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2) A 7.90 L container holds a mixture of two gases at 51 °C. The partial pressures of gas A and gas B, respectively, are 0.292 atm and 0.676 atm. If 0.110 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

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Final answer:

Using Dalton's Law of Partial Pressures, the total initial pressure of gases A and B is 0.968 atm. For a precise calculation of the final pressure after adding a third gas, detailed information such as the temperature, gas constant, and identity of the third gas is necessary.

Step-by-step explanation:

Dalton's Law of Partial Pressures

To determine the total pressure in a container after adding a third gas, we apply Dalton's Law of Partial Pressures. This law states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each component gas in the mixture. In the provided scenario, gas A exerts a partial pressure of 0.292 atm, and gas B has a partial pressure of 0.676 atm. The total initial pressure is therefore 0.292 atm + 0.676 atm = 0.968 atm.

To find the partial pressure of the third gas, we use the ideal gas law (PV=nRT), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. However, since the temperature and volume are constant, and R is a constant, the pressure exerted by the third gas is directly proportional to the number of moles of the gas. If we were given the gas constant and the temperature, we could calculate it exactly. Yet, in this question, we are not given these values, nor are we given what the third gas is.

Without the specific details needed to calculate the exact pressure of the third gas, we cannot calculate the final pressure in the container. More information is required to provide a complete answer.

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