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3 years ago

Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?

2 answers:

4 0

<span>Equation:2H2(g) + O2(g) → 2H2O(g)
</span><span>
Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.

</span>

mariarad [96]3 years ago

4 0

<h2>Equation:2H2(g) + O2(g) → 2H2O(g)</h2>

<h3>Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.</h3>

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Is chex mix an example of a pure substance or mixture?

irinina [24]

The answer is mixture

3 0

3 years ago

Consider the Gibbs energies at 25 ∘C.

zysi [14]

Answer:

A)ΔGrxn∘=55.7kJ/mol

B)Ksp=1.75×10^−10

C)ΔGrxn∘=70.0kJ/mol

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=ΔGf,products∘−ΔGf,reactants∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Cl(aq)−∘]−ΔGf,

AgCl(s)⇌Ag(aq)++Cl(aq)−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

ΔGrxn∘=55.7kJ/mol

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7kJ/mol=−(8.3145×10−3J/molK)(298.15K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Br(aq)−∘]−ΔGf,

ΔGrxn∘=[77.1kJ/mol+(−104.0kJ/mol)]−(−96.90kj/mol

ΔGrxn∘=70.0kJ/mol

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K

Ksp=5.45×10^−13

8 0

3 years ago

On lab day. Milania did an experiment with two solids: potassium iodide and silver nitrate. Milania measured and recorded the ma

Mkey [24]

D. It is a chemical change because solubility in water of Solid A is different than the solubility in water or both Potassium Iodide and Silver Nitrate

Explanation:

From the given table which summarizes findings about the experiment, we can conclude that It is a chemical change because solubility in water of Solid A is different than the solubility in water of both Potassium Iodide and Silver Nitrate.

A chemical change is a form of change in which new substances are produced.
A physical change is one in which the states of matter changes especially the form.

Now we that two soluble substances goes to insoluble product.

A substance is soluble when they go into solution with a solvent of the same type.
This change from solubility to insolubility of a compound is a typical chemical change.
Other forms of changes are just physical changes.

learn more:

Chemical change brainly.com/question/9388643

#learnwithBrainly

5 0

3 years ago

General or nonspecific observation is

drek231 [11]

It is an act or instance of viewing or noting a fact, or an occurrence for some scientific or other special purpose

8 0

3 years ago

Calculate the energy required to ionize a hydrogen atom to an excited state where the electron is initially in the n = 5 energy

SVETLANKA909090 [29]

. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:

E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ

Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ

2. The wavelength of a photon with this energy would be:

Energy = hc/wavelength

wavelength = hc/energy

wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m

Converting to nanometers gives: 91.16 nm

3. Repeat the calculation in 1, but using n=5.

4. Repeat the calculation in 2 using the energy calculated in 3.

7 0

3 years ago

Read 2 more answers