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).
A) James Cook.
B) He put his sailors on a strict diet to see if they would get scurvy.
C) Sauerkraut.
D) He told others of this diet and that none of his sailors died of scurvy.
E) Chemicals can be found almost anywhere and almost anyone can be a scientist.
<h3>Answer:</h3>
Mass = 96.47 g
<h3>
Solution:</h3>
Data Given:
M.Mass = 28.97 g.mol⁻¹
Moles = 3.33 mol
Mass = ??
Formula Used:
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting values,
Mass = 3.33 mol × 28.97 g.mol⁻¹
Mass = 96.4701 g
Rounding to four significant numbers,
Mass = 96.47 g
Convert each amount of grams into moles:
I: 23.24g x 1 mol / 126.90g = 0.1831 mol I
C: 2.198 x 1 mol / 12.01g = 0.1830 mol C
N: 2.562 x 1 mol / 14.01g = 0.1829 mol N
Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1
Therefore the empirical formula is ICN
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
