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labwork [276]
3 years ago
10

How do I convert moles to atmospheres

Chemistry
1 answer:
V125BC [204]3 years ago
5 0
P = Pressure (atm) V = Volume (L) n = moles R = gas constant = 0.0821 atm. L/mol. K T = Temperature (Kelvin) The correct units are essential. Be sure to convert whatever units you start with into the appropriate units when using the ideal gas law.
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Qué método utilizaría para separar los componentes de una mezcla de: limaduras de hierro, azufre en polvo y piedras?
Pachacha [2.7K]

Answer:

Traducción por favor

Explanation:

8 0
3 years ago
How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
netineya [11]

The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

 H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

4 0
4 years ago
Read 2 more answers
The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is
WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

  • sp^{3}, four electron domains, as in ethane \text{C}_2\text{H}_6
  • sp^{2}, three electron domains, as in ethene \text{C}_2\text{H}_4
  • sp, two electron domains, as in ethyne \text{C}_2\text{H}_2

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

  • Carbon atoms with a sp^{3} hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately 109.5\textdegree{}
  • Carbon atoms with a sp^{2} hybridization would demonstrate a triangular planar configuration with a bond angle of 120\textdegree{}
  • Carbon atoms with a sp hybridization would demonstrate a linear configuration with a bond angle of 180\textdegree{}

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

  • sp^{3} in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
  • sp^{2} in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.
3 0
4 years ago
Please help.
mojhsa [17]

Explanation: An element is represented in the form of _{A}^{Z}\textrm{X} where,

X = Symbol of the element

A = Atomic Mass of the element X

Z = Atomic Number of element X

Hence, For the element Justwondoricium,

Symbol = Jw

Atomic number = 120

Atomic Mass = 224

Now, Atomic number = Number of electrons = Number of Protons

Number of electrons = 120

Number of Protons = 120

and Atomic Mass = Number of neutrons + Number of protons

Number of neutrons can be calculated as we know atomic mass and number of neutrons, Putting the numbers we get

Number of neutrons = 224 - 120 = 104

The nearest noble gas to the element having atomic number 120 is Oganesson (Og), which has an atomic number of 118, so the next two electrons will be filled in the 8s orbital.

Electronic Configuration of Jw is [Og]8s^2

This electronic configuration lets us know about the location of the element in periodic table.

As the electron is entering the 8th shell, it belongs to the 8th period and as the last electron enters the s-orbital, it belongs to the S-block of the periodic table.

When the s-electrons are 1, it belongs to Group 1.

When the s-electrons are 2, it belongs to Group 2.

As this element has 2 s-electrons, it belongs to the Group 2.

8 0
4 years ago
What was Ernest Rutherford's experiment?
Dafna1 [17]

Answer:

C. He shot tiny alpha particles through a piece of gold foil.

Explanation:

In the year 1911, Ernest Rutherford performed the gold foil experiment which gave a deeper perspective to the structure of an atom.

He simply collided a thin gold foil with an alpha particle which he generated from a radioactive source. He discovered that most of the alpha particles passed through the thin gold foil but a few were deflected back. His discovery led to the proposition of the nuclear model of the atom.

6 0
4 years ago
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