जो भी व्यक्ति मेरा कमेंट पढ़ रही हैंरही है हम लोग एक दूसरे से अनजान है पर मैं फिर भी भगवान से प्रार्थना करता हूं करता हूं कि आपकी लाइफ में कोई टेंशन हो तो वह दूर हो जाए 🙏🙏
How much nitrogen produced in litre from 50 gm sodium azide. To calculate the volume of nitrogen gas produced from 50 grams of sodium azide (NaN3), we need to know the balanced chemical equation for the reaction where sodium azide decomposes into nitrogen gas (N2). The balanced chemical equation for the decomposition of sodium azide is: 2 NaN3(s) → 2 Na(s) + 3 N2(g) From the equation, we can see that 2 moles of NaN3 produce 3 moles of N2 gas. Now, let's calculate the number of moles of sodium azide: Molar mass of NaN3 (sodium azide) = 23 + 14 + 3 * 16 = 65 g/mol Number of moles of NaN3 = Mass of sodium azide / Molar mass of sodium azide Number of moles of NaN3 = 50 g / 65 g/mol ≈ 0.769 moles Now, according to the balanced chemical equation, 2 moles of NaN3 produce 3 moles of N2 gas. So, the number of moles of N2 gas produced will be: Number of moles of N2 = Number of moles of NaN3 × (3 moles of N2 / 2 moles of NaN3) Number of moles of N2 = 0.769 moles × (3/2) ≈ 1.1545 moles Finally, to convert moles of N2 gas to liters, we use the ideal gas law, which states: PV = nRT Where: P = pressure (typically at standard pressure, around 1 atm) V = volume in liters n = number of moles of gas R = ideal gas constant (approximately 0.0821 L.atm/(mol.K)) T = temperature in Kelvin (typically at standard temperature, 273.15 K) At standard conditions (STP), which are 1 atm pressure and 273.15 K temperature: V = nRT / P V = 1.1545 moles × 0.0821 L.atm/(mol.K) × 273.15 K / 1 atm V ≈ 31.12 liters So, approximately 31.12 liters of nitrogen gas will be produced from 50 grams of sodium azide at standard conditions.
Dear Lesics हिंदी, You have you have not uploaded 2nd parts of many videos. I request you to upload second parts of videos which you have told us. You make such an awesome videos then why you don't upload all parts . Please upload 🙏 🙂 😊 👍
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जो भी व्यक्ति मेरा कमेंट पढ़ रही हैंरही है हम लोग एक दूसरे से अनजान है पर मैं फिर भी भगवान से प्रार्थना करता हूं करता हूं कि आपकी लाइफ में कोई टेंशन हो तो वह दूर हो जाए 🙏🙏
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How much nitrogen produced in litre from 50 gm sodium azide.
To calculate the volume of nitrogen gas produced from 50 grams of sodium azide (NaN3), we need to know the balanced chemical equation for the reaction where sodium azide decomposes into nitrogen gas (N2).
The balanced chemical equation for the decomposition of sodium azide is:
2 NaN3(s) → 2 Na(s) + 3 N2(g)
From the equation, we can see that 2 moles of NaN3 produce 3 moles of N2 gas.
Now, let's calculate the number of moles of sodium azide:
Molar mass of NaN3 (sodium azide) = 23 + 14 + 3 * 16 = 65 g/mol
Number of moles of NaN3 = Mass of sodium azide / Molar mass of sodium azide
Number of moles of NaN3 = 50 g / 65 g/mol ≈ 0.769 moles
Now, according to the balanced chemical equation, 2 moles of NaN3 produce 3 moles of N2 gas. So, the number of moles of N2 gas produced will be:
Number of moles of N2 = Number of moles of NaN3 × (3 moles of N2 / 2 moles of NaN3)
Number of moles of N2 = 0.769 moles × (3/2) ≈ 1.1545 moles
Finally, to convert moles of N2 gas to liters, we use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (typically at standard pressure, around 1 atm)
V = volume in liters
n = number of moles of gas
R = ideal gas constant (approximately 0.0821 L.atm/(mol.K))
T = temperature in Kelvin (typically at standard temperature, 273.15 K)
At standard conditions (STP), which are 1 atm pressure and 273.15 K temperature:
V = nRT / P
V = 1.1545 moles × 0.0821 L.atm/(mol.K) × 273.15 K / 1 atm
V ≈ 31.12 liters
So, approximately 31.12 liters of nitrogen gas will be produced from 50 grams of sodium azide at standard conditions.
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Dear Lesics हिंदी,
You have you have not uploaded 2nd parts of many videos.
I request you to upload second parts of videos which you have told us.
You make such an awesome videos then why you don't upload all parts .
Please upload 🙏 🙂 😊 👍
Add second part of topic of differential and footings
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