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Answer:
See explanation.
Explanation:
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In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Thus, we proceed as follows:
Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

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A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
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In this item, I supposed, that we are determine the molar fraction of oxygen and carbon dioxide in the sample. This can be done by dividing their respective partial pressures by the total pressure of the sample.
O2 : mole fraction = (100.7 mmHg) / (763.00 mmHg) = 0.13
CO2 : mole fraction = (33.57 mmHg) / (763.00 mmHg) = 0.044
Answers: O2 = 0.13
CO2 = 0.044