Answer:
C₆H₁₂O₆ < NaCl ≈ CH₃OH < MgCl₂
Step-by-step explanation:
The formula for <em>osmotic pressure</em> (Π) is
Π = icRT
If T is constant,
Π = kic
<em>C₆H₁₂O₆:
</em>
Π = k × 1 × 1 = k
<em>MgCl₂:
</em>
Π = k × 3 × 1 = 3k
<em>CH₃OH:
</em>
Π = k × 1 × 2 = 2k
<em>NaCl:
</em>
Π = k × 2 × 1 = 2k
The order of osmotic pressures is
C₆H₁₂O₆ < NaCl ≈ CH₃OH < MgCl₂
The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.
Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr
Vapor pressure of pure water at 
=25.81mmHg
= 
According to Dalton's law of partial pressures,
Total pressure = Partial pressure of Oxygen gas + Partial pressure of water
749 Torr = Partial pressure of Oxygen gas + 25.81 Torr
Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr
Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr
Answer:
The concentration of chloride ion is 
Explanation:
We know that 1 ppm is equal to 1 mg/L.
So, the 
content 100 ppm suggests the presence of 100 mg of in 1 L of solution.
The molar mass of is equal to the molar mass of Cl atom as the mass of the excess electron in is negligible as compared to the mass of Cl atom.
So, the molar mass of is 35.453 g/mol.
Number of moles = (Mass)/(Molar mass)
Hence, the number of moles (N) of present in 100 mg (0.100 g) of is calculated as shown below:
So, there is 
of present in 1 L of solution.
I think the correct answer from the choices listed above is the last option. The family that contains disinfectants is the halogens. This group includes the elements fluorine, chlorine, bromine, iodine, and astatine. Disinfectants are chemical agents that inhibit or kill microorganisms on inanimate objects.
Answer is: empirical formula for compound is C₂H₆O.
m(CO₂) = 1,6003 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 1,6003 g ÷ 44 g/mol.
n(CO₂) = 0,0364 mol.
n(CO₂) = n(C).
m(C) = 0,0364 mol · 12 g/mol = 0,4368 g.
m(H₂O) = 0,9827 g.
n(H₂O) = 0,9827 g ÷ 18 g/mol.
n(H₂O) = 0,0546 mol.
n(H) = 2 · n(H₂O) = 0,1092 mol.
m(H) = 0,1092 mol · 1 g/mol = 0,1092 g.
m(O) = 0,8376 g - 0,4368 g - 0,1092 g = 0,2916 g.
n(O) = 0,2916 g ÷ 16 g/mol = 0,0182 mol.
n(C) : n(H) : n(O) = 0,0364 mol : 0,1092 mol : 0,0182 mol.
n(C) : n(H) : n(O) = 2 : 6 : 1.
