Two substances in a mixture differ in density and particle size. These properties can be used to

Answer:

Complete Question:  

Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.

In comparison to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is

ANSWER : lower, since the London dispersion forces among C2H5OH molecules surpass those among CH3OH molecules.

Explanation:

To clarify the answer provided, let’s begin by defining some concepts.

The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.  

The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.

The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.

Note: The London dispersion force for C2H5OH is more substantial than for CH3OH because C2H5OH has more electrons than CH3OH.

To achieve the cancellation of electrons, the oxidation half-reaction needs to be multiplied by 4 while the reduction half-reaction must be multiplied by 3. Explanation: The oxidation reaction accounts for the loss of electrons, increasing the oxidation state, while the reduction implies gaining electrons, leading to a decrease in oxidation state. The respective half-reactions illustrate this, confirming that multiplying the oxidation by 4 and the reduction by 3 achieves the desired effect.

Option d is the correct choice, as both belong to the alkali metals category (group one).

Let's represent molecules with symbols as follows:

                   C₂O₄  =  X
and
                    H₂O  =  Y
Then,
                                K [ Co (X)₂ (Y)₂ ]

Since Potassium (K) has an oxidation number of +1

To achieve neutrality, the oxidation number of the coordination sphere needs to equal -1.
Thus,
                                    [ Co (X)₂ (Y)₂ ]  =  -1
Given that,
           the O.N of X is -2
Therefore,
       O.N of (X)₂ equals -4

Additionally,
O.N of H₂O is zero since it remains neutral. Therefore,

                                    [Co - 4 + 0 ]  =  -1
Or,
                                    Co  =  -1 + 4

                                    Co  =  +3

Conclusion:
          The oxidation number for the coordination sphere is -1, and the oxidation state of copper is +3.