The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38×10−3 M. Part A Using Henry's law, calculate the molar concentration of O2 in the surface water of a mountain lake saturated with air at 20 ∘C and an atmospheric pressure of 665 torr . Express your answer using two significant figures. nothing

Answers

Answer 1

Answer:

Answer:

$1.21x10^(-3)$ M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x $10^(-3)$

H = $(0.21)/(1.38x10^(-3) )$

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = $1.21x10^(-3)$ M

Answer 2

Answer:

Final answer:

The molar concentration of O2 in the surface water of a mountain lake at 20 °C and an atmospheric pressure of 665 torr is approximately 1.21×10-3 M.

Explanation:

To calculate the molar concentration of O2 in the surface water of a mountain lake using Henry's law, we first need to understand how pressure affects the solubility of gases and vice versa.

As per Henry's law, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The partial pressure of O2 in air at sea level is 0.21 atm. This means that when the atmospheric pressure is 1 atm, the molar concentration of O2 is 1.38×10−3 M. At higher altitudes, the atmospheric pressure reduces. The given atmospheric pressure at the mountain lake is 665 torr, which is approximately 0.875 atm.

Using these values in Henry's law, the molar concentration of O2 can be calculated as:

C = P * x

where C is molar concentration, P is atmospheric pressure, and x is given solubility at 1 atm. Substituting the values:

C = (0.875 atm) * (1.38×10−3 M) = 1.21x10-3 M approximately

Learn more about Henry's Law here:

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Answers

Answer:

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Explanation:

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Sulfonation of naphthalene, C10H8, results in two products. One product is kinetically favored and predominates in the beginning of the reaction. Because the reaction is reversible, eventually the kinetically slower but thermodynamically favored product predominates. Draw the structure of these two products. (The naphthalene ring is already drawn for you. Do not change the double bond configuration in the given structures.)

Answers

Explanation:

The sulfonation of the naphthalene yield 2 products under different conditions:

When the reaction is carried at 80 °C, 1-naphthalenesulfonic acid is the major product because it is kinetically favoured product as arenium ion formed in the transition state corresponding to 1-naphthalenesulfonic acid is more stable due to better resonance stabilization.

When the reaction is carried at 160 °C, 2-naphthalenesulfonic acid is the major product as it is more stable than 1-naphthalenesulfonic acid because of steric interaction of the sulfonic acid group in 1-position and the hydrogen in 8-position.

The products are shown in image below.

All light travels at the same speed.
O True
O False

Answers

False

thank me later

Answer:

False

Explanation:

It depends on wixh kid of light is it

Example

Sun light difders from lamp light

How many atoms are in 2.3 moles Au?

Answers

Answer:

6.02×10^23atoms of au

Answer:

i think 3

Explanation:

A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature changes from 5° C to 30° C? (Neglect the expansion of the beaker, evaporation of alcohol and absorption of water vapor by alcohol.) The volume coefficient of expansion γγ for alcohol = 1.12 x 10-4 K-1

Answers

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

$v_0=500$
$\alpha =1.12* 10^(-4)$
$\Delta \epsilon = 25$

Now,

Increase in volume will be:

⇒ $\Delta V = \alpha* v_0* \Delta \epsilon$

By putting the given values, we get

$=1.12* 10^(-4)* 500* 25$

$=1.12* 10^(-4)* 12500$

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Three samples of a solid substance composed of elements A and Z were prepared. The first contained 4.31 g A and 7.70 g Z. The second sample was 35.9% A and 64.1% Z. It was observed that 0.718 g A reacted with Z to form 2.00 g of the third sample. Show that these data illustrate the law of definite composition.

Answers

Answer:

The percentage composition of the elements of the compound in the three samples is the same.

Explanation:

The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements in the same proportion by mass.

Sample A:

Mass of A = 4.31 g; mass of Z = 7.70 g

Total mass of sample = 12.01

Percentage mass of A in the sample = (4.31 * 100)/12.01 = 35.9 %

Percentage mass of Z in the sample = (7.70 * 100)/12.01 = 64.1 %

Sample B:

Percentage mass of A in the sample = 35.9 %

Percentage mass of Z in the sample = 64.1 %

Sample C:

Mass of A = 0.718 g; Total mass of sample = 2.00 g

mass of Z = mass of sample - mass of A = 2.00 g - 0.718 g = 1.282 g

Percentage mass of A in the sample = (0.718 * 100)/2.00 = 35.9 %