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Mekhanik [1.2K]
3 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]3 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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Consider the molecules H2O, H2S, H2Se, and H2Te. Which do you expect to have the highest boiling point, and why?
STatiana [176]

Answer:

a. H20,because it experiences hydrogen bonding.

5 0
2 years ago
Write the complete ionic equations, spectator ions and net ionic equation for the following.
gizmo_the_mogwai [7]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

 ZnBr₂ (aq) + AgNO₃ (aq)  →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq)  →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq)  → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq)   →    2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) HgCl₂ (aq) + KI (aq)  →

Chemical equation:

HgCl₂ (aq) + KI (aq)  → KCl + HgI₂

Balanced chemical equation:

HgCl₂ (aq) + 2KI (aq)  → 2KCl(aq) + HgI₂(s)

Ionic equation:

Hg²⁺(aq)  + Cl₂²⁻  (aq) + 2K⁺(aq) + 2I⁻ (aq)  →  HgI₂ (s) + 2K⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Hg²⁺(aq)  + 2I⁻ (aq) →   HgI₂ (s)

The Cl⁻ ((aq)  and K⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The HgI₂ (s) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

 

3) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq)  + OH₂²⁻  (aq) + 2Na⁺(aq) + SO₄²⁻ (aq)  →   CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq)   + SO₄²⁻ (aq)  →   CaSO₄(s)

The OH⁻ ((aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4 0
3 years ago
What is characteristic of a covalent bond?
dmitriy555 [2]
A.electrons are shared between two different nuclei
7 0
3 years ago
Acetylsalicylic acid (aspirin), HC9H7O4, is the most widely used pain reliever and fever reducer. Find the pH of 0.035 M aqueous
KatRina [158]

<u> </u> The pH of 0.035 M aqueous aspirin is 2.48

<u>Explanation:</u>

We are given:

Concentration of aspirin = 0.035 M

The chemical equation for the dissociation of aspirin (acetylsalicylic acid) follows:

               HC_9H_7O_4\rightleftharpoons H^++C_9H_7O_4^-

<u>Initial:</u>         0.035

<u>At eqllm:</u>    0.035-x        x         x

The expression of K_a for above equation follows:

K_a=\frac{[C_9H_7O_4^-][H^+]}{[HC_9H_7O_4]}

We are given:

K_a=3.6\times 10^{-4}

Putting values in above expression, we get:

3.6\times 10^{-4}=\frac{x\times x}{(0.035-x)}\\\\x=-0.0037,0.0033

Neglecting the value of x = -0.0037 because concentration cannot be negative

So, concentration of H^+ = x = 0.0033 M

  • To calculate the pH of the solution, we use the equation:

pH=-\log[H^+]

We are given:

H^+ = 0.0033 M

Putting values in above equation, we get:

pH=-\log(0.0033)\\\\pH=2.48

Hence, the pH of 0.035 M aqueous aspirin is 2.48

7 0
3 years ago
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