Answer is: glycerol because it is more viscous and has a larger molar mass.
Viscosity depends on intermolecular interactions.
The predominant intermolecular force in water and glycerol is hydrogen bonding.
Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).
Answer:
V₂ = 1.5 L
Explanation:
Given data:
Initial volume of balloon = 1.76 L
Initial temperature = 295 K
Final temperature = 253.15 K
Final volume = ?
Solution:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 1.76 L ×253.15 K / 295 K
V₂ = 445.54 L.K /295 K
V₂ = 1.5 L
Explanation:
Mass of the organic compound = 200g
Mass of carbon = 83.884g
Mass of hydrogen = 10.486g
Mass of oxygen = 18.640g
The mass of nitrogen = mass of organic compound - (mass of carbon + mass of hydrogen + mass of oxygen)
Mass of nitrogen = 200 - (83.884 + 10.486 + 18.64) = 200 - 113.01
Mass of nitrogen = 86.99g
The empirical formula of a compound is its simplest formula.
It is derived as shown below;
C H O N
Mass 83.884 10.486 18.64 86.99
molar
mass 12 1 16 14
Moles 83.884/12 10.486/1 18.64/16 86.99/14
6.99 10.49 1.17 6.21
Divide
by
lowest 6.99/1.17 10.49/1.17 1.17/1.17 6.21/1.17
6 9 1 5
Empirical formula C₆H₉ON₅
learn more:
Empirical formula brainly.com/question/2790794
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Answer: At equilibrium, the partial pressure of 
is 0.0330 atm.
Explanation:
The partial pressure of is equal to the partial pressure of 
. Hence, let us assume that x quantity of 
is decomposed and gives x quantity of and x quantity of .
Therefore, at equilibrium the species along with their partial pressures are as follows.
At equilibrium: 0.123-x x x
Now, expression for 
of this reaction is as follows.
![K_{p} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}\\0.0121 = \frac{x \times x}{(0.123 - x)}\\x = 0.0330](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPCl_%7B3%7D%5D%5BCl_%7B2%7D%5D%7D%7B%5BPCl_%7B5%7D%5D%7D%5C%5C0.0121%20%3D%20%5Cfrac%7Bx%20%5Ctimes%20x%7D%7B%280.123%20-%20x%29%7D%5C%5Cx%20%3D%200.0330)
Thus, we can conclude that at equilibrium, the partial pressure of is 0.0330 atm.
