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e-lub [12.9K]
3 years ago
5

Enter your answer in the provided box. The partial pressure of CO2 gas above the liquid in a bottle of champagne at 20°C is 4.5

atm. What is the solubility of CO2 in champagne? Assume the Henry's law constant is the same for champagne as it is for water: at 20°C, kH = 3.7 × 10−2 mol/(L·atm).
Chemistry
1 answer:
sineoko [7]3 years ago
6 0

<u>Answer:</u> The molar solubility of carbon dioxide gas is 0.17 M

<u>Explanation:</u>

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{CO_2}=K_H\times p_{liquid}

where,

K_H = Henry's constant = 3.7\times 10^{-2}mol/L.atm

p_{CO_2} = partial pressure of carbonated drink = 4.5 atm

Putting values in above equation, we get:

C_{CO_2}=3.7\times 10^{-2}mol/L.atm\times 4.5atm\\\\C_{CO_2}=0.17mol/L=0.17M

Hence, the molar solubility of carbon dioxide gas is 0.17 M

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If 3.8 moles of zinc metal react with 6.5 moles of silver nitrate, how many moles of silver metal can be formed, and how many mo
bulgar [2K]

<u>Answer:</u> 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

<u>Explanation:</u>

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = \frac{1}{2}\times 6.5=3.25moles of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal,  hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = \frac{2}{2}\times 6.5=6.5moles of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

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