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3 years ago

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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?

1 answer:

marysya [2.9K]3 years ago

7 0

Answer: can someone please translate so i can answer.

Explanation:

sorry

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A solution is prepared by mixing 250 mL of 1.00 M CH3COOH with 500 mL of 1.00 M NaCH3COO. What is the pH of this solution? (Ka f

Svetllana [295]

Answer:

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid, CH 3 COOH , a weak acid, and sodium hydroxide, NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not. If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal 7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over 7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq] + OH − (aq] → CH 3 COO − (aq] + H 2 O (l]

Notice that:

1 mole of acetic acid will react with: 1 mole of sodium hydroxide, shown here as hydroxide anions, OH − , to produce 1 mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

c = n / V ⇒ n = c ⋅ V

n acetic = 0.20 M ⋅ 25.00 ⋅ 10 − 3 L = 0.0050 moles CH3 COOH

and

n hydroxide = 0.10 M ⋅ 40.00 ⋅ 10 − 3 L = 0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

n acetic remaining = 0.0050 − 0.0040 = 0.0010 moles

The reaction will also produce 0.0040 moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH = p K a + log ( [ conjugate base ] / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total = V acetic + V hydroxide

V total = 25.00 mL + 40.00 mL = 65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ] = 0.0010 moles / 65.00 ⋅ 10 − 3 L = 0.015385 M

and

[ CH 3 COO − ] = 0.0040 moles / 65 ⋅ 10 − 3 L = 0.061538 M

The p K a of acetic acid is equal to 4.75

Thus the pH of the solution will be:

pH = 4.75 + log ( 0.061538 M / 0.015385 M )

pH = 5.35

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3 years ago

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How are lysosomes and vacuoles the same? How are<br> they different?

andriy [413]

Answer:

This is the answer they're looking for:

Lysosomes and vacuoles both deal with waste materials. Lysosomes break down waste materials, and vacuoles store waste materials in the cell temporarily before the cell get rids of them.

Explanation:

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Why can't astronomers take a picture of a black hole?

san4es73 [151]

Answer:

Gravity is so strong that even light cannot escape a black hole.

Explanation:

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Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

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3 years ago

How are the terms molar mass and atomic mass different from one another

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2.Molar mass is measured in grams per mole while atomic mass is “unitless.”

3.Atomic mass is measured via mass spectrometry while molar mass is computed via atomic weight.

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