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Mila
4 months ago
10

Equimolar samples of CH3OH(l) and C2H5OH(l) are placed in separate, previously evacuated, rigid 2.0 L vessels. Each vessel is at

tached to a pressure gauge, and the temperatures are kept at 300 K. In both vessels, liquid is observed to remain present at the bottom of the container at all times. The change in pressure inside the vessel containing CH3OH(l) is shown below.Compared to the equilibrium vapor pressure of CH3OH(l) at 300 K, the equilibrium vapor pressure of C2H5OH(l) at 300 K is

Chemistry
1 answer:
Alekssandra [3K]4 months ago
3 0

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.

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A solution with 117 grams of lithium hydroxide and another with 141 grams of hydrogen bromide are combined. They react according
Tems11 [2777]

Answer: C. 151 g

Solution: The balanced equation given is:

LiOH+HBr\rightarrow LiBr+H_2O

From this equation, the ratio of moles between all substances is 1:1. We have 117 grams of LIOH and 141 grams of HBr available and need to calculate the theoretical yield of LiBr.

We should convert each reactant’s grams into moles to identify the limiting reagent since the theoretical yield relies on it.

Molar mass for LiOH = 6.94 + 15.999 + 1.008 = 23.947 grams per mole

Molar mass for HBr = 1.008 + 79.904 = 80.912 grams per mole

To find the moles of each reactant, we divide their grams by their respective molar masses.

Moles of LiOH = 117gLiOH(\frac{1mol}{23.947}) = 4.89 mol

Moles of HBr = 141gHBr(\frac{1mol}{80.912g}) = 1.74 mol

As there are fewer moles of HBr, it is the limiting reactant. With a 1:1 mol ratio between HBr and LiBr, 1.74 moles of LiBr can be produced.

Molar mass of LiBr = 6.94 + 79.904 = 86.844 grams per mole

The mass of LiBr formed = 1.74molLiBr(\frac{86.844g}{1mol}) = 151 g LiBr

Based on calculations, the theoretical yield of LiBr is 151 g, hence the correct answer is C.


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4 months ago
If a particular ore contains 55.4 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosp
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