Answer:
Silver, 0.239 J/(g °C)
Explanation:
- The heat change is related to specific heat as given by the formula;
Heat change = mass of substance × specific heat × change in temperature
- Therefore; considering same amount of substance or equal masses and have the same initial temperature.
- The change in temperature will be inversely proportional to the specific heat.
- Therefore; the higher the specific heat lower the temperature change.
- Hence, the change in temperature will be highest for the substance with the lowest specific heat.
Therefore; the one that will increase in temperature the most is Silver
The electron configuration of V³⁺ is [Ar]
. The ion is paramagnetic because it has two unpaired electrons
<h3>
What is paramagnetic?</h3>
- A weak magnetic field supplied externally can weakly attract some materials, which then create internal magnetic fields that are directed in the same direction as the applied magnetic field. This phenomenon is known as paramagnetic.
- Diamagnetic materials, in contrast, are attracted to magnetic fields and produce induced magnetic fields that are directed in the opposite direction from the applied magnetic field.
- The majority of chemical elements and some compounds are considered to be paramagnetic materials.
- Paramagnetic materials have a relative magnetic permeability that is somewhat more than 1, which makes them attracted to magnetic fields.
- The applied field induces a linearly decreasing magnetic moment that is relatively weak.
- Modern experiments on paramagnetic materials are frequently done with a sensitive analytical balance since it typically requires a sensitive analytical balance to identify the effect.
To learn more about paramagnetic with the given link
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Answer:

Explanation:
Hello there!
In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

Therefore, the partial pressure of chloroform turns out to be:

Regards!
Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J