What is the only thing that changes in a chemical reaction

Kamila [148]

The equation of state for a hypothetical ideal gas is known as the ideal gas law, sometimes known as the general gas equation. i.e. PV = nRT or P1V1 = P2V2.

According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.
Robert Boyle, Gay-Lussac, and Amedeo Avogadro's observational work served as the basis for the ideal gas law. The Ideal gas equation, which simultaneously describes every relationship, is obtained by combining all of their observations into a single statement.
When applying the gas constant R = 0.082 L.atm/K.mol, pressure, volume, and temperature should all be expressed in units of atmospheres (atm), litres (L), and kelvin (K).
At high pressure and low temperature, the ideal gas law basically fails because molecule size and intermolecular forces are no longer negligible but rather become significant considerations.

Learn more about ideal gas law here:

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3 0

1 year ago

BRAINLIEST IF CORRECT

zhannawk [14.2K]

Answer:

1 a person sings into a microphone

2 the sound is converted into an electrical signal

3 the analog signal is converted into a digital signal made of many samples of the digital sound

4 the samples of the sound is converted into 1s and 0s

5 the digital pattern is converted to a medium such as a hard disk or flash drive

Explanation:

5 0

3 years ago

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Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of

Alenkinab [10]

Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n . R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

5 0

3 years ago

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The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

3 years ago

What percentage of radioactive substance remains after two half-lives

Ksju [112]

Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

3 0

3 years ago