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3 years ago

Which acid could not be prepared by treating a Grignard reagent with CO2?

Chemistry

1 answer:

Anni [7]3 years ago

6 0

Answer:

4-oxopentanoic acid.

Explanation:

In this case, we must remember that the Grignard reaction is a reaction in which carbanions are produced. Carboanions have the ability to react with CO2 to generate a new C-C bond and a carboxylate ion. Finally, the acid medium will protonate the carboxylate to produce the carboxylic acid group. 

The molecules that can follow the mechanism described above are the molecules: p-methylbenzoic acid, cyclopentane carboxylic acid and 3-methylbutanoic acid. (See figure 1)

In the case of 4-oxopentanoic acid, the possible carbanion will attack the carbonyl group to generate a cyclic structure and an alcohol group (1-methylcyclopropan-1-ol). Therefore, this molecule cannot be produced by this reaction. (See figure 2)

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A salt shaker filled with 17.0g sodium chloride contain how many moles?

kati45 [8]

The molecular formula for sodium chloride is NaCl. The sum of their atomic weights is (22.99 grams/mole + 35.45 grams/mole) = 58.44 grams/mole
take (17.0 grams)/(58.44 grams/mole), which equals 0.291 moles of NaCl.

8 0

4 years ago

In this activity, you will be observing a reaction where copper and silver nitrate react to form copper (II) nitrate and silver.

aalyn [17]

Answer:

Copper(II) nitrate and silver

Explanation:

2AgNO₃(aq ) + Cu (s) --> Cu(NO₃)₂(aq ) + 2Ag(s)

This can be called a redox reaction because silver nitrate is reduced and copper is oxidized. This can also be called a single replacement reaction because copper replaces silver in the substance silver nitrate.

4 0

3 years ago

An organic compound is analyzed, and it has twice as many hydrogen atoms as oxygen atoms. this compound is most likely a _______

Alborosie

For example, monosaccharides:

C₆H₁₂O₆
aldohexoses
ketohexoses

C₃H₆O₃
aldotrioses
ketotrioses

and many others

4 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

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

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

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

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

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

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

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

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

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

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

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 $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

Write the chemical symbols for three different atoms or atomic anions with 10 electrons.

anzhelika [568]

Answer:

i) Neon (Ne)

ii) Fluoride Ion (F⁻¹)

iii) Oxide Ion (O⁻²)

Explanation:

Ions are those charged species which are either positively charged (by loosing electrons) called as cations or negatively charged (by gaining electrons) called as anions.

In given examples, Neon is a neutral atom which has an atomic number 10. It contains 10 electrons in its neutral state with the electronic configuration 1s², 2s², 2p⁶.

Fluorine atom has an atomic number of 9. Therefore, it contains 9 electrons in its neutral state with an electronic configuration of 1s², 2s², 2p⁵. When Fluorine atom gains one electron it gets 10 electrons with electronic configuration of 1s², 2s², 2p⁶.

Oxygen atom has an atomic number of 8. Therefore, it contains 8 electrons in its neutral state with an electronic configuration of 1s², 2s², 2p⁴. When Oxygen atom gains two electron it gets 10 electrons with electronic configuration of 1s², 2s², 2p⁶ forming an Oxide Ion.

8 0

4 years ago