How to correctly solve this problem : 4.05Kg+567.95g+100.1g correct and best way

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.

The work done is 19.26 kJ.

Answer:

0.31%

Explanation:

For the chemical reaction:

I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻

0.043 L multiplied by 0.117 M of sodium thiosulfate gives 5.031x10⁻³ moles of S₂O₃²⁻

5.031x10⁻³ moles of S₂O₃²⁻ produces:

ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O

2.5156x10⁻³ moles of I₂ equates to moles of NaClO

2.5156x10⁻³ moles of NaClO times yields 0.187 g of NaClO

Thus, the mass percentage composition is:

= 0.31%

I hope this helps!

Density is defined as the "mass per unit volume" of an object.

Thus, for an object weighing 100 grams with a volume of 100 milliliters, the density calculates to 100 grams / 100 ml.

When weighing the object, if there is water on the scale's surface, it will contribute additional weight, making the object seem heavier than its actual mass. Consequently, you might mistakenly conclude that the density is GREATER than it truly is.

For instance, if there were 5 ml of water on the scale, with water's density being 1 gram per milliliter (1 g/ml), it would add 5 grams to the object's weight. Using the previous example, the object's mass appears as 105 grams instead of 100 grams. Thus, you would calculate:

density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml

Thus, the effect on density would be to misleadingly suggest it is greater.

I hope this is helpful!

Best of luck