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Margaret [11]
3 years ago
5

Which of the temperatures below is most likely to be the boiling point of water at 880 torr?view available hint(s)which of the t

emperatures below is most likely to be the boiling point of water at 880 torr?92°c105°c100°c?
Chemistry
1 answer:
Luden [163]3 years ago
6 0
Boiling point<span>  is the </span>temperature<span> at which the vapor pressure of the liquid equals the surrounding pressure.

Above boiling point point, liquid get converted into vapour.

Now, boiling point of water is 100 oC at room pressure. Room pressure is equal to 760 torr. Thus, at 100 oC, vapour pressure of water becomes equal to 760 torr.

Now, if external pressure is increased to 880 torr, more heat is to be supplied so that vapour pressure of water equals 880 torr.

So, at 880 torr, boiling point of water will be more than 100 oC. In present case, most like the boiling point of water is equal to 105 oC.


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A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve
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Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

                  PV = nRT

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                 = \frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}

                  = 0.0052 mol

Also,  No. of moles = \frac{mass}{\text{molar mass}}

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                  mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at 25^{o}C

                P_{\text{water vapor}} = 24 mm Hg

                                = \frac{24}{760} atm

                                = 0.03158 atm

Now,   P = \frac{756}{760} - 0.03158

              = 0.963 atm

Hence,   n = \frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}

                 = 0.0056 mol

So, mass of H_{2} = 0.0056 mol × 2

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Therefore, calculate the percentage yield as follows.

      Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100

                              = \frac{0.01013 g}{0.0104 g} \times 100            

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Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

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