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photoshop1234 [79]
3 years ago
9

Which class of organic compound is least likely to be used as an organic solvent? aldehyde ketone alkyl halide ether

Chemistry
2 answers:
Nady [450]3 years ago
5 0

Answer:

The organic compund that is the least likely to be used as an organic solvent is:

aldehyde

Explanation:

The reason behind this is that even though aldehydes are used in the perfumery industry. They are not very good solvents because the majority of them are hazardous. Only a minimal fraction of them can be tolerated by organisms and that can even make them have weak poisoning reactions. However, using them as solvents in bigger quantities will make them hazardous for organisms considering their high concentration of ketones.

Over [174]3 years ago
3 0

i think its a

Aldehyde

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Answer : The correct option is, (C) 2, 4 and 5.

Explanation :

Combustion reaction : It is a type of reaction in which a hydrocarbon react with an oxygen molecule to give carbon dioxide, water as a product.

For example : Methane react with oxygen to give carbon dioxide and water.

CH_4+2O_2\rightarrow CO_2+2H_2O

In the given list of chemical substances, CuO,Cu_2O,Li_2O are in oxide form. They can not be both reactant and product of a single combustion reaction.

In the given list, C_2H_5NH_2 is the only hydrocarbon which shows a combustion reaction. That means C_2H_5NH_2 react with O_2 to give CO_2,H_2O and N_2 as a product.

The balanced combustion reaction of C_2H_5NH_2 is,

20C_2H_5NH_2+40O_2\rightarrow 40CO_2+7H_2O+10N_2

Therefore, the correct answer is, (C) 2, 4, and 5.

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How GM food are produced?​
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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
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