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Leto [7]
2 years ago
8

Dead space is the portion of the respiratory system that: Select one: A.. must be filled with air before gas exchange can take p

lace. B. contains no alveoli and does not participate in gas exchange. C. includes the alveoli and capillaries surrounding the alveoli. D. receives oxygen but is unable to release carbon dioxide.
Chemistry
1 answer:
ololo11 [35]2 years ago
5 0

Answer: Option B

Explanation:

Dead space is the volume of sir in the body which is inhaled but it does not takes part in the gaseous exchange this is because it does not contains alveoli.

Not all the air that is inhaled inside the body is used in the process of respiration.

The dead space in the lungs is the region which helps in conducting airways, the air is not perfused to the alveoli.

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One hour of bicycle racing can require 500-900 kcal of energy, depending on the speed of the race, the terrain, and the weight o
diamong [38]

Answer:

0.467 kilograms of protein or carbohydrates

Explanation:

First, there is a need to understand that <em>1 kJ = 0.239 kCal</em>

Hence, 17 kJ = 17 x 0.239 = 4.063 kCal

The race requires 650.0 kCal/hr and has to last for 175 minutes.

175 minutes = 175/60 = 2.917 hrs

The total kCal requires for the race = 650 x 2.917 = 1,895.833 kCal

1 g of protein or carbohydrate food produces 17.0 kJ or 4.063 kCal of energy. Hence, the total g of protein or carbohydrate that will produce 1,895.833 kCal of energy would be;

                    1,895.833/4.063 = 466.609 g

<em>1 g = 0.001 kg</em>

466.609 g = 466.609 x 0.001 = 0.467 kg

<em>Hence, </em><em>0.467 </em><em>kilograms of proteins or carbohydrates must be consumed.</em>

4 0
1 year ago
What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product.
Mamont248 [21]

Answer:

78.46 grams of  2-bromopropane could be prepared from 25.5 g of propene

Explanation:

C_3H_6+HBr\rightarrow C_3H_7Br

Moles of propene = \frac{25.5 g}{39 g/mol}=0.6538 mol

According to reaction, 1 mole of propene gives 1 mole of propane.

Then 0.6538 moles of bromo-propane will give:

0.6538 mol\times 120 g/mol=78.46 g

78.46 grams of  2-bromopropane could be prepared from 25.5 g of propene.

5 0
2 years ago
Which substance is the oxidizing agent in this reaction? 2CuO+C→2Cu+CO2 Express your answer as a chemical formula.
Alja [10]
The oxidizing agent is the one that is reduced in the reaction. In this reaction, the charge of Cu falls from +2 to zero charge (neutral atom in the right side). Hence, CuO is the oxidizing agent. The reducing agent, the one being oxidized is carbon from zero charge to +4. The answer is CuO.
3 0
1 year ago
A certain element consists of two stable isotopes. The first has atomic mass of 7.02 amu and a percent natural abundance of 92.6
statuscvo [17]

<u>Answer:</u> The average atomic mass of the element is 6.95 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

We are given:

Mass of isotope 1 = 7.02 amu

Percentage abundance of isotope 1 = 92.6 %

Fractional abundance of isotope 1 = 0.926

Mass of isotope 2 = 6.02 amu

Percentage abundance of isotope 2 = 7.42 %

Fractional abundance of isotope 2 = 0.0742

Putting values in equation 1, we get:

\text{Average atomic mass of element}=[(7.02\times 0.926)+(6.02\times 0.0742)]

\text{Average atomic mass of element}=6.95amu

Hence, the average atomic mass of the element is 6.95 amu

8 0
1 year ago
Which statement is true of a reversible reaction at equilibrium?
Vedmedyk [2.9K]

Answer:

D.

The concentration of reactants and the concentration of products are constant.

Explanation:

pls mark as brainliest

7 0
1 year ago
Read 2 more answers
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