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MrRa [10]
2 years ago
5

En el aire que respiramos y en el aire comprimido de los tanques de los buzos se encuentra, además de oxígeno, el gas nitrógeno

(N2). Este gas entra y sale de los pulmones sin consecuencias para el organismo pero, a más de 30 m de profundidad del agua, este gas se disuelve en el torrente sanguíneo.
Cuando los buzos suben a la superficie abruptamente, la presión disminuye, lo que provoca que se reduzca la solubilidad del nitrógeno disuelto y en consecuencia este gas salga de la sangre en forma violenta. Este proceso puede provocar problemas de salud o incluso la muerte del buzo, por lo cual se recomienda que suban lentamente y haciendo pausas.
a) ¿Es la solubilidad una propiedad intensiva o extensiva?
Chemistry
1 answer:
marysya [2.9K]2 years ago
7 0

Answer: can someone please translate so i can answer.

Explanation:

sorry

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A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make
hoa [83]

Answer:

A buffer solution is prepared by adding 13.74 g of sodium acetate (NaC2H3O2) and 15.36 g of acetic acid to enough water to make 500 mL of solution.

Calculate the pH of this buffer.

Explanation:

The pH of a buffer solution can be calculated by using the Henderson-Hesselbalch equation:

pH=pKa+log\frac{[salt]}{[acid]}

The pH of the given buffer solution can be calculated as shown below:

6 0
3 years ago
Describe the procedure Bourouiba uses to study sneezes.
Gekata [30.6K]

Answer:

The Lydia Bourouiba is recorded is around more than the 100 sneezes and coughs cases.

Explanation:

Lydia Bouroubia was a mathematical physicist who does experiments on sneezes and coughs. She was a leading expert in coughs and sneezes. Sneezing is a phenomenal activity that occurs in our body to excrete the dust particles and germs.

The disease also spread through it. Lydia thought that her research will help people related to their health. It will help in stopping epidemics in public. It is a type of outbreak of disease at a bigger level in the public.

6 0
3 years ago
24 grams of CH4 was added to the above reaction. Calculate the theoretical yield of CO2. A. 66 grams B. 33 grams c. 132 grams. D
Sonbull [250]

Answer:

Option A. 66 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the mass of CH₄ that reacted and the mass of CO₂ produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4 = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Finally, we shall determine the theoretical yield of CO₂. this can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted to produce 44 g of CO₂.

Therefore, 24 g of CH₄ will react to produce = (24 × 44) /16 = 66 g of CO₂.

Thus, the theoretical yield of CO₂ 66 g

8 0
2 years ago
Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent
Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

4 0
3 years ago
What is the enthalpy of combustion (per mole) of C4H10 (g)? 
Artemon [7]
The balanced chemical reaction for the complete combustion of C4H10 is shown below:

                    C4H10 + (3/2)O2 --> 4CO2 + 5H2O

The enthalpy of formation are listed below:
          C4H10: -2876.9 kJ/mol
              O2:   none (because it is pure substance)
             CO2: -393.5 kJ/mol
             H2O: -285.8 kJ/mol

The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.

               ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
                       = -<em>126.1 kJ</em>

Thus, the enthalpy of combustion of the carbon is -126.1 kJ. 
5 0
2 years ago
Read 2 more answers
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