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Tju [1.3M]
3 years ago
6

Which answer choice correctly describes the gas law and physical changes represented by the image?

Chemistry
2 answers:
vazorg [7]3 years ago
8 0
A. The image shows Boyle’s law because volume increases as pressure increases
skelet666 [1.2K]3 years ago
5 0

Answer:

<h2>A. The image shows Boyle's Law because volume decreases as pressure increases.</h2>

Explanation:

According to Boyle's Law, at constant temperature, pressure and volume have inverse relation that is as pressure would increase, volume would decrease.

According to Charles Law, at constant pressure, the volume is directly proportional to temperature.

According to the given graph and diagram, the temperature is constant and volume and pressure have inverse relation.

Thus, option A is correct.

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Many students are tempted to say "human error", but this term is vague and lazy; any decent teacher will not accept it. Instead, think about specific things that happened during the lab exercise where the end results may have been affected.
To give an example one might find in a bio lab: perhaps a water bath's temperature was not monitored very carefully and you found that an enzyme's activity was greater than you expected. In that case, you could write something like,
"The temperature of the water bath during this exercise was not monitored carefully. It is possible that it was warmer or cooler than intended, and this would have affected the enzyme activity accordingly. The fact that our enzyme activity was found to be higher than expected leads me to believe that perhaps the water bath was too warm."
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Explanation:

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A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add
erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

  • 47.2 mL * 150 mM = 24.0 mM * V₂
  • V₂ = 295 mL

And <u>converting into L </u>becomes:

  • 295 mL * \frac{1 L}{1000mL} = 0.295 L

6 0
3 years ago
Given that H2(g) + F2(g) - &gt; 2HF(g) delta H rxn = -546.6 kJ 2H2(g)+ O2(g) - &gt; 2H2O(l) delta H rxn = 571.6 kJ
olasank [31]

Answer:

ΔHrxn = -521.6 kJ

Explanation:

To do this, let's write the equations by separate:

H₂ + F₂ -------> 2HF        ΔH = -546.6 kJ

2H₂ + O₂ -------> 2H₂O   ΔH = -571.6 kJ

For these reactions, we want to get the following reaction:

2F₂ + 2H₂O -------> 4HF + O₂     ΔHrxn = ?

To do this, all we have to do is take the first two reaction and put them in the way to obtain the third reaction. When we look the final reaction we can see that the water is on the reactants, when originally it was on the product, while Florine is doubled. So all we have to do is rewrite the first two reactions, duplicate the first reaction, and reverse the second reaction, and that way we will get the final reaction:

1) (H₂ + F₂ -------> 2HF ) x2      ΔH = -546.6 kJ x 2

2) (2H₂O --------> 2H₂ + O₂)   ΔH = -571.6 kJ x -1

---------------------------------------------------------------------------

1) 2H₂ +2F₂ -------> 4HF      ΔH = -1093.2 kJ

2) 2H₂O --------> 2H₂ + O₂   ΔH = +571.6 kJ      

Now we sum 1) and 2). In this way, hydrogen cancels out and we do the same with the enthalpy:

1) 2H₂ +2F₂ -------> 4HF      ΔH = -1093.2 kJ

2) 2H₂O --------> 2H₂ + O₂   ΔH = +571.6 kJ  

---------------------------------------------------------------

2F₂ + 2H₂O -------> 4HF + O₂     ΔHrxn = -1093.2 + 571.6 = -521.6 kJ

So the enthalpy of this reaction is

<em>ΔHrxn = -521.6 kJ</em>

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