Answer is: d). 1-propanol. Because between two molecules of 1-propanol can for form hydrogen bond, between hydrogen and oxygen in hydroxy group.
1) Chemical formula for propanal is CH₃-CH₂-CH=O. Propanal is a saturated three carbon aldehyde (have a carbonyl center).
2) Chemical formula for propane is CH₃-CH₂-CH₃. Propane is a three carbon alkane (acyclic saturated hydrocarbon).
3) Chemical formula for propanone is (CH₃)₂-C=O. Propanone or acetone is he simplest and smallest ketone.
4) Chemical formula for propanol is CH₃-CH₂-CH₂-OH. 1-propanol is a primary alcohol.
Hydrogen bond is
an electrostatic attraction between two polar groups that occurs when
a hydrogen atom (H), covalently bound to a
highly electronegative atom such as flourine (F), oxygen (O) and
nitrogen (N) atoms.
Answer:
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Explanation:
Let's consider the oxidation and reduction half-reactions and the global reaction.
Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻
Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)
Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)
The standard 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
E° = 0.771 V - 0.154 V = 0.617 V
The Nernst equation allows us to calculate the cell potential (E) under the given conditions.
![E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V](https://tex.z-dn.net/?f=E%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20logQ%5C%5CE%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20log%5Cfrac%7B%5BSn%5E%7B%2B4%7D%5D.%5BFe%5E%7B2%2B%7D%5D%7D%7B%5BSn%5E%7B2%2B%7D%5D.%5BFe%5E%7B3%2B%7D%20%5D%7D%20%5C%5CE%3D0.617V-%5Cfrac%7B0.05916%7D%7B2%7D%20log%5Cfrac%7B%280.13%29.%280.0037%29%7D%7B%280.0023%29.%280.11%29%7D%20%5C%5CE%3D0.609V)
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Answer:
34.6 cm³
Explanation:
<em>A chemistry student needs 55.0 g of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of carbon tetrachloride is 1.59 g/cm³. Calculate the volume of carbon tetrachloride the student should pour out. Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of carbon tetrachloride (m): 55.0 g
- Density of carbon tetrachloride (ρ): 1.59 g/cm³
Step 2: Calculate the required volume of carbon tetrachloride
Density is an intrinsic property of matter. It can be calculated as the quotient between the mass of the sample and its volume.
ρ = m/V
V = m/ρ
V = 55.0 g/(1.59 g/cm³)
V = 34.6 cm³
The chemistry student should pour 34.6 cm³ of carbon tetrachloride.