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Mekhanik [1.2K]
4 years ago
13

Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many

grams of formaldehyde are permissible in a 6.0-L breath of air having a density of 1.2 kg/m3?
Chemistry
1 answer:
Flauer [41]4 years ago
8 0

Answer : The amount of formaldehyde permissible are, 5.4\times 10^{-6}g

Explanation : Given,

Density of air = 1.2kg/m^3=1.2g/L     (1kg/m^3=1g/L)

First we have to calculate the mass of air.

\text{Mass of air}=\text{Density of air}\times \text{Volume of air}

\text{Mass of air}=1.2g/L\times 6.0L

\text{Mass of air}=7.2g

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 5.4\times 10^{-6}g

Thus, the amount of formaldehyde permissible are, 5.4\times 10^{-6}g

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I need for a test and explain please
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Answer:

0.200M H₃PO₄

0.600N H₃PO₄

pH = 1.46

Explanation:

The acid-base reaction of phosphoric acid (H₃PO₄) with LiOH is:

3 LiOH + H₃PO₄ → Li₃PO₄ + 3H₂O

<em>Where 3 moles of LiOH reacts per mole of H₃PO₄</em>

<em />

Moles of LiOH are:

0.030L× (0.5mol / L) = 0.0150 moles of LiOH

Moles of acid neutralized are:

0.0150 moles of LiOH × (1 mole H₃PO₄ / 3 moles LiOH) = 0.005 moles H₃PO₄

As volume of acid was 25mL, molarity is:

0.005mol H₃PO₄ / 0.025L =<em> 0.200M H₃PO₄</em>

Normality is:

0.200M × (3N H⁺ / 1M H₃PO₄) = <em>0.600N H₃PO₄</em>

H₃PO₄ dissolves in water thus:

H₃PO₄ ⇄ H₂PO₄⁻ + H⁺

Ka = 7.1x10⁻³ = [H₂PO₄⁻] [H⁺] / [H₃PO₄]

Where molar concentrations in equilibrium will be:

[H₂PO₄⁻] = X

[H⁺] = X

[H₃PO₄] = 0.200M - X

Replacing in Ka formula:

7.1x10⁻³ = [X] [X] / [0.200 - X]

1.42x10⁻³ - 7.1x10⁻³X = X²

0 = X² + 7.1x10⁻³X - 1.42x10⁻³

Solving for X:

X = -0.04M →False answer, there is no negative concentrations.

X = 0.0343M

As, [H⁺] = 0.0343M

pH = - log [H⁺],

<em>pH = 1.46</em>

6 0
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