For this problem, we should use the Henry's Law formula which is written below:
P = kC
where
P is the partial pressure of the gas
k is the Henry's Law constant at a certain temperature
C is the concentration
Substituting the values,
1.71 atm = (7.9×10⁻⁴<span> /atm)C
Solving for C,
C = 2164.56 molal or 2164.56 mol/kgwater
Let's make use of density of water (</span>1 kg/1 m³) and the molar mass of NF₃ (71 g/mol).<span>
Mass of NF</span>₃ = 2164.56 mol/kg water * 1 kg/1 m³ * 1 m³/1000000 mL * 150 mL * 71 g/mol = 23.05 g
Mass of water required = 64.42 g
<h3>Further explanation</h3>
Given
Reaction
6CO2 + 6H2O → C6H12O6 + 6O2
157.35 g CO2
Required
mass of water
Solution
mol CO₂ = mass : MW CO₂
mol CO₂ = 157.35 g : 44.01 g/mol
mol CO₂ = 3.575
From the equation, mol ratio CO₂ : H₂O = 6 : 6, so mol H₂O = mol CO₂=3.575
Mass H₂O = mol x MW H₂O
mass H₂O = 3.575 x 18.02
mass H₂O = 64.42 g
Answer:
Halogen, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts).
