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1 year ago

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

Chemistry

1 answer:

Kay [80]1 year ago

7 0

The equilibrium constant for this reaction at 350°C is D. 282.

<h3>Equilibrium constant</h3>

A dynamic chemical system approaches chemical equilibrium constant when enough time has passed and its composition no longer exhibits any discernible propensity to change further. The equilibrium constant of a chemical reaction is the value of its reaction quotient in this condition. The equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture for a specific set of reaction conditions. Understanding equilibrium constants is crucial for comprehending many chemical systems as well as biological processes like the transport of oxygen by hemoglobin in the blood and the maintenance of acid-base homeostasis in the human body. There are many different kinds of equilibrium constants, including stability constants, formation constants, binding constants, association constants, and dissociation constants.

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A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

Br2(g) + I2(g) ↔ 2IBr(g)

When the mixture has come to equilibrium, the concentration of iodine monobromide is

1.190 M. What is the equilibrium constant for this reaction at 350°C? Show step-by step explanation.

A) 3.55 × 10^3

B) 1.24

C) 1.47

D) 282

E) 325

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What Size container do you need to hold 0.0459 mol of N2 gas at stp

Allushta [10]

Hello!

You need to calculate the volume of the container. To calculate the volume of this amount of N₂ gas we need to make the assumption that N₂ behaves like an ideal gas.

1 mole of an ideal gas under Standard Temperature and Pressure occupies 22,4 L so the calculations are as follows:

$0,0459 mol N_2* \frac{22,4 L}{1 mol N_2}= 1,028 L$

So, the volume of the container should be 1,028 L or more.

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3 years ago

A student finds two unlabeled flasks of clear liquids. One is believed to be 0.1 m nacl and the other to be 0.1 m naclo3. What i

KonstantinChe [14]

Answer:

Add AgNO₃ solution to both unlabeled flasks: based on solubility rules, you can predict that when you add AgNO₃ to the NaCl solution, you will obtain AgCl precipitate, while no precipitate will be formed from the NaClO₃ solution.

Explanation:

<u>1. Adding AgNO₃ to NaCl solution:</u>

AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)

<u>2. Adding AgNO₃ to NaClO₃ solution</u>

AgNO₃ (aq) + NaClO₃ (aq) → AgClO₃ (aq) + NaNO₃ (aq)

<u />

<u>3. Relevant solubility rules for the problem.</u>

Although most salts containing Cl⁻ are soluble, AgCl is a remarkable exception and is insoluble.

All chlorates are soluble, so AgClO₃ is soluble.

Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.

Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.

Then, the precipiate will permit to conclude which flask contains AgCl.

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Un sistema químico balanceado para la reacción de HC1 y NaOH<br>

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3 years ago

Addition of water to an alkyne gives a keto‑enol tautomer product. Draw an enol that is in equilibrium with the given ketone

Marrrta [24]

Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.

<h3>What is the keto-enol means in tautomer?</h3>

They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.

The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.

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2 years ago