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jeka94
3 years ago
13

What effect does rapid exhalation of CO2 during exercise have on the concentration of H2CO3 in the blood?

Chemistry
2 answers:
nydimaria [60]3 years ago
7 0

Answer:

The concentration of H2CO3 decreases.

Explanation:

Respiration may be defined as the process of exchange of gases between the alveoli, tissues and external environment. The oxygen is inhaled and carbon dioxide is exhaled out during respiration process.

Rapid exhalation of CO2 means the amount of oxygen is more than the carbon dioxide. The concentration of H2CO3 will decrease as less amount of carbon dioxide is available in the body. The concentration of carbon dioxide is directly proportional to the concentration of H2CO3. Hence, the concentration of H2CO3 will decrease.

Thus, the correct answer is option (1).

BARSIC [14]3 years ago
3 0

If there is a constant loss of CO_{2}~(carbon~dioxide) the concentration of H_{2} CO_{3}~(carbonic~acid) will be affected and decrease since CO_{2} is a main component of H_{2}CO_{3}.


Hope it helped,


BioTeacher101

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Answer:

The answer is

<h2>25 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}

From the question

mass of CCl4 = 40g

Density = 1.6 g/mL

The volume is

volume =  \frac{40}{1.6}

We have the final answer as

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5 0
3 years ago
Calculate the molarity of an HCl solution if 23.88 mL of it reacts with 6.5287 grams of sodium carbonate (106 g/mol) according t
Ivanshal [37]

Answer:

5.158 mol/L

Explanation:

To find the molarity, you need to use the formula:

Molarity (M) = moles / volume (L)

You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.

<u>Steps 1 - 2:</u>

2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂

6.5287 g Na₂CO₃         1 mole            2 moles HCl
--------------------------  x  -------------  x  -------------------------  =  0.12318 mole HCl
                                      106 g           1 mole Na₂CO₃

<u>Step 3:</u>

23.88 mL / 1,000 = 0.02388 L

<u>Step 4:</u>

Molarity = moles / volume

Molarity = 0.12318 mole / 0.02388 L

Molarity = 5.158 mole/L

**mole/L is equal to M**

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