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Tema [17]
1 year ago
11

What would be the vital capacity be for a man that has a tidal volume of 500mL, a residual volume of 1100 mL, an expiratory rese

rve of 1100 mL and an inspiratory reserve of 3000 mL
Chemistry
1 answer:
dezoksy [38]1 year ago
3 0

The vital capacity will be 4600ml.

<h3>What is vital capacity?</h3>

The highest amount of air a person can inhale following their maximal exhalation is known as their vital capacity. It is equivalent to the total of the inspiratory, tidal, and expiratory reserve volumes. It roughly corresponds to Forced Vital Capacity. A wet or conventional spirometer can assess a person's vital capacity.

Normal people have a 3 to 5-liter vital capacity.

It enables simultaneous inhalation of the greatest possible volume of clean air and exhalation of the greatest possible volume of stale air. So, by increasing gaseous exchange between the body's various tissues, it improves the amount of energy available for bodily function.

VC = TV₊IRV₊ERV

where,

VC = Vital capacity

TV = Tidal volume

IRV = inspiratory reserve volume

ERV = expiratory reserve volume

VC = 500 ₊ 3000 ₊ 1100

VC = 4600ml

Therefore, the vital capacity will be 4600ml.

To know more about  vital capacity refer to: brainly.com/question/14877276

#SPJ4

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The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com
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Answer:

(a) Ksp=4.50x10^{-7}

(b) Ksp=1.55x10^{-6}

(c) Ksp=2.27x10^{-12}

(d) Ksp=1.05x10^{-22}

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)

Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L}   )^2\\\\Ksp=4.50x10^{-7}

(B) Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)

Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}

(C) NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)

Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}

(D) La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)

Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}

Best regards.

7 0
3 years ago
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