. O2 gas is slightly soluble in water. How does this affect the reported moles of KClO3 decomposed in the mixture? Explain.
2.
a. When the water level inside the gas collecting test tube is higher than it is outside, is the O gas pressure greater or less than atmospheric pressure? Explain.
b. When you equilibrate the gas pressures in Question 2a, does the volume of Ogas increase or decrease? Explain.
c. Suppose you did not equilibrate the gas pressure to atmospheric pressure, but assumed the pressure of the O gas equals atmospheric pressure. Are the reported moles of evolved Ogas greater or less than actual? Explain.
a. That has no effect on the calculated moles of O2 gas in the collection vessel. In the experiments I've seen one weighs the KClO3 test tube both before and after heating to expel O2 gas. That different in mass of the two weighings tells you how much weight was lost and that translates into the mols O2 gas evolved.
b. The water level in the collection bottle is higher than the water level in the tank holding the bottle. Therefore, the pressure outside the bottle is higher than the pressure inside the bottle and the volume of O2 gas is less than when the water level is lowered by equilibration.
c. See answer a above.
1. Well, it seems like O2 gas really wants to get cozy with water. It's slightly soluble in the H2O, like a moody teenager hoping for a cuddle buddy. So, when you have a mixture of water and KClO3, some of that O2 gas is going to dissolve into the water. Now, when you go to measure the moles of KClO3 decomposed in the mixture, it might get a little sneaky. The dissolved O2 gas won't be accounted for in the measurement, so the reported moles of KClO3 decomposed could be higher than the actual amount. It's like counting the number of cookies you ate without realizing your dog snuck a few of them. Sneaky gas!
2a. When the water level inside the gas collecting test tube is higher than outside, it's like the O gas is having a pool party, but it forgot to invite atmospheric pressure. The pressure inside the test tube is greater because of the extra weight from the water pushing down on it. So, the O gas pressure is higher than atmospheric pressure, just like when you accidentally leave your windows closed during a burrito-eating marathon.
2b. Now, let's get our gas pressure levels to equalize, shall we? When you let the gas pressures inside and outside the test tube reach equilibrium, the volume of O gas has a little identity crisis. It's not sure whether it wants to be larger or smaller, so it stays the same. It's like that one friend who can never make a decision and just sticks with the status quo.
2c. Ah, the dangers of assumptions. If you don't equilibrate the gas pressure to atmospheric pressure and simply assume that the pressure of O gas equals atmospheric pressure, you're in for some inflated moles of evolved O gas. It's like pretending you're the heavyweight champion of the world when you really just have a slightly above-average arm-wrestling game. Those reported moles will be greater than the actual amount because you're not accounting for the excess pressure caused by the water. Assumptions: they'll play tricks on you faster than a mischievous clown bot!
1. The solubility of O2 gas in water will affect the reported moles of KClO3 decomposed in the mixture. When O2 gas dissolves in water, it forms an equilibrium with the dissolved oxygen. This means that some of the O2 gas will dissolve into the water, while some will remain in the gas phase.
When the moles of KClO3 decompose, they produce a certain amount of O2 gas. However, since some of this O2 gas is soluble in water, a portion of it will dissolve into the water instead of being collected as gas. As a result, the reported moles of KClO3 decomposed will be less than the actual moles, because some of the O2 gas will not be accounted for in the gas collection.
2.
a. When the water level inside the gas collecting test tube is higher than it is outside, the O2 gas pressure will be less than atmospheric pressure. This is because the water exerts a pressure on the gas, reducing the overall pressure experienced by the gas. The hydrostatic pressure exerted by the water column is added to the atmospheric pressure on the water surface. Therefore, the pressure inside the tube will be lower.
b. When the gas pressures are equilibrated, the volume of O2 gas does not change. Equilibrating the gas pressures means that the pressure inside the tube will equal the atmospheric pressure, so there is no net pressure difference to cause a change in volume.
c. If you assume that the pressure of the O2 gas equals atmospheric pressure without equilibrating, the reported moles of evolved O2 gas will be greater than the actual amount. This is because the actual pressure of the gas inside the tube is lower than atmospheric pressure due to the hydrostatic pressure of the water column. Therefore, assuming atmospheric pressure would overestimate the amount of gas collected and thus lead to an overestimation of the moles of O2 gas.
1. O2 gas being slightly soluble in water will affect the reported moles of KClO3 decomposed in the mixture. When KClO3 decomposes, it produces O2 gas as one of the products. However, since O2 gas is slightly soluble in water, a portion of the O2 gas will dissolve in the water, reducing the amount of O2 gas that is collected in the gas collecting test tube. This means that the measured moles of O2 gas will be lower than the actual moles of O2 gas produced during the decomposition of KClO3. Therefore, the reported moles of KClO3 decomposed will also be lower than the actual moles.
2a. When the water level inside the gas collecting test tube is higher than it is outside, the O2 gas pressure is less than atmospheric pressure. This can be explained by Dalton's law of partial pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas. In this case, the partial pressure of the water vapor inside the gas collecting test tube adds to the atmospheric pressure, resulting in a higher total pressure outside the tube. Therefore, the O2 gas pressure inside the tube is lower than the atmospheric pressure.
2b. When you equilibrate the gas pressures mentioned above, the volume of O2 gas remains unchanged. This is because the volume of a gas is independent of the pressure or the presence of other gases at constant temperature. Equilibrating the gas pressures ensures that the total pressure both inside and outside the gas collecting test tube is equalized, but it does not affect the volume of the O2 gas.
2c. If you did not equilibrate the gas pressure to atmospheric pressure and assumed the pressure of the O2 gas equals atmospheric pressure, the reported moles of evolved O2 gas would be greater than the actual moles. This is because the pressure of the O2 gas inside the gas collecting test tube is lower than atmospheric pressure (as explained in 2a). By assuming the pressure to be equal to atmospheric pressure, you would overestimate the moles of O2 gas based on the incorrect assumption. Therefore, the reported moles of O2 gas would be greater than the actual moles.