Fecl3 reacts with only one of the two compounds aspirin and salicylic acid with which part of a molecule does fecl3 react

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

O sea,

Dado que:

Para :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de son:

Moles de = 0.004 moles

Para :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de son:

Moles de = 0.012 moles

Según la reacción:

1 mol de reacciona con 1 mol de

Por lo tanto,

0.012 mol de reacciona con 0.012 mol de

Moles disponibles de = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de reacciona con 1 mol de y produce 1 mol de

0.004 mol de reacciona con 0.004 mol de y genera 0.004 mol de

Los moles restantes de son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, , después de la reacción es:

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.

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

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:

• A. Which element, X or Z, has a higher molar mass?

Explanation:

Heating the original compounds intensely to remove all oxygen causes chemical decomposition reactions:

• 2XClO₃ (solid) → 2XCl + 3O₂ (gas)

• 2ZClO₃ (solid) → 2ZCl + 3O₂ (gas)

By measuring the initial mass of each sample and the mass remaining after heating, the student can compute the oxygen gas mass released:

• Mass of oxygen released = initial sample mass minus residue mass

Using this oxygen mass, she can calculate how many moles of oxygen were present in each sample:

• Moles of oxygen = oxygen mass (g) divided by molar mass of oxygen

Next, the moles of the original sample are determined:

• Each mole of XClO₃ or ZClO₃ has 3 moles of oxygen atoms.

So, dividing the moles of oxygen released by 3 gives the moles of the sample.

Applying the formula molar mass = mass / moles, the student finds the molar masses of XClO₃ and ZClO₃.

Thus, this data allows answering question A: Which of X or Z has the higher molar mass?