The melting point of water is 0°C at 1 atm pressure because under these conditions:

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.

M = 81.50g, mm = m/n
n =???
PV = nRT --> n = PV/RT
n = (1.75)(4.92)/(.0821)(307)
n = 8.61/25.20 =.342
--> mm = m/n = 81.5/.342 = 238.58

Answer:

The reaction will proceed in the forward direction, resulting in the reduction of NiO to Ni

Explanation:

Step 1: Provided data

Kp = 6.0 * 10²

Pressure of CO = 150 torr

Total pressure remains under 760 torr

Step 2: The balanced reaction equation:

NiO(s) + CO(g) ⇆ Ni(s) + CO2(g)

Step 3: Determine P(CO) and P(CO2)

For this equilibrium, the Kp expression is:

Kp = P(CO2)/P(CO) = 6.0*10^2

Given that NiO and Ni are solids, they do not affect the Kp

⇒ with P(CO) = 150 torr

⇒ P(CO) = 760 - 150 = 610 torr

Step 4: Calculate the reaction quotient

Q = 610/150 = 4.1

Since Q is significantly less than Kp, there are more reactants than products. Some reactants will convert into products, driving the reaction rightward.

Thus, the reaction will proceed forward, leading to the reduction of NiO to Ni.

We need to calculate the volume of Gold, assuming its mass matches that of copper.

Given information:

Density of Copper = 8.96 g/ml.
Volume of Copper = 141 ml.
Mass of Gold = Mass of Copper.
Density of Gold = 19.3 g/ml.

To find copper's mass, we use the density equation:
Density = mass/volume.

To find mass of copper:
Mass of copper = Density of Copper * Volume of Copper.
Mass of copper = 8.96 g/ml * 141 ml = 1263.36 g.
Thus,
Mass of gold = Mass of copper = 1263.36 g.
Now, using the density formula for gold to get its volume:
Volume of gold = Mass of gold / Density of gold.
Volume of gold = 1263.36 g / 19.3 g/ml = 65.46 mL.

Consequently, the volume of gold required to match the mass of copper is 65.46 mL.