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ss7ja [257]
2 years ago
13

Calculate the final temperature of a sample of carbon dioxide of mass 16.0 g that is expanded reversibly and adiabatically from

500 cm3 at 298.15 K to 2.00 dm3. 1) If you want to calculate the work, dW=-pdV -> W = -nRTln(V2/V1). Is this expression ok for this problem? If yes, then what T should I use? 1) Why the Charles law V1/T1=V2/T2 does not hold? If you increase V you do work and so you expect to reduce U and therefore T. But according to charles law, in order to keep the proportion, if you increase V, you also increase T.
Chemistry
1 answer:
Pie2 years ago
4 0

For your first question, that equation only works if your situation is occurring at a constant temperature. Your original question is such a situation - everything occurs at 298.15 K. Therefore, you can use this value in the equation to calculate work. For your second question, Charles' Law describes how the volume of gas changes as you heat or cool it, PROVIDED PRESSURE AND MOLES OF GAS REMAIN CONSTANT THE WHOLE TIME. In your original question above, temperature stays constant while volume changes. However, what they don't tell you is that this necessarily requires a change in either pressure or moles of gas. Because the question works with the same sample the of gas the whole time (i.e. moles are constant), it is pressure that is changing (and this change will occur according to Boyle's Law, since temperature and moles are held constant). Hope that clarifies things!

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3 years ago
What is water oxidation number
-Dominant- [34]

Answer:

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6 0
3 years ago
Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown
defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

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O : 0.368 : 16 =0.023

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Mn : O =

\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2

The empirical formula : MnO₂

8 0
3 years ago
The pKa of lactic acid is 3.9. A lactate buffer will be useful from pH values ________. The pKa of lactic acid is 3.9. A lactate
Vedmedyk [2.9K]

Answer:

Explanation:

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This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since  buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .

From the Henderson-Hasselbach equation we have that

pH = pKa + log [A⁻]/[HA]

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0.1 ≤  [A⁻]/[HA] ≤ 10

Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.

Now we are equipped to answer our question:

pH range = 3.9 +/- 1 = 2.9 through 4.9

7 0
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frez [133]

Answer:

That is a good question... Maybe Because it is artificial and the makers do not know how to mimic nature.

Explanation:

5 0
3 years ago
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