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3 years ago

I need help.... How does the kinetic energy of molecules explain both Boyle’s and Charles’ Laws?

Chemistry

2 answers:

yulyashka [42]3 years ago

7 0

Answer:

Kinetic Molecular Theory states that gas particles are in constant motion and exhibit perfectly elastic collisions.

Explanation:

Kinetic Molecular Theory states that gas particles are in constant motion and exhibit perfectly elastic collisions. Kinetic Molecular Theory can be used to explain both Charles' and Boyle's Laws. The average kinetic energy of a collection of gas particles is directly proportional to absolute temperature only.

Aleks [24]3 years ago

4 0

Answer:

Explanation:

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How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward

fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

5 0

3 years ago

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t

topjm [15]

Answer:

$\boxed{3}$

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

$\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )$

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

$\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}$

Data:

$n_{f} = 2$

λ = 657 nm

Calculation:

$\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}$

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3 years ago

. What classification should this reaction have? Cu + 2AgNO3 Cu(NO3)2 + 2Ag synthesis decomposition single replacement double di

Monica [59]

<span>What classification should this reaction have?
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</span><span>single replacement</span>

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3 years ago

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r-ruslan [8.4K]

Answer:

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Artyom0805 [142]

The answer is 1 3 and 5 just took the test !!!!

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