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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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Convert 15.2 moles of K to atoms of K.
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Answer:

\boxed {\boxed {\sf 9.15*10^{24} \ atoms \ K}}

Explanation:

To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.

This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

Multiply by the given number of moles: 15.2

15.2 \ mol \ K *\frac {6.022*10^{23} \ atoms \ K }{ 1 \ mol \ K }

The moles of potassium cancel.

15.2 *\frac {6.022*10^{23} \ atoms \ K }{ 1 }

The denominator of 1 can be ignored.

15.2 * {6.022*10^{23} \ atoms \ K }{

Multiply.

9.15344*10^{24} \ atoms \ K

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.

9.15*10^{24} \ atoms \ K

In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>

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How many grams of oxygen gas occupy 12.3 L of space at 109.4 kPa and 15.4oC?
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17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.

<h3>How to calculate mass?</h3>

The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.

However, the number of moles of the gas must be calculated first as follows:

PV = nRT

Where;

  • P = pressure = 1.0796941atm
  • V = volume = 12.3L
  • n = number of moles
  • T = temperature = 288.4K
  • R = gas law constant = 0.0821 Latm/molK

1.079 × 12.3 = n × 0.0821 × 288.4

13.27 = 23.68n

n = 13.27/23.68

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Learn more about mass at: brainly.com/question/19694949

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