Answer:
Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?
a. Al (s)
b. H2O (l)
c. HCN (g)
d. CH3COOH (l)
e. C2H6 (g)
Explanation:
Entropy is the measure of the degree of disorderness.
In solids, the entropy is very less compared to liquids and gases.
The entropy order is:
solids<liquids<gases
Among the given substances, water in liquid form has a strong intermolecular H-bond.
So, it has also less entropy.
Next acetic acid.
Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.
HCN has slight H-bonding in IT.
Hence, the entropy order is:
Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)
Answer:
Explanation:
When filling a burette for a titrant, adjust the burette so that the opening is near or below the eye leve preferably over the sink.
Then, use a funnel to add the titrant into the burette.
The titrant should be filled almost to the zero mark.
Answer : Below are three ways described in brief for improving the quality of indigenous soap.
1) Add virgin coconut oil which serves as a softener and betel leaf extract. It is used to improve the quality of soap as an antimicrobial agent.
2) Add activated charcoal black granules or powder to the soap, it removes maximum dirt from the skin.
3) Adding a filler which will increase the shelf life of usage of the soap will also improve its quality.
Answer:
Al
Explanation:
4 Al + 3 O₂ → 2 Al₂O₃
You need to figure out which one has the smaller mole ratio. Convert both substances from grams to moles.
(10.0 g Al)/(26.98 g/mol) = 0.3706 mol Al
(19.0 g O₂)/(32.00 g/mol) = 0.5938 mol O₂
Now, use the mole ratios of reactant to product to see which substance produces the least amount of product.
(0.3706 mol Al) × (2 mol Al₂O₃/4 mol Al) = 0.1853 mol Al₂O₃
(0.5938 mol O₂) × (2 mol Al₂O₃/3 mol O₂) = 0.3958 mol Al₂O₃
Since aluminum produces the least amount of product, this is the limiting reagent.
Answer:
1.18 V
Explanation:
The given cell is:
Half reactions for the given cell follows:
Oxidation half reaction: 
Reduction half reaction: 
Multiply Oxidation half reaction by 2 and Reduction half reaction by 3
Net reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAl%5E%7B3%2B%7D%5D%5E2%7D%7B%5BFe%5E%7B2%2B%7D%5D%5E3%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +1.21 V
n = number of electrons exchanged = 6
Putting values in above equation, we get:
