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sergey [27]
3 years ago
8

What happened to the concentration of the product of H2O when the reaction shifts to the left

Chemistry
1 answer:
lilavasa [31]3 years ago
6 0

I might need a diagram for this, but I have a vague idea of what you are talking about.

If H20 is going left it means the temperature is going lower.

The molecules will condense to slowly become ice

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Draw the organic and inorganic products for the following acid/base reaction as well as the charges.
Reptile [31]
<span>The product would be :
H3O+ Benzaldehyde -----> C6H5CHO.
It will form a neutral molecule. So the charge is neutral.
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4 years ago
What is true about most of the wavelengths found in the electromagnetic spectrum?
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D, you can't see ultraviolet rays and other similar frequency waves with the naked eye.
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3 years ago
Define an arrhenius base and describe properties of bases. use an example to explain how an arrhenius base will behave in water.
murzikaleks [220]

An Arrhenius base is a molecule that when dissolved in water will break down to yield an OH^- or hydroxide in solution.

<h3>What is Arrhenius base?</h3>

An Arrhenius base is a compound that increases the OH− ion concentration in aqueous solution.

An Arrhenius base is a substance that, when dissolved in an aqueous solution, increases the concentration of hydroxide, or OH^-, ions in the solution.

Bases Properties

Arrhenius bases that are soluble in water can conduct electricity.

Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.

Bases also change the colour of indicators. Red litmus turns blue in the presence of a base (see figure below), while phenolphthalein turns pink.

Some bases react with metals to produce hydrogen gas.

Acids (pH < 7.0) react with bases (pH > 7.0) to produce a salt and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The resulting mixture will have a more neutral pH.

An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. In other words, it increases the number of H+ ions in the water. In contrast, an Arrhenius base dissociates in water to form hydroxide ions, OH^-.

Example, sodium hydroxide, is added to an aqueous solution. NaOH dissociates into sodium, Na^+, and hydroxide, OH^-, ions.

Learn more about the Arrhenius bases here:

https://brainly.in/question/8273595

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4 0
2 years ago
A chemical reaction occurring in a cylinder equipped with a moveable piston produces 0.601 mol of a gaseous product. If the cyli
netineya [11]

Answer:

4.33 L

Explanation:

Assuming ideal behaviour and that all 0.300 moles of gas reacted, we can solve this problem using Avogadro's law, which states that at constant temperature and pressure:

  • V₁n₂ = V₂n₁

Where in this case:

  • V₁ = 2.16 L
  • n₂ = 0.601 mol
  • V₂ = ?
  • n₁ = 0.300 mol

We <u>input the given data</u>:

  • 2.16 L * 0.601 mol = V₂ * 0.300 mol

And <u>solve for V₂</u>:

  • V₂ = 4.33 L
3 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
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