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Vitek1552 [10]
3 years ago
5

A sample of N2O3(g) has a pressure of 0.046 atm . The temperature (in K) is then doubled and the N2O3 undergoes complete decompo

sition to NO2(g) and NO(g). Find the total pressure of the mixture of gases assuming constant volume and no additional temperature change. Enter your answer numerically, in terms of atm.
Chemistry
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

0.184 atm

Explanation:

The ideal gas equation is:

PV = nRT

Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.

So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.

The decomposition occurs:

N₂O₃(g) ⇄ NO₂(g) + NO(g)

So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.

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According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than  112°C is caused by reusing samples leads to errors.

A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).

However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above  112°C.

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If 45.8 grams of potassium chlorate decomposes, how many grams of oxygen gas can be produced? 2KClO3 → 2KCl + 3O2
natta225 [31]
To do this problem, we must first look at the balanced chemical equation for the decomposition of potassium chlorate: 

<span>2KClO3 --> 2KCl + 3O2 </span>

<span>We can take the given amount of grams, and use the molar mass of KClO3 to convert to moles. Then, we can use the stoichiometric ratios to relate moles of KClO3 to moles of O2. </span>

<span>(39.09)+(35.45)+(3*15.99)= 122.51 g/ mol = molar mass of KClO3 </span>
<span>45.8 g KClO3/ 122.51 g/ mol KClO3 = .374 moles KClO3 </span>
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<span>Once we have moles of O2, we can convert to grams of O2. </span>

<span>(2*15.99)= 31.98 g/mol = molar mass of O2 </span>
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