A 110.0-g sample of metal at 82.00°C is added to 110.0 g of H2O(l) at 27.00°C in an insulated container. The temperature rises t

Answer:

Ir(NO2)3

Explanation:

The molar mass is 330.2335, in case that's also required.

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:

In all listed reactions, ΔH°rxn does not correspond to the ΔH°f of the resulting product.

Explanation:

The standard enthalpy of formation (ΔH°f) signifies the enthalpy change that occurs when 1 mole of a product is created from its basic elements in their standard states.

1/2 O₂(g) + H₂O(g) ⟶ H₂O₂(g)

ΔH°rxn does not equal ΔH°f of the product, since H₂O(g) is a compound rather than an element.

Na⁺(g) + F⁻(g) ⟶ NaF(s)

ΔH°rxn is not the same as ΔH°f of the product because Na and F are not in their standard states (Na(s); F₂(g)).

K(g) + 1/2 Cl₂(g) ⟶ KCl(s)

ΔH°rxn is not equal to ΔH°f of the product due to K being outside its standard state (K(s)).

O₂(g) + 2 N₂(g) ⟶ 2 N₂O(g)

ΔH°rxn does not match ΔH°f of the product as 2 moles of N₂O are produced.

In none of the above cases does ΔHrxn match ΔHf of the product.

Answer: 0.0007 moles of are released when the temperature rises.

Explanation:

To determine the moles, we utilize the ideal gas law, as follows:

where,

P = gas pressure = 1.01 bar

V = gas volume = 1L

R = gas constant =

• Calculated moles at T = 20° C

The gas temperature = 20° C = (273 + 20)K = 293K

Substituting values into the equation gives:

• Calculated moles at T = 25° C

The gas temperature = 25° C = (273 + 25)K = 298K

Substituting values into the equation gives:

• Released moles =

Therefore, 0.0007 moles of are released when the temperature increases from 20° C to 25° C.