The methylene chloride solution is filtered through sodium sulfate to:

Chemistry

1 answer:

Vlad [161]2 years ago

5 0

Answer:

C. Dry the methylene chloride by removing water

Explanation:

Anhydrous sodium sulfate is known for its high capacity to absorb water, for this reason it is widely used in laboratories as a drying agent.

Sodium sulfate is a neutral molecule so it cannot be used to neutralize and is very stable, so it is difficult to precipitate organic molecules.

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Is there an N−Cl bond in solid ammonium chloride?

Tems11 [23]

When ammonia is reacted with HCl it abstracts proton from acid and forms Ammonium Ion and Chloride Ion.

NH₃ + HCl → ⁺NH₄ + Cl⁻ (simply Written NH₄Cl)
Structure,
The structure of Ammonium Chloride is among those structures which contains all three types of bonding's, i.e.

Ionic Bond

Covalent Bond

Coordinate Covalent Bond

Three Hydrogen atoms previously bonded with Nitrogen are covalent in nature. The new incoming proton from HCl forms co-ordinate covalent bond with Nitrogen and Chloride Ion containing negative charge make Ionic Bond with the positive Ammonium Ion. In question, if the line between Nitrogen and Chlorine atom is assumed covalent then it is incorrect. Structure is shown below,

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2 years ago

A __________ reaction occurs when a substance, usually containing carbon, reacts with oxygen to produce energy in the form of he

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3 years ago

Write a balanced equation for the decomposition reaction described, using the smallest possible integer coefficients. When ammon

MAXImum [283]

Answer:

NH₄Cl ------> NH₃ + HCl

Explanation:

Ammonium Chloride =(NH₄Cl

Ammonia = NH₃

Hydrochloric Acid = HCl

NH₄Cl ------> NH₃ + HCl

In decomposition reaction, the reactant is breaking down into smaller parts. In this case, all of the coefficients are 1. The reaction is already balanced.

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1 year ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$