H2O2(I)
C6H6(O)
CO2(I)
C2H6(O)
HNO3(I)
<span>B) phosphodiester </span> is the correct answer
The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)

<u>Answer:</u> The number of moles of gas present is 0.276 moles
<u>Explanation:</u>
To calculate the number of moles of gas, we use the equation given by ideal gas:
PV = nRT
where,
P = Pressure of the gas = 725 mm Hg
V = Volume of the gas = 7.55 L
n = number of moles of gas = ?
R = Gas constant = 
T = Temperature of the gas = 
Putting values in above equation, we get:

Hence, the number of moles of gas present is 0.276 moles