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BlackZzzverrR [31]
1 year ago
5

During normal, relaxed respiration, about 500ml of air enters and leaves the lungs with each respiratory cycle. this is called t

he ___________________.
Chemistry
1 answer:
Marianna [84]1 year ago
6 0

During normal respiration, about 500ml of air enters and leaves the lungs with each respiratory cycle. This is called the<u> tidal volume</u>.

When a person is relaxed, the normal amount of air such a person breathes in and out is called the tidal volume. It is usually measured in millimeters. For the average adult male, it is 500ml, and the tidal volume of average adult female 400ml.

In order to regulate oxygen intake and expulsion of carbon dioxide, the lungs act as buffers in order to absorb the maximum amount of oxygen possible for respiration and other metabolic functions in the body.

When the tidal volume is above or below the 500ml mark, it could signal the presence of underlying pathological conditions like bronchitis, emphysema and asthma.

To find out more about tidal volume, visit:

brainly.com/question/17439101

#SPJ4

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Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass
denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the <u>ppm formula</u>:

ppm=\frac{mg~of~solute}{Litters~of~solution}

If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of F^-, because the molarity formula is:

M=\frac{mol}{L}

In other words:

0.0320~M=\frac{mol}{1~L}

mol=0.032~M*1~L=0.032~mol

1~L~of~Solution=0.0320~mol~of~solute

If we use the <u>atomic mass</u> of F  (19 g/mol) we can convert from mol to g:

0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g

Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

0.607~g\frac{1000~mg}{1~g}=607~mg

Finally we can <u>divide by 1 L</u> to find the ppm:

ppm=\frac{607~mg}{1~L}=~607~ppm

<u>We will have a concentration of 607 ppm.</u>

I hope it helps!

4 0
3 years ago
a student performed an experiment, using a cocktail peanut, before it was burned the peanut half weighed .353 g. After burning t
tekilochka [14]
Not sure but it should be on google
5 0
3 years ago
Based upon the mass of baking soda (NaHCO3) and using an excess of HCl in this experiment, you will 1) determine the mass of CO2
weqwewe [10]

Equation of reaction

NaHCO3 +  HCl  --------->  NaCl  +  H2O  +  CO2(g)

1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.

Assume 1.0g of CO2 was gotten as a product of the experiment

2) For the theoretical yield

The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow

Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3

Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles

1 mole of NaHCO3 yielded 1 mole of CO2

0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2

Mass of CO2 = Number of moles * Molar Mass

Mass of CO2 = 0.0238 * 44 = 1.0472g

Theoretical yield of CO2 = 1.0472 grams

3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%

Percentage yield of CO2 = 1.0/1.0472 *100

Percentage yield of CO2 = 95.49%

8 0
3 years ago
Read 2 more answers
To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7
Cerrena [4.2K]

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

7 0
2 years ago
) A children’s liquid cold medicine has a density of 1.23 g/mL. If a child is to take 2.5 tsp in a dose, what is the mass in gra
vekshin1

The relation between density and mass and volume is

Density=\frac{Mass}{volume}

the dose required is 2.5 tsp

each tsp contain 5mL

So dose required in mL = 2.5 X 5 = 12.5 mL

the mass will be calculated using following formula

Mass=DensityXvolume

Mass=1.23\frac{g}{mL}X12.5mL= 15.38g

The mass of dose in grams will be 15.38 g


5 0
3 years ago
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