All categories

3 years ago

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

1 answer:

4 0

<span>Humans breathe in oxygen and breathe out carbon dioxide.During the process of breathing, humans convert sugar into energy. Carbon dioxide is a waste product of this process. Carbon dioxide is released into the blood, travels to the lungs and is exhaled. Because carbon dioxide is a weak acid, the more carbon dioxide in the blood, the more acidic the blood becomes.</span>

You might be interested in

Which of the following is NOT an example of a Heterogeneous Mixture?

yuradex [85]

Answer:

pretty sure its table salt

5 0

2 years ago

Read 2 more answers

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl

4vir4ik [10]

Well, we need to find the ratio of Al to the other reactant.

Al:HCl = 1:3

--> this means that for every 1 Al used, you have to use 3 HCl.

6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.

The ratio of HCl:AlCl = 3:1

13/3 = 4.3333...

The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.

Hope this helps!!! :)

6 0

2 years ago

In a ceiling fan, N, electrical energy is converted to M, Kinetic energy. Some of the energy is "wasted" as heat. What is someth

Kruka [31]

Answer: It's D

Explanation:

I just did this question and i picked C but it's actually D

6 0

3 years ago

Read 2 more answers

Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo

Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

$n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4$

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

$left\ over=0g$

Regards.

7 0

2 years ago

A gas with a volume of 250 mL at a temperature of 293K is heated to 324K. What is the new volume of the gas?

ANEK [815]

Answer:

The new volume of the gas is 276.45 mL.

Explanation:

8 0

3 years ago