Answer:
One of the main components of an airbag is the gas that fills it. As part of the design process, you need to determine the exact amount of nitrogen that should be produced. Calculate the number of moles of nitrogen required to fill the airbag. Show your work. Assume that the nitrogen produced by the chemical reaction is at a temperature of 495°C and that nitrogen gas behaves like an ideal gas. Use this fact sheet to review the ideal gas law.
Answer:
0.06654345229738384 moles of chromium.
<u>Answer:</u> The increase in pressure is 0.003 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure = ?
= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![801^oC=[801+273]K=1074K](https://tex.z-dn.net/?f=801%5EoC%3D%5B801%2B273%5DK%3D1074K)
= final temperature = ![(801+1.00)^oC=802.00=[802+273]K=1075K](https://tex.z-dn.net/?f=%28801%2B1.00%29%5EoC%3D802.00%3D%5B802%2B273%5DK%3D1075K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28800J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B1074%7D-%5Cfrac%7B1%7D%7B1075%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D3%5Ctimes%2010%5E%7B-3%7Datm%5C%5C%5C%5CP_2%3De%5E%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1.003atm)
Change in pressure = 
Hence, the increase in pressure is 0.003 atm
The grams of ethane present in a sample containing 0.4271 mole is 12.84 g
<h3>Description of mole </h3>
The mole of a substance is related to it's mass and molar mass according to the following equation
Mole = mass / molar mass
With the above formula, we can obtain the mass of ethane. Details below
<h3>How to determine the mass of ethane</h3>
The following data were obtained from the question:
- Mole of ethane = 0.4271 mole
- Molar mass of ethane = 30.067 g/mol
- Mass of ethane =?
The mass of ethane present in the sample can be obtained as follow:
Mole = mass / molar mass
Cross multiply
Mass = mole × molar mass
Mass of ethane = 0.4271 × 30.067
Mass of ethane = 12.84 g
Learn more about mole:
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What are the names of the two magnetic poles?
Ur answer is
North and South
