A substance is 35.7% carbon by mass. How much carbon could be recovered from 769 g of the substance? 1. 180 mol 2. 258 mol 3. 22

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

The half-reactions are:

Anode oxidation half reaction:  

Cathode reduction half reaction:  

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature =

n = electrons exchanged in oxidation-reduction = 2

= standard electrode potential for the cell = +0.63 V

= cell potential for the reaction =?

= concentration of Zn²⁺ = 3.5 M

=

Now substituting all known values into the equation, we arrive at:

So, the resulting cell potential for this reaction is 0.50 V

Answer:

The response is provided below.

Explanation:

Numerous aspects can influence the actual results of titration. These factors vary from human error to misjudging measurements, a researcher's interpretation of color changes, and improper techniques during the experimental procedure.

Thus, to mitigate these errors, researchers must engage thoroughly throughout experimentation, and employing gross readings can assist in reducing mistakes when determining the final titre value.

Answer:

3.816 × 10⁻³ M

Explanation:

A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?

We can derive the following relations:

• The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
• Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.

The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:

The molarity of Cu²⁺ is:

Vapor pressure refers to the force exerted by vapor or gas molecules above the surface of a liquid. It is inversely related to the concentration of solute particles; an increase in solute concentration results in a decrease in vapor pressure, and vice versa. For (a), it dissociates into two particles. In (b), the total count of particles from dissociation becomes 1 + 2, totaling three. For (c), dissociation yields 1 + 3 for a total of four particles. (d) Since sucrose is a covalent compound, it does not break apart into ions, so it remains as one particle. For (e), dissociation results in 1 + 1, equating to two particles.