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podryga [215]
2 years ago
9

An experiment in chm 2045 requires students to prepare a 1.0 M aqueous solution of potassium phosphate.Jennifer fills a 1.0 lite

r volumetric flask to the calibration line with water. She then weighs out 212.3 g of potassium phosphate and adds it to the volumetric flask.Joe weighs out 212.3 g of potassium phosphate and adds it to a 1.0 liter volumetric flask. He then fills the volumetric flask to the calibration line with water.Which student has correctly prepared a 1.0 M aqueous solution of potassium phosphate?
Chemistry
1 answer:
horrorfan [7]2 years ago
6 0

Answer:

Joe correctly prepared the solution.

Explanation:

When comparing both methods, we see that the mass of potassium phosphate weighed is the same for Jennifer and Joe, so that's not a factor.

The difference lies in the fact that Jennifer added the solid to 1.0 liters of water, so<u> the final volume is higher than 1.0 L</u>, that's why t<u>he concentration will be lower than 1.0 M</u>.

The final volume of Joe's solution is 1.0 L, that's why he prepared the solution correctly.

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Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

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or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
2 years ago
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HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.

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yes 2.5 is correct for plato!

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