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1 year ago

Thermal Energy and Kinetic Molecular Theory Quick Check

1 answer:

4 0

The Kinetic Molecular Theory is a scientific model that states atoms in a compound are found in a constant state of motion (movement).

<h3>What is the Kinetic Molecular Theory?</h3>

The Kinetic Molecular Theory is a scientific model that states atoms in a compound are found in a constant state of motion (movement).

Thermal energy refers to the movement of particles and therefore both concepts are interrelated.

In conclusion, the Kinetic Molecular Theory is a scientific model that states atoms in a compound are found in a constant state of motion (movement).

Learn more about the Kinetic Molecular Theory here:

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Potassium-40 is a radioactive isotope that decays into a single argon-40 atom and other particles with a half-life of 1:25 billi

stiks02 [169]

Answer:

0.147 billion years = 147.35 million years.

Explanation:

It is known that the decay of a radioactive isotope isotope obeys first order kinetics.

Half-life time is the time needed for the reactants to be in its half concentration.

If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

The half-life of Potassium-40 is 1.25 billion years.

For, first order reactions:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.

Also, we have the integral law of first order reaction:

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).

[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).

At the time needed to be determined:

8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.

If we start with 100% Potassium-40:

∴ The remaining concentration of Potassium-40 ([A] = 88.88%).

and that of argon-40 produced from potassium-40 decayed = 11.11%.

That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).

∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.

8 0

4 years ago

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 K is given below. 2 H2(g) + O2(g) → 2

DENIUS [597]

Answer:

(a) ΔG° = -474 kJ/mol; E° = 1.23 V

(b) ΔH° negative; ΔS° negative

(c) Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

Explanation:

Let's consider the following reaction.

2 H₂(g) + O₂(g) → 2 H₂O(l)

with an equilibrium constant K = 1.34 × 10⁸³

(a) Calculate E° and ΔG° at 298 K for the fuel-cell reaction.

We can calculate the standard Gibbs free energy (ΔG°) using the following expression:

ΔG° = - R × T × lnK

ΔG° = - 8.314 × 10⁻³ kJ . mol⁻¹.K⁻¹ × 298 K × ln 1.34 × 10⁸³ = -474 kJ/mol

To calculate the standard cell potential (E°) we need to write oxidation and reduction half-reactions.

Oxidation: 2 H₂ ⇒ 4 H⁺ + 4 e⁻

Reduction: O₂ + 4 e⁻ ⇒ 2 O²⁻

The moles of electrons (n) involved are 4.

We can calculate E° using the following expression:

$E\°=\frac{0.0591V}{n} .logK\\E\°=\frac{0.0591V}{4} .log1.34 \times 10^{83}=1.23V$

(b) Predict the signs of ΔH° and ΔS° for the fuel-cell reaction. ΔH°: positive negative ΔS°: positive negative

The standard Gibbs free energy is related to the standard enthalpy (ΔH°) and standard entropy (ΔS°) through the following expression:

ΔG° = ΔH° - T.ΔS°

Usually, the major contribution to ΔG° is ΔH°. So, if ΔG° is negative (exergonic), ΔH° is expected to be negative (exothermic).

The entropy is related to the number of moles of gases. There are 3 gaseous moles in the reactants and 0 in the products, so the final state is predicted to be more ordered than the initial state, resulting in a negative ΔS°.

(c) As temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same?

The maximum amount of work obtained depends on the standard Gibbs free energy.

wmax = ΔG° = ΔH° - T.ΔS°

Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

5 0

4 years ago

What is the answer for those questions plz ,help

marysya [2.9K]

Answer:

1) 1.51 × 10²³ particles of Mg

2) 0.54 × 10⁻³ moles

Explanation:

Given data:

Number of moles of Mg = 0.250 mol

Number of representative particles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³ particles

0.250 mol × 6.022 × 10²³ particles / 1 mol

1.51 × 10²³ particles of Mg

Given data:

Number of moles of lead = ?

Number of atoms of lead = 3.25×10²⁰ atoms

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³ atoms

1 mol × 3.25×10²⁰ atoms/ 6.022 × 10²³ atoms

0.54 × 10⁻³ moles

6 0

3 years ago

How many joules of heat are absorbed to raise the temperature of 435 grams of water at 1 atm from 25°C to its boiling point, 100

Lilit [14]

The temp would be 37.0c

3 0

3 years ago

A mixture of Ar, He, and O2 has a total pressure of 1,015 kPa. Argon exerts a partial pressure of 152 kPa and Helium exerts a pa

arsen [322]

This one is easy. It's just addition and subtraction. Add the partial pressure of Argon and the partial pressure of Helium:
152 kPa
+305 kPa
 457 kPa

Next, subtract the result from the total pressure:

1015 kPa
- 457 kPa
558 kPa

8 0

3 years ago