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noname [10]
3 years ago
7

If the atmospheric pressure on the top of a mountain is 0.2396 atm, what is the pressure in Kilopascal

Chemistry
2 answers:
arsen [322]3 years ago
4 0

Answer:

24.28 kPa

Step-by-step explanation:

You want to convert atmospheres to kilopascals.

The conversion factor is 101.325 kPa = 1 atm

p = 0.2396 atm × (101.325 kPa/1 atm)

p = 24.28 kPa

Ket [755]3 years ago
4 0

Answer:

The pressure in Kilopascal is 24.28 kPa

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

Then

x=\frac{c*b}{a}

Being 1 atmosphere = 101.325 kilopascal, the rule of three applies as follows: if 1 atm equals 101.325 kPa, 0.2396 atm how many kPa do they equal?

Pressure=\frac{0.2396atm*101.325 kPa}{1 atm}

Pressure=24.28 kPa

<u><em>The pressure in Kilopascal is 24.28 kPa</em></u>

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Answer:

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A chemist dissolves 249.mg of pure perchloric acid in enough water to make up 380.mL of solution. Calculate the pH of the soluti
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Answer:

pH = 2.18

Explanation:

Perchloric acid (HClO₄) is a strong acid. This means that in an aqueous solution it completely dissociates into H⁺ and ClO₄⁻ species.

First we <u>convert 249 mg HClO₄ into moles</u>, using its <em>molecular weight</em>:

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<em>Because it is a strong acid</em>, 2.49 mmol HClO₄ = 2.49 mmol H⁺

We <u>calculate the molar concentration of H⁺</u>:

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Finally we <u>calculate the pH of the solution</u>:

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4 0
3 years ago
A certain second-order reaction (B→products) has a rate constant of 1.30×10−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 224 s
soldier1979 [14.2K]

Answer:

       \large\boxed{\large\boxed{0.291M}}

Explanation:

By definition one <em>half-life</em> is the time to reduce the initial concentration to half.

For a <em>second order reaction </em>the rate law equations are:

              \dfrac{d[B]}{dt}=-k[B]^2

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The <em>half-life</em> equation is:

            t_{1/2}=\dfrac{1}{k[A]_0}

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           224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}

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Otherwise I would look at whether the data is reliable, for example whether  the author cites the sources for their data
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