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3 years ago

Why do the oil and vinegar separate into layers when they are stirred together, but completely mix when lecithin is stirred in?

Chemistry

1 answer:

Sedaia [141]3 years ago

4 0

Answer:

See explanation

Explanation:

Vinegar and oil have different densities, as such we expect the two substances to separate when mixed. However, oil constitutes the organic phase(water insoluble) while vinegar constituents the aqueous phase since it is water soluble. Therefore, the two substances separate into layers when left to stand for a while.

The structure of lecithin makes it a good emulsifier. Hence, it can be able to cause oil and vinegar to mix, constituting a single phase when lecithin is stirred in.

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I believe it’s Hydrogen

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The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen,

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Answer : The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}$

$\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}$

$\text{Rate of formation of }CO=+\frac{d[CO]}{dt}$

$\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}$

The rate of reaction expression is:

$\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}$

As we are given that:

$+\frac{d[CO]}{dt}=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}$

$+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s$

$\frac{d[H_2]}{dt}=3\times 0.35M/s$

$\frac{d[H_2]}{dt}=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s

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