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12 days ago

Why did we use a mixture of ethanol and water to perform the reaction between the oil and naoh?

Chemistry

1 answer:

alisha [964]12 days ago

4 0

Ethanol, with the formula C2H5OH, is also referred to as Ethyl alcohol.

Explanation:

The interaction between oil and sodium hydroxide (NaOH) is recognized as the Saponification process.

A combination of ethanol and water yields a fully homogeneous solution, miscible in all proportions.

We use a mixture of ethanol and water for the Saponification because it prevents the fat from reacting with atmospheric oxygen.

This mixture is advantageous as it exhibits lower polarity than water, aiding in the dissolution of non-polar fats, thereby facilitating reaction with sodium hydroxide.

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VMariaS [1037]

Answer:

Endothermic: water formation from ice and a cold instant ice pack Explanation:

8 0

1 day ago

Read 2 more answers

Calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid.

VMariaS [1037]

The question is incomplete,the complete question:

Determine the molality of a 10.0% (by weight) solution of hydrochloric acid in water:

a) 0.274 m

b) 2.74 m

c) 3.05 m

d) 4.33 m

e) the solution's density is necessary for calculations

Answer:

The molality for a 10.0% (by weight) hydrochloric acid solution is 3.05 mol/kg.

Explanation:

The solution is a 10.0% (by weight) hydrochloric acid mix.

This means there are 10 grams of HCl in 100 grams of the solution.

Amount of HCl = 10 g

Total mass of solution = 100 g

Total mass of solution = Mass of solute + Mass of solvent

Mass of solvent (water) = 100 g - 10 g = 90 g

Calculate moles of HCl = $\frac{10 g}{36.5 g/mol}=0.2740 mol$

Mass of water converted to kilograms = 0.090 kg

Molality = $\frac{0.2740 mol}{0.090 kg}=3.05 mol/kg$

<strongTherefore, the molality of a 10.0% (by weight) hydrochloric acid solution is 3.05 mol/kg.

6 0

7 days ago

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Tems11 [846]

Response:

ΔH = -793.6 kJ

Reasoning:

The ΔH for this particular reaction can be calculated through Hess's law, which allows one to combine the ΔH values of the half-reactions to find the overall ΔH:

The half-reactions are as follows:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The calculation involves summing (4) + 4×(3) - (2) - 2×(1) - 2×(5):

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369.2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356.9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483.6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

<pThus, ΔH is:

ΔH = -639.1 kJ -369.2 kJ -356.9 kJ +483.6 kJ +88.0 kJ

ΔH = -793.6 kJ

I trust this clarifies things!

5 0

3 days ago

2CH4(g)⟶C2H4(g)+2H2(g)

Alekssandra [968]

Answer: The enthalpy change for the reaction is, 201.9 kJ

Explanation:

Based on Hess’s law of constant heat summation, the energy released or absorbed in a chemical reaction stays constant, regardless of whether the process unfolds in one step or multiple steps.

This principle implies, that chemical equations can be treated analogously to algebraic expressions, allowing addition or subtraction to create the needed equation. Thus, the overall enthalpy change corresponds to the summation of the individual enthalpy changes of the reactions occurring in between.

The balanced equation for $CH_4$ appears as follows,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced reactions are outlined as follows,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Next, we will multiply the first reaction by 2, reverse the second, and reverse and halve the third and fourth reactions before combining them. This gives us:

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$