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Stella [2.4K]
3 years ago
15

A scuba diver that ascends to the surface too quickly can experience decompression sickness, which occurs when nitrogen that dis

solves in the blood under high pressure, forms bubbles as the pressure decreases during the ascent. Therefore, an understanding of the gas laws is an important part of a scuba diver's training. In fresh water, the pressure increases by 1 atm every 34 ft below the water surface a diver descends.
If a diver ascends quickly to the surface from a depth of 102 ft without exhaling, by what factor will the volume of the diver's lungs change upon arrival at the surface?
Chemistry
1 answer:
Butoxors [25]3 years ago
5 0

Answer:

The factor that will change the volume of the diver's lungs upon reaching the surface is 4

Explanation:

Given data:

Pressure increases 1 atm = 101.325 kPa

34 ft = 10.3632 m

Depth of 102 ft = 31.0896 m

Question: What factor will the volume of the diver's lungs change upon arrival at the surface, V₂/V₁ = ?

The pressure at 31.0896 m:

P_{1} =101.325+(\frac{31.0896}{10.3632} *101.325)=405.3kPa

The factor will the volume of the diver's lungs change upon arrival at the surface:

\frac{V_{2}}{V_{1} } =\frac{P_{1}}{P_{2}} =\frac{405.3}{101.325} =4

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BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: K_{b} = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: K_{w} = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

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