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tamaranim1 [39]
3 years ago
10

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air

is 2.67 ✕ 10-4 M at sea level and 25°C, what is the solubility of O2 at an elevation of 12,000 ft where the atmospheric pressure is 0.657 atm? Assume the temperature is 25°C, and that the mole fraction of O2 in air is 0.209 at both 12,000 ft and at sea level.
Chemistry
1 answer:
IceJOKER [234]3 years ago
8 0

Answer:

1.75\cdot 10^{-4} M

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

S = k_H p^o

Where k_H is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

S_1 = 2.67\cdot 10^{-4} M

Also, at sea level, we have an atmospheric pressure of:

p = 1.00 atm

Given mole fraction:

\chi_{O_2} = 0.209

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

p^o = \chi_{O_2} p

Then the equation becomes:

S_1 = k_H \chi_{O_2} p

Solve for k_H:

k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

p = 0.657 atm

Apply Henry's law using the constant we found:

S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M

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Answer:

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Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Initial rate of the reaction = R = 4.0\times 10^5 M/s

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The initial rate of the reaction when concentration of hydrogen gas is doubled : R'

[H_2]'=2[H_2]

R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3

R'=8\times k\times [N_2][H_2]^3

R'=8\times R=8\times 4\times 10^5 M/s=3.2\times 10^6 M/s

Initial rate of the reaction when concentration of hydrogen gas is doubled will be 3.2\times 10^6 M/s.

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Propane can be turned into hydrogen by the two-step reforming process. In the first step, propane and water react to form carbon
JulsSmile [24]

Answer:

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

Explanation:

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In the first step, propane and water react to form carbon monoxide and hydrogen. The balanced chemical equation is:

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In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide. The balanced chemical equation is:

CO(g) + H₂O(g) ⇒ H₂(g) + CO₂(g)

In order to get the net chemical equation for the overall process, we have to multiply the second step by 3 and add it to the first step. Then, we cancel what is repeated.

C₃H₈(g) + 3 H₂O(g) + 3 CO(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g) + 3 H₂(g) + 3 CO₂(g)

C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)

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What is formed when hydrobromic acid, hbr, and calcium hydroxide, ca(oh)2, are combined?
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balancing the elements: <span>

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6 0
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Which of the following is NOT a compound?<br> Table salt<br> Carbon dioxide<br> Water<br> Gold
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Answer: Gold.

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3 0
2 years ago
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If 16.5g of c6h14o2 are reacted vwith .499 mol of o2. How many moles of co2 should be produced
Fed [463]

Answer:

moles of carbon dioxide produced are 410.9 mol.

Explanation:

Given data:

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Moles of CO₂ = ?

First of all we will write the balance chemical equation.

2C₆H₁₄O₂  +  17O₂  →   14CO₂  +  12H₂O

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Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.

                 O₂      :     CO₂

                 17       :      14

                499     :      14/17× 499 = 410.9 moles

          C₆H₁₄O₂   :   CO₂

                    2    :      14

                 1947 :     14/2× 1947 =  13629 moles

Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.

5 0
3 years ago
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