What alkene reacts the fastest with HBr?

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.

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

Answer:

B, D

Explanation:

We need to recognize that the ice will rise in temperature from -6.5 ºC to 0 ºC for it to change into water.

Let's define q₁ as the heat needed to warm the ice to 0ºC, and q₂ as the heat for the transition from solid to liquid.

The calculation for q₁ is as follows:

q₁ = s x m x ΔT, where s represents the specific heat of ice (2.09 J/gºC), m is the mass, and ΔT is the temperature difference.

For q₂, the enthalpy of fusion is computed as:

q₂ = C x ΔT

with C indicating the specific heat for the phase transition, denoted as AH in kJ/mol.

All necessary data for computing q₁, q₂, and the total heat change (q₁ + q₂) is provided.

q₁ = 25.0 g x (2.09 J/gºC) x (0 - (-6.5 ºC))

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

Combining these values gives us qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ.

Now we can answer the question:

(a) False, AH refers to the heat capacity during melting.

(b) True, as we concluded earlier.

(c) False, there’s only one phase transition from solid (ice) to liquid.

(d) True based on our calculations above.

(e) False, according to our findings.

Answer:

Heat is absorbed.