in any solution will be calculated as:
Mass percent = (Mass of substance/Total mass of solution)×100%
According to the question,
Mass of water (H₂O) = 18.02 g/mol
Mass of solvent (CdCl₂.4H₂O) = 183.32 + 4(18.02) = 255.4 g/mol
On putting values, we get
Mass percent = (18.02 / 255.4) × 100% = 28.2%
Hence mass percent of water is 28.2%.
To know more about mass percent, visit the below link:
brainly.com/question/13896694
Moles of MgS2O3 = 223/molar mass of MgS2O3
= 223/136.42
= 1.634 moles.
Hope this helps!
