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kakasveta [241]
3 years ago
7

The partial pressure of CO2 inside a bottle of soft drink is 4.0 atm at 25°C. The solubility of CO2 is 0.12 mol/L. When the bott

le is opened, the partial pressure drops to 3.0 ✕ 10-4 atm. What is the solubility of CO2 in the open drink? Express your answer in grams per liter. 
Chemistry
1 answer:
navik [9.2K]3 years ago
4 0
We can solve this problem by using Henry's law. 
Henry's law states that the amount of dissolved gas is proportional to its partial pressure.
C=kP
C is <span>the solubility of a gas.
</span><span>k is Henry's law constant.
</span><span>P is the partial pressure of the gas.
</span>We can calculate the constant from the first piece of information and then use Henry's law to calculate solubility in open drink.
0.12=4k
k=0.03
Now we can calculate the solubility in open drink.
C_o=kP_o&#10;
C_o=0.03 \cdot  3\cdot 10^{-4}=0.09\cdot 10^{-4} \frac{mol}{L}
Now we need to convert it to g/L. One mol of CO2 is 44.01<span>g. 
</span>The final answer is:
C_o=0.09\cdot 10^{-4}\cdot 44,01=3.4\cdot 10^{-4} \frac{g}{L} 


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Taking into account the definition of density,  assuming all other conditions remain the same, increasing mass will cause an object's density to increase.

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Density is an intensive property since it does not vary with the amount of substance.

Since density is the relationship between the mass and the volume of a substance, its calculation is defined as the quotient between the mass of a body and the volume it occupies:

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2 years ago
1. 100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.
stealth61 [152]

Answer:

The volume of water to be added is 0.175 liters of water

Explanation:

The given concentration of the nitric acid = 55% (M/M)

The mass of the nitric acid solution = 100 gm

The concentration solution is to diluted to = 20% (M/M)

The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution

Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get

Let "x" represent the volume of the resulting solution, we have;

20% of x = 55 g of nitric acid

∴ 20/100 × x = 55 g

x = 55 g × 100/20 =  275 g

The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid

The mass of extra water to be added = 275 g - 100 g = 175 g

Volume = Mass/Density

The density of water ≈ 1 g/ml

∴ The volume of water to be added that gives 175 g of water =  175 g/(1 g/ml) = 175 ml. = 0.175 l

The volume of water to be added = 0.175 liters of water.

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3 years ago
What volume of Co2 (carbon (iv) oxide)
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Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL

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