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Kobotan [32]
3 years ago
14

he solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at

2.5 atm and 25 °C is ________ g/L.
Chemistry
1 answer:
Tema [17]3 years ago
8 0

Answer:

0.1025 g/L

Explanation:

From the question above,

The solubility of oxygen gas in water at 25°c and 1.0 atm pressure is 0.041g/L

Henry's law states at a constant temperature the solubility of a gas is proportional to the pressure

Since the temperatures are both similar (25°c) then, the solubility of oxygen in water at 2.5 atm can be calculated as follows

= 0.041/1.0 × 2.5

= 0.041×2.5

= 0.1025 g/L

Hence the solubility of oxygen in water at 2.5 atm and 25°c is 0.1025 g/L

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To answer the question above, substitute the given values to the given equation,
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Solving for t gives t = 9,987.38 years or approximately equal to 9,990 years. Thus, the answer is letter C.












4 0
3 years ago
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KMnO4<br><br> Assign oxidation numbers to each element in this compound.
zysi [14]

K=+1

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That’s it!!!

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Explanation:

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3 0
3 years ago
You have 1 mole of a gas at STP. If you apply the ideal gas law what is the approximate volume of the gas?
goldfiish [28.3K]

Answer:

A) 22.4L

Explanation:

we know, ideal gas law states

PV=nRT

V=nRT/P

At STP,

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V=(1*0.082*273.15)/ 1

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7 0
3 years ago
You drop an unknown substance that weighs 8.3g into a graduated cylinder with 6ml of water. The water rises to 8ml when you drop
valkas [14]

Answer:

<h3>The answer is 4.15 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of object = 8.3 g

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volume = 8 - 6 = 2 mL

So we have

density =  \frac{8.3}{2}  \\

We have the final answer as

<h3>4.15 g/mL</h3>

Hope this helps you

7 0
4 years ago
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