Boiling Point

I have all of the answers except this one........... A photon interacts with a ground state electron in a hydrogen atom and is absorbed. The electron is ejected from the atom and exhibits a de Broglie wavelength of 5.908×10−10 m. Determine the frequency (in hz) of the interacting photon.

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Explain the interplay between enthalpy (H) and entropy (S) changes taking place during ligand binding. someone help please

I have a diff problem .......Titanium (Ti) can be produced by the reaction of metallic sodium (Na) with titanium tetrachloride vapor (TiCl4). The byproduct of this reaction is sodium chloride (NaCl). Calculate the amount of titanium produced (in kg) when a reactor is charged with 73.0 kg of TiCl4 and 10.0 kg of Na.

Sulfur vapor is analyzed by photoelectron spectroscopy (PES). Measurements determine that photoelectrons associated with the 1st ionization energy of sulfur move with de Broglie wavelength λ=5.091 A˚. What is the maximum wavelength (in meters) of radiation capable of ionizing sulfur and producing this effect?

The gas-phase reaction between trichloromethane, CHCl3, and chlorine, Cl2, has time-independent stoichiometry and can be represented as follows:

Equation 1 CHCl3(g) + Cl2(g) = CCl4(g) + HCl(g)

Under certain experimental conditions, the experimental rate equation was found to be:

Equation 2 J = kR[CHCl3][Cl2]^1/2

A three-step mechanism has been proposed for Reaction 1:

Equation 3 k1 Cl2 ↔ 2Cl• k-1

Equation 4 k2 CHCl3 + Cl• → HCl + CCl3•

Equation 5 k3 CCl3 + Cl• → CCl4

(i) Explain why the form of the experimental rate equation indicates that Reaction 1 cannot be an elementary reaction.

(ii) With reference to the three-step mechanism (Equations 3–5 ), and assuming that the second step (Equation 4) is rate-limiting, derive the chemical rate equation for this mechanism and then compare it with the experimental rate equation given in Equation 2.

(iii)The activation energy for the forward reaction, Ef , of step 1 (Equation 3) is 243.4 kJ mol^-1. Given this information, determine the activation energy for the reverse reaction, Er, and comment on the significance of the value (one sentence only).

You wish to increase the carbon content of a slab of steel by exposing it to a carburizing atmosphere at elevated temperature. The carbon concentration in the steel before carburization is 359.5 ppm and is initially uniform through the thickness of the steel. The atmosphere of the carburizing furnace maintains a carbon concentration of 6695.0 ppm at the surface of the steel. Calculate the time required to carburize steel so that the concentration of carbon at a depth of 38.0 x 10-2 cm is one half the value of the carbon concentration at the surface. The diffusion coefficient of carbon in steel is 3.091 x 10-7 cm2/s at the carburizing temperature. Express your answer in hours.

DATA: Error Function Values erf(ξ

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copper 2 carbonate + hot dilute sulfuric acid will give what

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What are the names of the salts produced by hydrochorlic acid?

CuCO3 + H2SO4 --> CuSO4 + H20 + CO2

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I want to prepare copper sulphate using sulfuric acid. I have a sample of copper metal and a sample of copper carbonate. If I'm to use one of the materials which should it be and why

/write and balance three euations that have the abulity to produce hydrogen gas as a product

You may often come across a question "What gas turns limewater cloudy?" The answer to this question is well known. Carbon dioxide is the only gas that turns lime water cloudy. You may be wondering what is lime water used for. Limewater is an aqueous solution of slaked lime and you will find it in antacids, medicines and lotions. But one of its most noteworthy property is that it is used to absorb carbon dioxide from the air. In this article, we have answered all the questions related to the reaction of lime water and .

What happens when Lime Water reacts to Carbon Dioxide?

Carbon dioxide reacts with limewater to form calcium carbonate, which precipitates out of the solution. The carbon dioxide and limewater react to produce water in addition to the calcium carbonate. Calcium carbonate is chalk, and when it is produced, it precipitates and solid particles of chalk appear. The appearance of this solid makes the liquid appear ‘milky’. The white milky suspension/precipitate is caused by the formation of calcium carbonate. The characteristic carbon dioxide test, is checking that the limewater is milky. This is because chalk is precipitating in the limewater. Bubbling carbon dioxide through the solution for an extended period of time makes the solution become clear and colorless. This happens as the carbon dioxide forms acidic carbonic acid when it dissolves in the water, the carbonic acid (H2CO3) reacts further with the calcium carbonate. This chemistry is important in understanding how hard water is formed and then limescale is formed in kettles and hot water boilers.

Written as an Equation

In its equation form that makes:

How to test for carbon dioxide?

Now, we will answer how to test for carbon-dioxide. One of the most effective ways to test for carbon dioxide gas is the limewater test. When carbon dioxide reacts with lime water (calcium hydroxide solution), a white precipitate of calcium carbonate is produced. The solution of calcium hydroxide is limewater and if carbon dioxide bubbles through the limewater, it turns cloudy white or milky.

How does carbon dioxide turns lime water milky?

When limewater which is a solution of calcium hydroxide, reacts with the carbon dioxide gas, an insoluble solid known as calcium carbonate is generated. The equation of this reaction is given below:

Now, the question arises why the solution turns milky. Well, the answer is simple, The reason for the milky solution is that calcium carbonate which is produced as a result of this reaction is a white precipitate. This nature of calcium carbonate also helps us to test for the presence of carbon dioxide gas. All you have to do is to bubble the gas through a solution of calcium hydroxide. If the gas is carbon dioxide, then the solution will turn milky. If not then the gas which is subjected to the test is not carbon dioxide. If you continue to bubble the carbon dioxide gas through limewater, you will witness another acid-base reaction that will dissolve the precipitate to generate soluble calcium hydrogen carbonate. The equation of this reaction is given below:

The reaction between copper(II) oxide and sulphuric acid is a classic chemistry experiment often used to demonstrate how acids react with metal oxides.

In this reaction, a base (the metal oxide) reacts with an acid to produce a salt and water. This specific reaction is widely used in schools to prepare crystals of copper(II) sulphate.

Reactants and Products

To understand this reaction, we must first identify the physical states and colours of the chemicals involved:

• Copper(II) oxide (CuO): This is the reactant. It is a black solid powder. It is a base because it can neutralise an acid.
• Sulphuric acid (H₂SO₄): This is the acid. It is typically used as a clear, colourless aqueous solution (dissolved in water).
• Copper(II) sulphate (CuSO₄): This is the product (the salt). In solution, it is a characteristic bright blue colour.
• Water (H₂O): This is the liquid by-product.

Chemical Equation

We can write this reaction in two ways.

Word Equation: Copper(II) oxide + Sulphuric acid → Copper(II) sulphate + Water

Balanced Symbol Equation: The symbol equation including state symbols is:

What Type of Reaction Is It?

This reaction is classified as a neutralisation reaction.

Neutralisation occurs when an acid reacts with a base. In this case, the copper(II) oxide acts as the base (specifically a metal oxide) and the sulphuric acid provides the hydrogen ions (H⁺).

The general formula for this type of neutralisation is:

Practical Overview & Observations

If you were to carry out this reaction in a laboratory to create copper sulphate crystals, you would follow these steps:

1. Heat the acid: Warm dilute sulphuric acid in a beaker (do not boil it).
2. Add the base: Add the black copper(II) oxide powder to the acid a little at a time, stirring constantly.
3. Observe: The mixture will turn from colourless to blue.
4. Excess: Continue adding copper oxide until no more reacts. You will see black powder sitting at the bottom of the beaker. This ensures all the acid has been neutralised.
5. Filter: Filter the mixture to remove the excess black copper oxide. You are left with a clear blue solution of copper(II) sulphate.

The most important visual change is the black solid disappearing and the solution turning a transparent blue.

Summary

• Reactants: Black copper(II) oxide solid and colourless sulphuric acid.
• Products: Blue copper(II) sulphate solution and water.
• Type: Neutralisation (Acid + Metal Oxide).
• Observation: Black powder dissolves; solution turns blue.