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3 years ago

Draw the products of the complete hydrolysis of a ketal. draw all products of the reaction.

Chemistry

2 answers:

Advocard [28]3 years ago

8 0

Following reaction shows the complete hydrolysis of ketal. The products formed are 1) Diol and 2) Ketone. Mechanism is shown below,

ser-zykov [4K]3 years ago

4 0

Answer:

The main products obtained from complete hydrolysis of a ketal are ketone (90%) and hydrate (10%), we also obtained as byproducts two alcohols (R $_{1}$ OH and R $_{2}$ OH). See attached image

Explanation:

To understand this question, one has to know what a ketal is and also what is complete hydrolysis.

Ketal: functional group present in organic chemistry that has a central carbon attached to 2 R groups and 2 OR.

Hydrolysis: reaction of water with a molecule

We can observe in the diagram that this reaction is found in equilibrium, meaning that depending on the concentration of the diverse substrates and products, the reaction will favor the direction towards one or the other.

Note: In the last part of the reaction we can observe that the equilibrium favors the production of ketone over that of a hydrate. This occurs since a ketone is more stable than that of a hydrate (about 90% will be ketone while 10% will be a hydrate).

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3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

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3 years ago

Which of the following is true about the concept of half-life?

Reptile [31]

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3 years ago

Read 2 more answers

Calculate the vapor pressure of a solution made by dissolving 11.1 g Ca(OH)2 in 1 102 g of water at 25 °C. Vapor pressure of pur

Levart [38]

Answer:

23.15 mmHg.

Explanation:

To solve this question you need to understand that part of Raoult's law in your Chemistry textbook. So, let us delve right into the solution of the question.

We are given parameters such as the mass of Ca(OH)2 to be = 11.1 grams, mass of solvent = 102 grams, the Vapor pressure of pure water= 23.76 mm Hg, temperature = 25°C and vapour pressure of the solution= ??.

The molar mass of Ca(OH)2= 74 g/mol.

The first thing to do is to find the number of moles of Ca(OH)2 and that of water from the formula below;

Mass/ molar mass = Number of moles.

===> Number of moles, n= 11.1/ 74.

Number of moles = 0.15 moles Ca(OH)2.

===> Number of moles, n= 102/ 18.

Number of moles, n= 5.67 moles of water.

Next, we add the two moles together to find the solvent mile fraction since vapour pressure is proportional to mole fraction.

Then;

0.15 + 5.67 = 5.82.

Therefore, 5.67/ 5.82= 0.97.

Hence the vapor pressure of a solution = 0.97 × 23.76.

vapor pressure of a solution = 23.15 mmHg.

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4 years ago

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Board instances involving eligibility or disciplinary issues that are addressed by a board-approved order public records and accessible on the BON's web page.

Board Position Statements are a way to give nurses guidance on topics that are important to the Board and relate to the safety of the public, but they do not have the legal force of law. In the context of the position statement's overall intent, each position statement is intended to offer direction. Board position statements are examined every year to determine their applicability and accuracy in light of current practice, the Nursing Practice Act, and Board regulations. In January 2022, the Board conducted its most recent evaluation.

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Learn more about BON's web page here:

brainly.com/question/16982080

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2 years ago