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Kryger [21]
2 years ago
5

The normal boiling point of fingernail polish remover is 56℃. what is the vapor pressure of fingernail polish remover at the sal

on, which is kept at a temperature of 22℃ for the comfort of customers?
Chemistry
1 answer:
Elenna [48]2 years ago
7 0

The normal boiling point of fingernail polish remover is 56℃  the vapor pressure of fingernail polish remover at the salon 0.027 atmospheric pressure

<h3>What is atmospheric pressure?</h3>

The pressure exerted by the vapors due to the atmosphere as it is surrounded by it only is known as vapor pressure and its unit is atm.

As air have small molecules which forms multiple layers and create pressure.

vapor pressure = {1/T1 - 1/T2}

T1 = 56 c

T2 = 22 c

substituting the values,

vapor pressure = [ 1/ 56 - 1 / 22 ]

                         = 0.027 atm.

Therefore, the vapor pressure of nail paint remover is 0.027 atm

Learn more about vapor pressure , here:

brainly.com/question/8696772

#SPJ4

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Answer :  The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

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The expression for rate of reaction will be :

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

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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,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}

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

\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}

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\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}

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As,  

-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s

and,

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+\frac{d[H]}{dt}=0.36mol/L.s

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

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