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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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3. (10 Points) You buy a helium balloon of 14 Liters for your friend's birthday and
vesna_86 [32]

Answer: picture shows work for # 1,2,4,5,7

Explanation:

number 3: as the pressure in the volume decreases, the volume increases causing it to expand and eventually blow.

number 6: because the temperature and the amount of gas don’t change, these terms don’t appear in the equation. What Boyle’s law means is that the volume of a mass of gas is inversely proportional to its pressure. This linear relationship between pressure and volume means doubling the volume of a given mass of gas decreases its pressure by half.

hope this helps :))

7 0
2 years ago
| A solution containing 4.48 ppm KMnO4 exhibits
Artist 52 [7]

Answer:

Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹

Explanation:

First convert Concentration from ppm inM or mol/l

⇒ Molar mass of KMnO₄ = 158.03 g

⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l

⇒ Molarity = \frac{4.48 X10^{-3} }{158.03X 1(lit)} = 2.83 x 10⁻⁵ molar

Absorbance (A) = - log(T)     ( T = % transmittance)

                          = - log(0.859)

                          = 0.06

According to Lambert Beer's law

     

                 ε = \frac{A}{C X l}

      or,      ε = \frac{0.06}{2.83 X 10^{-5}X1 cm }

      or,      ε = 2120.14 cm⁻¹M⁻¹

Where

    ε = Molar absorptivity

    A = absorbance

    C = Molar concentration of KMnO₄ solution

     l = length  

6 0
3 years ago
Which of the following is the best explanation for why it is important to follow lab safety guidelines?
I am Lyosha [343]

Answer:

Following laboratory safety guidelines minimizes the chance of lab accidents.

Explanation:

8 0
2 years ago
One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su
Tpy6a [65]

Answer : The percent purity of Fe_2O_3 in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

Fe_2O3+3CO\rightarrow 2Fe+3CO_2

First we have to calculate the mass of Fe.

\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}

Molar mass of Fe = 55.8 g/mole

\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole

Now we have to calculate the moles of Fe_2O_3

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of Fe_2O_3

So, 3.15\times 10^4mole of Fe produced from \frac{3.15\times 10^4}{2}=15750 mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3

Molar mass of Fe_2O_3 = 159.69 g/mole

\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg

Now we have to calculate the percent purity of Fe_2O_3 in the original sample.

Mass of original sample = 2.86\times 10^3kg

\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100

\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%

Therefore, the percent purity of Fe_2O_3 in the original sample is 87.94 %

3 0
2 years ago
How to go this problem
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Dccccccccccccccccccccccsssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss
4 0
3 years ago
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