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.
Since phosphoric acid is H3PO4, which is known from PO4, with a charge of 3- so 3 hydrogen would balance it out, and sodium hydroxide is NaOH, it can be assumed that it results in H3(OH)3 + Na3PO4.
Answer:
The answer to your question is P1 = 1.4 atm
Explanation:
Data
Volume 1 = V1 = 735 ml
Pressure 1 = P1 = ?
Volume 2 = V2 = 1286 ml
Pressure 2 = P2 = 0.8 atm
Process
To solve this problem use Boyle's law.
P1V1 = P2V2
-Solve for P1
P1 = P2V2 / V1
-Substitution
P1 = (0.8 x 1286) / 735
-Simplification
P1 = 1028.8 / 735
-Result
P1 = 1.4 atm
Water's specific heat capacity is 4200 J/Kg°C
95-28=67
72.5grams in kg is 0.0725kg
Energy = 67×0.0725×4200
Energy = 20,401.5 J or 20.4015 kJ
Answer: 60.7 g of 
will be formed.
Explanation:
To calculate the moles :
The balanced chemical reaction is
is the limiting reagent as it limits the formation of product and 
is the excess reagent.
According to stoichiometry :
6 moles of produce = 4 moles of
Thus 2.68 moles of will produce=
of
Mass of 
Thus 60.7 g of will be formed by reactiong 60 L of hydrogen gas with an excess of
