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Alexxx [7]
3 years ago
8

The solubility of O2 in water is approximately 0.00380 g L-1 of water when the temperature is 25.0°C and the partial pressure of

gaseous oxygen is 760. torr. The oxygen gas above the water is replaced by air at the same temperature and pressure, in which the mole fraction of oxygen is 0.210. What will the solubility of oxygen in water be under these new conditions?
Chemistry
1 answer:
alexandr402 [8]3 years ago
6 0

Answer:

The correct answer is 0.00080 gram per liter.

Explanation:

Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.  

Based on Henry's law,  

Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas

Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas

= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr

= 0.00012 mol/L/atm

= 0.00012 M/atm

Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure

= 0.210 × 1 atm

= 0.210 atm

Now putting the values in Henry's law equation we get,  

Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,

= 0.000025 mol/L × 32 gram/mol

= 0.00080 gram per liter

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<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

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AB\rightarrow A+B

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AB+CD\rightarrow AD+CB

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In stomach, an acid is present known as hydrochloric acid and to neutralize its effect, antacid is taken which has CaCO_ as a component.

The reaction between HCl and CaCO_3 is a type of neutralization reaction and it is a type of double displacement reaction.

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3 years ago
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lyudmila [28]

Answer:

If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3

Explanation:

Step 1: Data given

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