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Luba_88 [7]
3 years ago
11

Human blood contains a buffer of carbonic acid (H_2 CO_3) and bicarbonate anion (HC〖O_3〗^-) in order to maintain blood pH betwee

n 7.35 and 7.45, as a value higher than 7.8 or lower than 6.8 can lead to death. In this buffer, hydronium and bicarbonate anion are in equilibrium with carbonic acid. Furthermore, the carbonic acid in the first equilibrium can decompose into CO_2 gas and water, resulting in a second equilibrium system between carbonic acid and water. Because CO_2 is an important component of the blood buffer, its regulation in the body, as well as that of O_2 , is extremely important. The effect of this can be important when the human body is subjected to strenuous condition.
Carbonic acid buffer:
H_2 CO_3 (aq)+ H_2 O(l) ↔H_3 O^+ (aq)+HC〖O_3〗^- (aq)
1. Write the equilibrium constant expression K_c for the carbonic acid buffer in the blood.

Chemistry
1 answer:
kirill115 [55]3 years ago
6 0

Kc = [H3O+][HCO3-] / [H2CO3]

Remember that Kc is products over reactants. Also, you do not include liquid water in a Kc expression, since liquid water has no concentration.

You might be interested in
Determining Density and Using Density to Determine Volume or Mass
Shalnov [3]

corrected question:

Determining Density and Using Density to Determine Volume or Mass

(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³

(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?

(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

Answer:

density = \frac{mass}{volume}

ρ=m/v ,m=ρv,    v=m/ρ

(a)m=1*10g  , v=7.36cm³

    ρ=10/7.36 =1.36g/cm³

(b) m=65g, ρ=0.791 g/mL.

   v= 65/0.791 =82.17g/mL

(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³

    m=19..32*8=154.56g/cm³

(d) mass of copper=374.5g , v=41.8cm³

   ρ=374.5/41.8 =8.96g/cm³

 mass of ethanol=15g,  density of ethanol=0.789g/mL

v=15/0.789 =19.01mL

volume of mecury=25mL, density of mercury=13.6g/mL

m=25*13.6=340g

4 0
3 years ago
An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam
MAXImum [283]

Explanation:

The given data is as follows.

       Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence,   moles of NaOH in 97 ml = \frac{0.1090 \times 97}{1000}

                                                       = 10.573 \times 10^{-3} mol

Moles of HCl in 21.00 ml = \frac{0.2060 \times 21}{1000}

                                         = 4.326 \times 10^{-3} mol

Therefore, total moles of NaOH that reacted are as follows.

           10.573 \times 10^{-3} mol - 4.326 \times 10^{-3} mol      

                = 6.247 \times 10^{-3} mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

                  \frac{6.247 \times 10^{-3}}{3}

                    = 2.082 \times 10^{-3} mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

                    2.082 \times 10^{-3} mol \times 40 g/mol

                         = 83.293 \times 10^{-3} g

                         = 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore,       40 g NaOH = 122 g benzoic acid

So,            0.0832 g NaOH = \frac{122 g}{40 g} \times 0.0832 g

                                             = 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

                      \frac{0.253 g}{0.3471 g} \times 100

                           = 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0
3 years ago
The total pressure in a mixture of gases is equal to the partial pressures of
baherus [9]

Answer:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

Explanation:

Dalton's law of partial pressures state that, in a mixture of gases, the total pressure is equal to the sum of the partial pressure exerted by each gas of the mixture. The equation is:

Total pressure = Partial pressure Gas 1 + Partial pressure Gas 2 + .... + Partial pressure Gas n

To complete the sentence we can say:

"The total pressure in a mixture of gases is equal to the sum of partial pressures of each gas"

5 0
3 years ago
Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation
nikdorinn [45]

<u>Answer:</u> The isotopic symbol of barium is _{56}^{138}\textrm{Ba} and that of strontium is _{38}^{89}\textrm{Sr}

<u>Explanation:</u>

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

  • For the given fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

A = 139

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is _{56}^{139}\textrm{Ba}

  • For the given fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is _{38}^{89}\textrm{Sr}

Hence, the isotopic symbol of barium is _{56}^{138}\textrm{Ba} and that of strontium is _{38}^{89}\textrm{Sr}

7 0
3 years ago
Why is it that the decomposition of nitroglycerin can result in so
patriot [66]

Answer:

with the molecular formula C3H5(ONO2)3, has a high nitrogen content (18.5 percent) and contains sufficient oxygen atoms to oxidize the carbon and hydrogen atoms while nitrogen is being liberated, so that it is one of the most powerful explosives known.

Explanation:

NTG reduces preload via venous dilation, and achieves modest afterload reduction via arterial dilation. These effects result in decreased myocardial oxygen demand. In addition, NTG induces coronary vasodilation, thereby increasing oxygen delivery.

3 0
3 years ago
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