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cricket20 [7]
3 years ago
6

The most important chemical regulator of respiration is question 2 options: carbon dioxide. sodium ions. potassium ions. oxygen.

nitrogen.
Chemistry
2 answers:
Alenkinab [10]3 years ago
8 0
CARBON DIOXIDE is the most important chemical regulator of respiration. Respiration primarily is the process of gas (carbon dioxide and oxygen) exchange between the air and the person's blood. There is a signal to increase ventilation when: there is to much carbon dioxide in the blood and if there is a decrease in the pH levels of the blood.

Another important chemical regulator is oxygen. There is a signal to increase ventilation when there is a drop of oxygen levels in the body.
anzhelika [568]3 years ago
4 0
The most important chemical regulator of respiration is carbon dioxide. 
Breath control is accomplished chemically, mainly using Carbon dioxide and oxygen chemoreceptors. The most important chemical regulator of respiration is either carbon dioxide or oxygen, since the chemical regulation of breathing is different in healthy and sick individuals. 
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Alex Ar [27]
What do you need help with exactly can you explain ? I think I understand what you mean cause I just had this assignment in chemistry
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3 years ago
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What are rarefactions and compressions of sound waves?
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Compression and rarefaction. However instead of crests and troughs, longitudinal waves have compressions and rarefactions. A compression is a region in a longitudinal wave where the particles are closest together. A rarefaction is a region in a longitudinal wave where the particles are furthest apart.
6 0
3 years ago
A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
ivanzaharov [21]

Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

4 0
3 years ago
Fill in the [?]:<br> 3,721 nm = [?]m<br> Give your answer in standard form.
swat32

0.000003721

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A 25-mL sample of 0.160M solution of NaOH is titrated with 17 mL of an unknown solution of
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V ( NaOH ) = 25 mL in liters : 25 / 1000 => 0.025 L

M ( NaOH ) = 0.160 M

V ( H2SO4) = 17 mL / 1000 => 0.017 L

M ( H2SO4) = ?

number of moles NaOH:

n = M x V = 0.160 x 0.025 => 0.004 moles NaOH

Mole ratio :

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ----------- 1 mole H2SO4
0.004 moles NaOH ----- ? moles H2SO4

moles H2SO4 = 0.004 x 1 / 2

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M ( H2SO4) = n / V

M = 0.002 / 0.017

 = 0.117 M H2SO4

Answer B

hope this helps!

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