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Tema [17]
2 years 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]2 years 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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skad [1K]
Answer:

Bi (Bismuth)
4 0
3 years ago
Read 2 more answers
If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit
Blizzard [7]

<u>Answer:</u>

<u>For a:</u> The volume of the box is 217.5 mL

<u>For b:</u> The volume of the box is 0.2175 L

<u>Explanation:</u>

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

V=10.00\times 7.25\times 3.00=217.5cm^3

  • <u>For a:</u>

To convert the volume of cuboid into milliliters, we use the conversion factor:

1mL=1cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL

Hence, the volume of the box is 217.5 mL

  • <u>For b:</u>

To convert the volume of cuboid into liters, we use the conversion factor:

1L=1000cm^3

So,

\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L

Hence, the volume of the box is 0.2175 L

3 0
4 years ago
7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.
Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

3 0
3 years ago
Explain why lithium is a strong reducing agent, whereas fluorine is a strong oxidizing agent.
Tom [10]
Lithium is a good reducing agent because it is electropositive [it rapidly gains electrons]
fluorine is  good oxidizing agent electronegative [it loses electrons fastly]
3 0
3 years ago
How do you convert from grabs to moles of a substance
Lorico [155]

The correct question is as follows:

How do you convert from grams to moles of a substance

1. Divide by the molar mass

2. Subtract the molar mass

3. Add the molar mass

4. Multiple by the molar mass

Answer: In order to convert from grams to moles of a substance divide by the molar mass.

Explanation:

The number of moles of a substance is the mass of substance in grams divided by its molar mass.

The formula to calculate moles is as follows.

Moles = \frac{mass}{molar mass}

This means that grams are converted to moles when grams is divided by molar mass.

Thus, we can conclude that in order to convert from grams to moles of a substance divide by the molar mass.

8 0
3 years ago
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